day 3 - caiet de informatică și tic · exista in jur de 14 tipuri de date varchar2 –lungine...

DAY 3

Instructor: Sanda Popescu

SECTIUNE 1/ DATABASE PROGRAMMING

INTRODUCTION TO FUNCTION SINGLE ROW FUNCTION

Exista 2 tipuri de functii:

• SINGLE ROW FUNCTION – care returneaza un singur rezultat pe

linie

• MULTIPLE ROW FUNCTION – care returneaza un singur rezultat pe un

anumit grup de linii

Posibilitati de transformare a datelor pentru a se potrivi unor situatii particulare

Single-row functions

Multiple-row functions:

Se afiseaza cel mai

mare salariu, cel

mai mic salariu si

media salariilor

angajatilor.

Exemple:

Se foloseste atunci cand avem de

efectuat calcule care nu sunt

legate de un tabel al bazei de date

Are o singura coloana numita

DUMMY si o singura linie cu

valoarea X

Exemplu de utilizare a

operatorului de

concatenare || cu functia

INITCAP

SUBSTR(sir_sursa,poz_inc,nr_car)

Ex. : SUBSTR(‘Test extragere’,2,3)=> va returna “est”

SUBSTR(‘Test extragere’,-5,4)=> va returna “ager” de la sf.catre inceput

CONCAT(sir1,sir) / unde sir1, sir2 pot fi si nume de coloane

CONCAT(‘Test’,’concatenare’ => Test concatenare

LPAD- completeaza in partea stanga cu un caracter specificat

RPAD- completeaza in partea dreapta cu un caracter specificat

Both -implicit

INSTR(sir_sursa,caracter,poz_de_unde_incep_cautarea,a_cata_aparitie_o_caut)

Returneaza pozitia unui caracter intr-un sir

INSTR(‘Mississippi’,’i’,1,3) =>8

INSTR(‘Mississippi’,’s’,8,3) =>0

REPLACE (string1,string_to_replace,[replacement_string])

select replace('Oracle Internet Academy','Oracle','Test') as test

from dual =>

select replace('Oracle Internet Academy','Oracle') as

test

from dual =>

TRY IT / SOLVE IT

ROUND(45.926,2) => 45.93

ROUND(45.926,-1) => 50

round(45.35,-3) => 0

round(65.35,-3) => 0

round(45.35,4) => 45.35

Observatie: ROUND

SI TRUNC pot fi

utilizate si pe tipul de

data calendaristica

select trunc(45.35,-3) test

from dual => 0

3. select trunc(45.35,1) test

from dual => 45.3

1. select trunc(45.35,3) test

from dual => 45.35

2. Implicit 0 zecimale

select trunc(45.35,-1) test

from dual => 40

TRY IT / SOLVE IT

DATE CALENDARISTICE ( DATE)

secolul curent este selectat in

functie de data serverului bazei

de date

Intern, anul dintr-o data

calendaristica este memorata

pe 4 cifre : 2 cifre pentru secol

si 2 cifre pentru an

Observatie : numai functia Months_Between returneaza un

numar, restul functiilor returneaza date calendaristice

select add_months(sysdate,6) test

from dualSYSDATE= ’21-03-07’

select TRUNC(sysdate,'YEAR') test

from dual

select TRUNC(sysdate,‘MONTH') test

from dual

Observatii: TRUNC si ROUND se

pot aplica datelor calendaristice

select TRUNC(sysdate,‘DAY') test

from dual =>

Observatie: anumite sisteme de

calcule recunoasc numele lunilor

(sau a zilelor sapatamanii numai in

lb.romana)

select round(sysdate,'YEAR') TEST

from dual

SYSDATE= ’21-03-07’

select round(sysdate,‘MONTH') TEST

from dual => Se analizeaza ziua (21 >15) si

se rotunjeste luna

select round(TO_DATE('21-08-2007','DD-MM-RRRR'),'YEAR') TEST

from dual se analizeaza luna si se rotunjeste anul

select trunc(TO_DATE('21-08-2007','DD-MM-RRRR'),'YEAR') TEST

from dual

select trunc(TO_DATE('21-08-2007','DD-MM-RRRR'),‘MONTH') TEST

from dual

select round(TO_DATE('21-08-2007','DD-MM-RRRR'),‘MONTH') TEST

from dual se analizeaza ziua si se rotunjeste luna

TRY IT / SOLVE IT

SECTIUNE 2

OBSERVATII

Exista in jur de 14 tipuri de date

VARCHAR2 – lungine vb.maxim 4000 de caractere

CHAR – lungime fica maxim 2000 de caractere

NUMBER- nr.reale, cu precizie 38 de cifre zecimale

DATE – data calendaristica ,format DD-MON-RR (YY)

CONVERSII EXPLICITE

• formatul model tb. sa fie inclus in ‘ ‘ si este case-sensitive

• valorile de date din formatul model tb sa fie separate prin virgula

• orice format de date valid poate fi inclus

• utilizeaza un element fm pentru a ignora blank-urile sau a zerourile din fata numarului

• utilizeaza un element sp pentru a scrie un numar asa cum se citeste

• utilizeaza un element th pentru a specifica un numar ordinal (specific lb. engleze)

• utilizeaza “ “ pentru a adauga un string in formatul model

select TO_CHAR(sysdate,'"ziua"dd,"luna",month,"anul",year') TEST

from dual

select TO_CHAR(sysdate,'"ziua "ddsp "luna" mm "anul" yyyy') TEST

from dual

select TO_CHAR(sysdate,'dd-mon-yy') TEST

from dual

NOTA: daca doresc sa introduc un

comentariu (un sir…) atunci folosesc “ ”

select TO_CHAR(4352.53,'99.9eeee') TEST

from dual

select TO_CHAR(4352.53,'09999.9999') TEST

from dual

select TO_CHAR(-4352.53,'9999.999PR') TEST

from dual

select TO_CHAR(-4352.53,'9999.999MI') TEST

from dual

select TO_NUMBER(TO_CHAR(sysdate,'DDMMYY')) test

from dual

SYSDATE =22.03.2007

OBS.: daca formatul ar fi fost fmddmmyy =>2237 s-ar fi

ignorat zerourile din fata fiecarei componente(zi,luna,an)

SELECT TO_NUMBER('453.88') TEST

from dual

SELECT TO_NUMBER('453') TEST

from dual => formatul standard poate lipsi daca

numarul nu are zecimale

Oracle afiseaza # (??) in locul intregului numar ale carui cifre

depasesc numarul total de cifre furnizat in formatul model si

rotunjeste numerele la un anumit nr. de zecimale ca in formatul

model

SELECT TO_NUMBER('45000','99') TEST

from dual

select TO_DATE('22Mai 2007','fxDDMonth RRRR') test

from dual

select TO_DATE('22-MaR-2007','DD-Mon-RR') test

from dual

select TO_DATE('22,MaRtie,2007','fxDD,Month,rrrr') test

from dual

select TO_DATE('22032007','fxDDMMRRRR') test

from dual

select TO_CHAR(TO_DATE('22032007','fxDDMMRRRR'),'Month ddsp year') test

from dual

select TO_CHAR(TO_DATE('22032007','fxDDMMRRRR'),'Month ddth year') test

from dual

SELECT TO_CHAR(SYSDATE,'Month') test

from dual

SELECT TO_CHAR(SYSDATE,‘yyyy') test

from dual

SELECT TO_CHAR(SYSDATE,‘DD') test

from dual

Extrage luna,an, zi dintro

data calendaristica

SELECT TO_CHAR(SYSDATE,‘year') test

from dual

OBSERVATII OPERATII DATE CALENDARISTICE

SYSDATE =22.03.2007

select (sysdate+60) test

from dual

select TO_DATE('31-12-07','fmdd-mm-yy')-sysdate test

from dual

select (TO_DATE('31-12-07','fmdd-mm-yy')-sysdate)/7 test

from dual

select TO_DATE('31-12-07','fmdd-mm-yy')+sysdate test

from dual

RECAPITULARE FUNCTII

Formatele pentru an RR si YY

Tabelul ilustreaza

formatul RR sau

RRRR

Format YY

sau YYYY –

acelasi secol al

anului curent

retine

Daca formatul de date

specificat este YYYY,

secolul pentru valoarea

returnata va corespunde

cu secolul anului curent.

Deci, daca anul este 1995

si tu folosesti formatul YY

sau YYYY iar anul curent

este 2007, atunci se va

afisa 2095, iar pe formatul

este RRRR atunci va afisa

1995

Example of RR Date Format

To find employees hired prior to 1990, use the RR

format, which produces the same results whether the

command is run in 1999 or now:

SELECT last_name, TO_CHAR(hire_date, ’DD-Mon-YYYY’)

FROM employees

WHERE hire_date < TO_DATE(’01-Jan-90’, ’DD-Mon-RR’);

The following command, on the other hand, results in no rows being

selected because the YY format

interprets the year portion of the date in the current century (2090).

SELECT last_name, TO_CHAR(hire_date, ’DD-Mon-yyyy’)

FROM employees

WHERE TO_DATE(hire_date, ’DD-Mon-yy’) < ’01-Jan-1990

TRY IT / SOLVE IT

FUNCTII GENERALE

Functii care permit conversia

valorilor Null

These functions work with any data type and pertain

to using nulls.

• NVL (expr1, expr2)

• NVL2 (expr1, expr2, expr3)

• NULLIF (expr1, expr2)

• COALESCE (expr1, expr2, ..., exprn

ATENTIE: la tipul de date

Trebuie sa fie tipuri compatibile

de date

NVL – poate fi utilizata

pentru conversia unei

valori NULL intr-un

numar inainte de a intra in

calculul expresiei.

Daca un operator din

cadrul expresiei este

NULL atunci rezultatul va

fi NULL

NVL2(EXP1,EXP2,EXP3)

Daca exp1 IS NOT NULL atunci functia returneaza EXP2

altfel returneaza EXP 3

EXP1 poate fi orice tip de date

EXP2 si EXP3 pot fi orice tip de date cu exceptia lui LONG

OBS: tipul de date returnat este intotdeauna la fel cu tipul de

date al EXP2, mai putin cazul in care returneaza VARCHAR2

SELECT last_name, salary, commission_pct,

NVL2(commission_pct,salary+(salary*commission_pct),salary) income

FROM copy_employees

UPDATE copy_employees

SET commission_pct=0.3

where lower(last_name)='king'

Daca exp1 IS NOT NULL atunci functia returneaza EXP2

altfel returneaza EXP 3

NVL2(EXP1,EXP2,EXP3)

Daca exp1=exp2 atunci returneaza NULL

altfel returneaza exp1

O extensie a functiei NVL cu diferenta ca pot exista mai multe valori

Daca prima expresie IS NOT NULL atunci functia va returna prima expresie

altfel cauta prima valoare diferita de NULL


COALESCE(commission_pct,salary+(salary*commission_pct),10) test

FROM employees


COALESCE(to_char(commission_pct),NULL,'Atentie la compatibilitatea de tip') test

FROM employees

TRY IT / SOLVE IT

Conditional Expressions• Provide the use of IF-THEN-ELSE logic within a SQL statement

Use two methods:

– CASE expression

–DECODE functionConditional Expressions

Two methods used to implement conditional processing (IF-THEN-ELSE logic)

within a SQL statement are the CASE expression and the DECODE function.

Note: The CASE expression is new in the Oracle9i Server release.

The CASE expression complies with ANSI SQL; DECODE is

specific to Oracle syntax

Atentie la compatibilitatea de tip

DECODE este proprietate ORACLE

CASE –standard ANSI SQL

Using the CASE Expression

Facilitates conditional inquiries by doing the work of

an IF-THEN-ELSE statement:

SELECT last_name, job_id, salary,

CASE job_id WHEN ’IT_PROG’ THEN 1.10*salary

WHEN ’ST_CLERK’ THEN 1.15*salary

WHEN ’SA_REP’ THEN 1.20*salary

ELSE salary END "REVISED_SALARY"

FROM employees;

The DECODE FunctionThe DECODE function decodes an expression in a way similar to the IF-THEN-ELSE

logic used in various languages. The DECODE function decodes expression after

comparing it to each search value. If the expression is the same as search, result is

returned.

SELECT last_name, job_id, salary,

DECODE(job_id, ’IT_PROG’, 1.10*salary,

’ST_CLERK’, 1.15*salary,

’SA_REP’, 1.20*salary,

salary) /* salary valoare implicita

REVISED_SALARY /* alias

FROM employees;

Using the DECODE Function

In the preceding SQL statement, the value of JOB_ID is tested. If JOB_ID is IT_PROG,

the salary increase is 10%; if JOB_ID is ST_CLERK, the salary increase is 15%; if

JOB_ID is SA_REP, the salary increase is 20%. For all other job roles, there is no

increase in salary.

The same statement can be expressed in pseudocode as an IF-THEN-ELSE statement:

IF job_id = ’IT_PROG’ THEN salary = salary*1.10

IF job_id = ’ST_CLERK’ THEN salary = salary*1.15

IF job_id = ’SA_REP’ THEN salary = salary*1.20

ELSE salary = salary

TRY IT / SOLVE IT

SECTIUNILE 3 - 4

JOIN(JONCTIUNE)

REALIZAREA RAPOARTELOR UTILIZAND

DATE DIN TABELE DIFERITE

Oracle Proprietary

• Equijoin

• Non-equijoin

• Outer join

• Self joinSQL: 1999

• Cross joins

• Natural joins

• Using clause

• Full or two sided outer joins

Pentru a conecta impreuna datele

din mai multe tabele, poti folosi 2

seturi de comenzi sau sintaxe:

In interogarea datelor din mai mult decat o tabela folosind

sintaxele proprietate ORACLE foloseste o conditie de join in

clauza WHERE

Formatul de baza al unei declaratii de join

Pentru claritate se recomanda prefixarea coloanelor cu numele tabelei.

Combinatia dintre numele tabelei si numele coloanei elimina ambiguitatile

de nume. Ce se intimpla daca 2 tabele au acelasi nume de coloana?

(Ex.#id)

Use a join to query data from more than one table.

• Write the join condition in the WHERE clause.

• Prefix the column name with the table name when

the same column name appears in more than one table.

To join n tables together, you need a minimum of n-1

join conditions. For example, to join three tables, a

minimum of two joins is required.

EMPLOYEES

DEPARTMENTS

Conditia de join se va bazaintotdeauna pe o coloanacu aceleasi tipuri de date sicu aceeasi semnificatie, deobicei pe o FK provenitadintro relatie intre 2 tabele

Department_id este FK inemployees si a provenit dinrelatia M:1 cu departments

Manager_id este FK provenit din relatia recursiva

Combina liniile care au valori echivalente pentru coloanele specificate

operatorul =

• Use table prefixes to qualify column names that

are in multiple tables.

• Improve performance by using table prefixes.

• Distinguish columns that have identical names but

reside in different tables by using column aliases.

• Simplify queries by using table aliases.

SELECT e.employee_id, e.last_name, e.department_id, d.department_id, d.location_id

FROM employees e , departments d

WHERE e.department_id = d.department_id;

Guidelines

• Table aliases can be up to 30 characters in length, but shorter is better.

• If a table alias is used for a particular table name in the FROM clause, then that table alias

must be substituted for the table name throughout the SELECT statement.

• Table aliases should be meaningful.

• The table alias is valid only for the current SELECT statement.

select first_name,last_name,department_namefrom employees e, departments dwhere e.department_id=d.department_id

ATENTIE!

Se poate si asa, dar nu se recomanda

• Foloseste prefixarea pentruetichetarea numelor de coloanecare sunt in tabele multiple

• Imbunatateste performanta

• Coloane distincte care au numeidentice dar apartin unor tabelediferite

select first_name,last_name,department_namefrom employees , departments where employees.department_id=departments.department_id

select employees.first_name,employees.last_name,departments.department_namefrom employees , departments where employees.department_id=departments.department_id

1. Sa se afiseze numele angajatilor si departamentele unde lucreaza fiecare angajat

Sau fara prefixarea coloanelor in clauza SELECT

2. Sa se afiseze numele angajatilor care lucreaza in departamentul 90

Conditie suplimentara de restrictionare a liniilor afisate

Join cu 3 tabele

Daca in clauza Where conditia este invalida(saulipseste WHERE) atunci serverul Oracle returneazaprodusul cartezian al celor 2 tabele

Pentru a evita produsul cartezian, includeintotdeauna o conditie valida in clauza WHERE

Nr.linii returnate =

nr.linii tabela 1*nr.linii tabela 2

d_songs

d_themes

Produs cartezian

Obs.: Where p.id IN(47,48) arestrictionat numarul de linii la 16, altfelnumarul de linii afisate ar fi fost 8*6=48

TRY IT / SOLVE IT

Se utilizeaza atunci cand datele dintro tabela nu corespund exact cu datele

din alta tabela.(se gasesc intre niste limite de valori din cealalta tabela)

Ex: O companie isi plateste angajatii in functie de timpul

lucrat masurat in ore si minute, incrementand la fiecare 15

minute.

O tabela memoreaza orele si minutele lucrate, iar intro alta

tabela avem modul de plata in functie de timpul lucrat.

Daca un angajat lucreaza 40 de minute, atunci va fi platit cu

45 de centi (intervalul [31,45] )

De vreme ce nu exista o potrivire

exacta intre cele 2 coloane atunci

operatorul = nu poate fi folosit.

Desi pot fi folositi operatorii <= si

>=, se recomanda operatorul

BETWEEN..AND

select p.first_name,p.salary,t.grade_levelfrom employees p,job_grades twhere p.salary between t.lowest_sal and t.highest_sal

Sunt mai mult de 10 linii care ar trebui afisate.

sau

TRY IT / SOLVE IT

Un Outer Join este folosit

pentru a afisa atat liniile care

au valori corespondente in

alta tabela (valori

echivalente) plus liniile care

nu au valori corespondente

in alta tabela. Mentioneaza ce

tabela poate avea date lipsa

cu un semn (+) dupa numele

tabelei careia ii lipsesc date.

Se vor afisa departamente cu sau fara angajati

Semnul (+) la angajati =>

lipsesc date din angajati

Se vor afisa angajati cu sau fara departamente

Semnul (+) la departamente => lipsesc date din departamente

Atentie in Order By => NULL este primul la ordonarea descrescatoare

TRY IT / SOLVE IT

select worker.last_name||' workes for '||manager.last_name AS " SELF JOIN"from EMPLOYEES worker, employees managerWHERE worker.manager_id=manager.employee_idorder by worker.last_name

TRY IT / SOLVE IT

Joining Tables Using SQL: 1999 Syntax

SELECT column1,column2,……

FROM table1 NATURAL JOIN table2

SELECT table1.column, table2.column

FROM table1 CROSS JOIN table2

TRY IT / SOLVE IT

NATURAL JOIN cauta toate coloanele cu acelasi nume si returneaza liniilecare au valori echivalente. Ce se intimpla daca exista 2 coloane cu acelasinume dar au tipuri de date diferite? Acest lucru va returna o EROARE.Pentru a evita aceasta situatie clauza join poate fi modificata cu clauzaUSING, care va specifica care coloana este folosita pentru equijoin.

ATENTIE: se pot folosi alias-uri de tabele la USING sipot fi prefixate si coloanele, dar NU aveti voie saprefixati coloana din USING

o interogare care va returna numele si id_ultimelor joburi pentru acei

angajati care au avut inainte un rol in companie

Job_history

employeesATENTIE

Toate coloanele au fost

prefixate cu exceptia coloanei

employee_id

Atunci cand coloanele au nume diferite, sau daca trebuie sa se

foloseasca alt operator decat =, cum ar fi operatori de comparatie < , > ,

Between ..And (non-equijoin), sau trebuie SELF JOIN (aceeasi tabela

este referita de 2 ori si trebuie sa facem prefixare)

Write a query that will return the names and salary grades for employees.

Coloanele care intra in join OBLIGATORIU trebuie sa fie de acelasitip (valori echivalente), prin USING definim coloana care stim sigurca indeplineste aceasta conditie. (Natural Join le ia pe toate care auacelasi nume, dar nu verifica tipul datelor si va rezulta eroare)

TRY IT / SOLVE IT

In ANSI 99 SQL , un join intre 2 sau mai multe tabele care returneaza

doar liniile care au corespondenta se numeste un “inner join”

Cand un join returneaza atat liniile care au corespondenta cat si liniile care

nu au corespondenta => outer join

Atentie: termenii LEFT, RIGHT, FULL sunt asociati cu

numele tabelei din declaratie in clauza FROM si SELECT

Se vor afisa angajati cu sau fara departamente

employees e LEFT OUTER JOIN departments d=>

lipsesc date din departmentsSintaxa ORACLE

Se vor afisa departamente cu sau fara angajati

employees e RIGHT OUTER JOIN departments d=>

lipsesc date din employeesSintaxa ORACLE

1. Departamentele fara angajati

1. OUTER JOIN (Oracle) - se pune plus pe tabela deficitara adica EMPLOYEES)

SELECT e.last_name, d.department_id, d.department_name

FROM employees e, departments d

WHERE e.department_id(+)=d.department_id

2. LEFT OUTER JOIN (SQL) - Ultima tabela este deficitara

SELECT e.last_name,d.department_id,d.department_name

FROM departments d

LEFT OUTER JOIN employees e

ON (e.department_id=d.department_id)

3. RIGHT OUTER JOIN (SQL) - Prima tabela este deficitara


FROM employees e

RIGHT OUTER JOIN departments d


2. Angatii fara departamente

1. OUTER JOIN (Oracle) - se pune plus pe tabela DEPARTMENTS


FROM employees e,departments d

WHERE e.department_id=d.department_id(+)

2. LEFT OUTER JOIN (SQL) - Tabela de la sfarsit este deficitara


FROM employees e

LEFT OUTER JOIN departments d


3. RIGHT OUTER JOIN (SQL) - Prima tabela este deficitara


FROM departments d

RIGHT OUTER JOIN employees e


TRY IT / SOLVE IT

RECAPITULARE

JOIN

Equijoin (Oracle)SELECT event_id, p.song_id, t.cd_number

FROM d_play_list_items p, d_track_listings t

WHERE p.song_id = t.song_id;

Natural Join (ANSI)SELECT event_id, song_id , cd_number

FROM d_play_list_items NATURAL JOIN d_track_listings

Natural Join este un equijoin care leaga cele doua tabele pe toate coloanele cu nume comun din cele doua tabele. Coloanele comune celor doua tabele trebuie sa aiba acelasi tip altfel rezulta eroare

Nu e nevoie sa se foloseasca alias pt. tabele…

Equijoin (Oracle) / Natural Join (ANSI)

Equijoin (Oracle) / Join…USING (ANSI)




Join cu clauza USING (ANSI)SELECT event_id, song_id , cd_number

FROM d_play_list_items JOIN d_track_listings

USING (song_id)

In clauza USING se trec coloanele pe care se face join-ul. Este

utila atunci cand tabelele au mai multe coloane cu nume

comun dar nu dorim in conditia de join sa folosim decat o

parte din aceste coloane.

Equijoin (Oracle) / Join…ON (ANSI)




Join cu clauza ON (ANSI)SELECT event_id, song_id , cd_number

FROM d_play_list_items p JOIN d_track_listings t

ON (p.song_id=t.song_id);

In clauza ON se pot specifica orice fel de conditii. Se pot lega

tabele care nu au coloane cu acelasi nume

Cartesian Product (Oracle) /

Cross Join (ANSI)

Cartesian Product (Oracle) – e ca un equijoin la care s-

a omis conditia de joinSELECT first_name, last_name, department_name

FROM employees, departments;

Cross join (ANSI) – are acelasi efect ca si Cartesian

Product, adica "leaga" fiecare inregistrare din prima

tabela cu fiecare inregistrare din a doua tabelaSELECT first_name, last_name, department_name

FROM employees CROSS JOIN departments;

Nonequijoin (Oracle) / Join…ON (ANSI)

Equijoin (Oracle)SELECT e.last_name, e.salary, j.grade_level

FROM employees e, job_grades j

WHERE e.salary

BETWEEN j.lowest_sal AND j.highest_sal

Join cu clauza ON (ANSI)SELECT e.last_name, e.salary, j.grade_level

FROM employees e JOIN job_grades j

ON (e.salary BETWEEN j.lowest_sal AND j.highest_sal)

Self Join (Oracle) / Join…ON (ANSI)

Equijoin (Oracle)SELECT e.last_name||' '||e.first_name AS "Angajat",

m.kast_name||' '||m.first_name AS "Manager"

FROM employees e, employees m

WHERE e.manager_id=m.employee_id

Join cu clauza ON (ANSI)SELECT e.last_name||' '||e.first_name AS "Angajat",

m.kast_name||' '||m.first_name AS "Manager"

FROM employees e JOIN employees m

ON (e.manager_id=m.employee_id)

LEFT-OUTER JOIN

Oracle Syntax

SELECT e.last_name, e.department_id,

d.department_name


WHERE e.department_id = d.department_id (+);

Afiseaza toti angajatii (e…) CU SAU FARA departamente (d…,

plusul e pe partea cu departamentul)

Join cu clauza ON (ANSI)SELECT e.last_name, e.department_id,

d.department_name

FROM employees e LEFT OUTER JOIN departments d

ON (e.department_id=d.department_id);

Afiseaza toate informatiile din tabela din STANGA (employees).

RIGHT-OUTER JOINOracle Syntax

SELECT e.last_name, e.department_id,

d.department_name


WHERE e.department_id(+) = d.department_id;

Afiseaza toate departamentele (d…) CU SAU FARA angajati (e…,

plusul e pe partea cu angajatii)

Join cu clauza ON (ANSI)SELECT e.last_name, e.department_id,

d.department_name

FROM employees e RIGHT OUTER JOIN departments d


Afiseaza toate informatiile din tabela din DREAPTA (departments).

FULL-OUTER JOIN

Oracle Syntax

NU EXISTA FULL OUTER JOIN IN

SINTAXA ORACLE PENTRU CA NU SE

POATE PUNE + PE AMBELE PARTI ALE

SEMNULUI DE EGALITATE

Join cu clauza ON (ANSI)SELECT e.last_name, e.department_id, d.department_name

FROM employees e FULL OUTER JOIN departments d


Afiseaza toate informatiile din ambele tabele.

Q1: Evaluate this SQL statement:

SELECT e.employee_id, e.last_name, e.first_name, d.department_name FROM employees e, departments d WHERE e.department_id = d.department_id AND employees.department_id > 5000 ORDER BY 4; Which clause contains a syntax error?

A: AND employees.department_id > 5000

Q2: Which of the following best describes a natural join?

A: A join between two tables that includes columns that share the same name, datatypes

and lengths

Q3: Which SELECT clause creates an equijoin by specifying a column name common to both tables?

A: USING clause

Q4 : Which of the following statements is the simplest description of a nonequijoin?

A: A join condition containing something other than an equality operator

Q5: Which query will retrieve all the rows in the EMPLOYEES table, even if there is no match in the DEPARTMENTS table?

A: SELECT e.last_name, e.department_id, d.department_name FROM employees eLEFT OUTER JOIN departments d ON (e.department_id = d.department_id);

Q6: Evaluate this SELECT statement: SELECT p.player_id, m.last_name, m.first_name, t.team_name FROM player p LEFT OUTER JOIN player m ON (p.manager_id = m.player_id) LEFT OUTER JOIN team t ON (p.team_id = t.team_id);

Which join is evaluated first? A: the self-join of the player table (*)

Q7: The EMPLOYEE_ID column in the EMPLOYEE table corresponds to the EMPLOYEE_ID column of the ORDER table. The EMPLOYEE_ID column in the ORDER table contains null values for rows that you need to display. Which type of join should you use to display the data?

A: outer join (*)

TRY IT / SOLVE IT

Raspandirea datelor

SELECT column,group_function(column),……….. FROM table WHERE condition GROUP BY column ORDER BY column

OBS.: toate functiile grup ignora valorile NULL

SELECT AVG(salary), MAX(salary),MIN(salary),

SUM(salary)

FROM employees

WHERE job_id LIKE '%PROG%';AVG(SALARY) MAX(SALARY) MIN(SALARY)

6400 9000 4200

FIRST_NAME LAST_NAME SALARY HIRE_DATE JOB_ID

DEPART

MENT_ID

Steven King 24000 17-Jun-87 AD_PRES 90

Lex De Haan 17000 13-Jan-93 AD_VP 90

Alexander Hunold 9000 03-Jan-90 IT_PROG 60

Bruce Ernst 6000 21-May-91 IT_PROG 60

Diana Lorentz 4200 07-Feb-99 IT_PROG 60

Kevin Mourgos 5800 16-Nov-99 ST_MAN 50

Randall Matos 2600 15-Mar-98 ST_CLERK 50

Peter Vargas 2500 09-Jul-98 ST_CLERK 50

Eleni Zlotkey 10500 29-Jan-00 SA_MAN 80

Write a query that shows the average, maximum, and minimum salaries for all employees with

jobs in the programming department.

TRY IT / SOLVE IT

Vei folosi COUNT(*) atunci cand vrei sa numeri toate liniile , inclusiv acele linii care pot avea NULL in una sau mai multe coloane

Cuvantul cheie DISTINCT este folosit pentru a returna valorile distincte sau combinatiile distincte intro interogare;

Uneori este necesar sa includem in calcule si

valorile NULL dar, toate functiile grup ignora

valorile NULL.

Pentru a putea include in calcule aceste valori

este nevoie ca aceste valori NULL sa fie

inlocuite cu valori compabile cu tipul de date

utilizat.

ATENTIE: la tipul de date

Trebuie sa fie tipuri compatibile de date

NVL cu functiile de grup

0.85:4=0.2125

0.85:20=0.0425

Numai 4 angajati (din

cei 20 inregistrati) au

commision_pct


DEPART

MENT_ID










What is the oldest hire date, and the most recent hire date

for all employees in the programming department?

SELECT MIN(hire_date), MAX(hire_date)

FROM employees

WHERE job_id like ‘%PROG%’;

MIN(HIRE_DATE) MAX(HIRE_DATE)

03-Jan-90 07-Feb-99

!!!! MAX(hire_date)

afiseaza data ultimului angajat

(cel mai nou angajat)

SELECT COUNT(*)

FROM employees

WHERE department_id = 60;

How many employees are there in department 60?

COUNT(*)

3


DEPART

MENT_ID










Sa se afiseze numarul departamentelor distincte

SELECT COUNT(DISTINCT department_id)

FROM employees; COUNT(DISTINCTDEPARTMENT_ID)

7

TRY IT / SOLVE IT

Obs: poti folosi clauza GROUP BYpentru a imparti liniile dintro tabela in grupuri mai mici iar functiile grup vor returna informatii pe fiecare grup in parte

Se afiseaza salariul mediu pe fiecare departament in ordine crescataoare a departamentelor (se grupeaza liniile pe departamente si se aplica AVG() )

Sa se afiseze salariul maxim pe fiecare departament

OBSERVATII

1. In GROUP BY poate sa apara o coloana care nu este listata in clauza SELECT

2. Functiile grup necesita ca fiecare coloana listata in clauza SELECT

care nu are functie grup TREBUIE sa fie listata in GROUP BY

Rata de absolvire pe fiecare

oras a studentilor

RETINETI:

Nu se poate

utiliza un

alias de

coloana in

clauza

GROUP BY.

Cateodata, ai nevoie sa imparti grupurile in grupuri mai mici. De exemplu, tu poti grupa angajatii dupa departament si in cadrul fiecarui departament grupezi dupa job.

Criteriile de divizare se iau in ordinea aparitiilor din clauza GROUP BY

select to_char(avg(salary),'99999.99'),department_idfrom employeesgroup by department_id

select max(avg(salary))

from employees

select department_id, max(avg(salary))

from employees

group by department_id


DEPART

MENT_ID










Sa se afiseze salariul mediu pe fiecare departament

DEPARTMENT_ID AVG(SALARY)

10 4400.00

20 9500.00

50 3500.00

60 6400.00

80 10033.33

90 19333.33

110 10150.00

(null) 7000.00

SELECT department_id,

AVG(salary)

FROM employees

GROUP BY department_id;


DEPART

MENT_ID










Show the department_id, job_id and the total

salary for each job within the department (using

full table instead of partial table above) .


job_id, SUM(salary)

FROM employees

GROUP BY department_id,

job_id;

DEPART

MENT_ID JOB_ID

SUM

(SALARY)

(null) SA_REP 7000

10 AD_ASST 4400

20 MK_MAN 13000

20 MK_REP 6000

50 ST_MAN 5800

50 ST_CLERK 11700

60 IT_PROG 19200

80 SA_MAN 10500

80 SA_REP 19600

90 AD_VP 34000

90 AD_PRES 24000

110 AC_MGR 12000

110AC_ACCOUNT 8300

Clauza WHERE restrictioneaza liniile selectate

Clauza HAVING restrictioneaza grupurile

Nu ai voie sa folosesti functii grup in WHERE

Daca doriti salariul maxim pe departamente,

dar numai pentru departamentele care au mai

mult de 1 angajat? In clauza WHERE nu ai

voie sa folosesti functiile grup.Pentru a

restrictiona grupurile trebuie sa folosesti

clauza HAVING

Observatii:

OBS

Se recomanda ca ordinea clauzelor intro declaratie sa fie:

CLAUZE


AVG(salary)

FROM employees

WHERE AVG(salary) > 10000

GROUP BY department_id;

This results in an error:

“group function not

allowed here”

SELECT department_id, AVG(salary)

FROM employees

HAVING AVG(salary) > 10000

GROUP BY department_id DEPARTMENT_ID AVG(SALARY)

80 10033.33333

90 19333.33333

110 10150

Show the department_id, and average salary for all departments where the average

salary is greater $10,000. (Use full employees table found in HTML DB)

SELECT department_id, AVG(salary)

FROM employees

GROUP BY department_id

HAVING MAX(salary) > 10000

DEPARTMENT_ID AVG(SALARY)

20 9500

80 10033.33333

90 19333.33333

110 10150

Show the department_id, and average salary for all departments where the maximum salary is

greater $10,000. (Use full employees table found in HTML DB)

SELECT job_id, SUM(salary)

FROM employees

WHERE job_id NOT LIKE '%PROG%'

GROUP BY job_id

HAVING SUM(salary) > 10000

ORDER BY SUM(salary) JOB_ID SUM(SALARY)

SA_MAN 10500

ST_CLERK 11700

AC_MGR 12000

MK_MAN 13000

AD_PRES 24000

SA_REP 26600

AD_VP 34000

Show the job id and the total salary for all jobs

that aren’t in programming where the total

salary is greater than $10,000. Arrange the data

by the total salary in ascending order.

TRY IT / SOLVE IT

Daca ai nevoie intro interogare(cerere) sa compari cu date din aceeasi tabela sau din alta

tabela, atunci folosesti subcererea.

Exemplu: Sa se afiseze toti angajatii care s-au nascut in aceeasi zi cu ”X”. In acest caz

trebuie sa stii in ce zi s-a nascut X si sa restrictionezi liniile in functie de data nasterii a lui

X=> trebuie sa combini 2 interogari(cereri), plasand o cerere in interiorul altei cereri

SUBQUERY=INNER QUERY plasarea unei cereri in interiorul altei cereri

OUTER QUERY = cerere externa care foloseste informatia de care ai nevoie returnata de

cererea interna

WHERE

HAVING

FROM

INSERT

REGULI

1. Subcererea este inclusa in

paranteze

2. Subcererea este plasata in partea

dreapta a operatorului de

comparare

3. Cererea externa si interna pot cauta

date din diferite tabele

4. O singura clauza ORDER BY poate

fi utilizata intro declaratie si trebuie

sa fie ultima.O subcerere nu poate

avea o proprie clauza ORDER BY

5. Numarul de subcereri depinde de

dimensiunea bufferului cererii

folosite (atentie aceste reguli le veti

gasi in quizz-uri)

ATENTIE !

la numarul liniilor returnate de subcerere si la operator

Daca o subcerere va returna NULL sau nicio linie, cererea externa va

considera rezultatul NULL si va folosi acest rezultat in clauza WHERE

Atunci cererea externa nu va

returna nicio linie

Rezultatul subcererii

este NULL, pt ca

Grant nu are

departament.

REGULI

1. Returneaza doar O SINGURA LINIE

2. Subcererea este inclusa in

paranteze

3. Subcererea este plasata in partea

dreapta a operatorului de

comparare

4. Cererea externa si interna pot cauta

date din diferite tabele

5. O singura clauza ORDER BY poate

fi utilizata intro declaratie si trebuie

sa fie ultima.O subcerere nu poate

avea o proprie clauza ORDER BY

6. Numarul de subcereri depinde de

dimensiunea bufferului cererii

folosite (atentie aceste reguli le veti

gasi in quizz-uri)

Sa se afiseze toti angajatii care lucreaza in departamentul “Marketing”

Se poate utiliza mai mult decat o subcerere pentru

a returna informatii cererii externe

Subcereri in clauza HAVING

Care departamente au salariul minim mai mare decat cel mai

mic salariu din departamentul 50?

TRY IT / SOLVE IT

1997

2000

2002

1999

2000

2001

1998

2004

Restrictioneaza liniile bazate pe lista

de valori returnate de la subcerere ANY

ANY CEL PUTIN UNA

< MAX

Compara o valoare cu fiecare valoare returnata de subcerere

(> ALL => mai mare decat toate ) >MAX

Pentru a evita erorile, cand exista o sansa ca valoarea NULL sa fie returnata de o cerere interna,

foloseste IN (=ANY). Aceasta este o alegere sigura, deoarece daca nu exista o valoare in setul de

rezultate, atunci cu siguranta nu sunt valori care sa satisfaca clauza WHERE

EVITA ASTFEL DE SITUATII NOT IN (100,101,NULL) => foloseste in subcerere IS NOT NULL

ECHIVALENTE

1. = ANY IN

2. < ANY < MAX (DACA EXISTA CEL PUTIN UNA MAI MICA DECAT MAX )

3. > ANY > MIN (DACA EXISTA CEL PUTIN UNA MAI MARE DECAT MIN )

4. < ALL < MIN ( MAI MICA DECAT TOATE)

5. > ALL > MAX

6. NOT IN <> ALL

Exemplu de subcerere multiple-row cu operatorul ANY (oricare)

Se returneaza CD-urile al

caror an este mai mic decat

cel putin un an al CD-

urilor produse de “The

Music Man”.

<ANY <MAXIM

>ANY >MINIM

=ANY IN (exemplificare pe slide-ul

urmator)

Exemplu de cerere multiple-row cu “=ANY”

Rezultatul

subcererii

Cele doua

comenzi

returneaza

acelasi lucru.

Exemplu de subcerere multiple-row cu operatorul ALL

Se vor returna CD-urile al caror

an este mai mare decat toti anii

cd-urilor producatorului “The

Music Man”.

<ALL <MINIM

>ALL >MAXIM

=ALL

ALL compara

valorile cererii

principale cu

fiecare valoare

returnata de

subcerere.

Valorile NULL si subcererile multiple-row cu operator IN

Subcererea returneaza si

valoarea NULL (pentru

angajatul KING, care nu are

manager_id).

Se afiseaza angajatii care

sunt manageri.

Daca utilizam operatorul IN si

subcerea returneaza NULL

cererea principala va returna

acele valori care corespund

valorilor nenule din subcerere.

Valorile NULL si subcererile multiple-row cu operator NOT IN

Daca utilizam operatorul

NOT IN si subcererea

returneaza NULL, atunci nu

vom obtine date, deoarece

NOT IN este echivalent cu

<>ALL.

Valorile NULL si subcererile multiple-row cu operator ANY

Daca utilizam operatorul ANY

si subcerea returneaza NULL

cererea principala va returna

acele valori care corespund

valorilor nenule din subcerere.

Desi subcererea

returneaza si

valoarea NULL

obtinem angajatii

care sunt manageri.

Valorile NULL si subcererile multiple-row cu operator ALL

Daca subcererea

returneaza NULL si

utilizam operatorul ALL,

nu vom obtine rezultate.

< ANY < MAX (DACA EXISTA CEL PUTIN UNA MAI MICA DECAT MAX )

VERIFICA NUMARUL DE LINII RETURNATE DE SUBCERERE

ASIGURA-TE CA OPERATORUL UTILIZAT ESTE CORECT

DACA AI DUBII IN CE PRIVESTE NUMARUL LINIILOR

RETURNATE FOLOSESTE MULTIPLE ROW SUBQUERY


DEPART

MENT_ID










From

subquery

SELECT last_name

FROM employees

WHERE salary <

(SELECT salary

FROM employees

WHERE last_name =

'Matos');

LAST_NAME

Vargas

Sa se afiseze angajatii care castiga mai putin decat Matos

SELECT job_id, AVG(salary)

FROM employees

GROUP BY job_id

HAVING AVG(salary) = (SELECT

MIN(AVG(salary))

FROM employees

GROUP BY job_id);MIN(AVG(SALARY))

2925

JOB_ID AVG(SALARY)

ST_CLERK 2925

From Subquery

Result of full query

Find the job with the lowest average salary.(Use the full employees table)

This will work:

SELECT employee_id, last_name

FROM employees

WHERE salary IN

(SELECT MIN(salary)

FROM employees

GROUP BY department_id)

Why will the following query not work?

SELECT employee_id, last_name

FROM employees

WHERE salary =

(SELECT MIN(salary)

FROM employees

GROUP BY department_id)

Single row operator

IN, ANY, ALL are

multiple-row

comparison operators

Why does the following query

not work?

SELECT last_name

FROM employees

WHERE employee_id NOT IN

(SELECT manager_id

FROM employees);

Subquery returns

NULL values

This one will work:

SELECT last_name

FROM employees

WHERE employee_id NOT IN

(SELECT manager_id

FROM employees

WHERE manager_id IS NOT NULL);

Sa se afiseze maximul salariului mediu si departamentul unde se castiga

select avg(salary),department_id

from employees

having avg(salary)=(

select max(avg(salary))

from employees

group by department_id)


select department_id, max(avg(salary))

from employees


FORMA CORECTA

Q: Below find the structure of the CUSTOMERS and SALES_ORDER tables:

CUSTOMERSCUSTOMER_ID NUMBER NOT NULL, Primary KeyCUSTOMER_NAME VARCHAR2 (30) CONTACT_NAME VARCHAR2 (30)CONTACT_TITLE VARCHAR2 (20)ADDRESS VARCHAR2 (30)CITY VARCHAR2 (25)REGION VARCHAR2 (10)POSTAL_CODE VARCHAR2 (20)COUNTRY_ID NUMBER Foreign key to COUNTRY_ID column of the COUNTRY table PHONE VARCHAR2 (20)FAX VARCHAR2 (20) CREDIT_LIMIT NUMBER(7,2)

SALES_ORDERORDER_ID NUMBER NOT NULL, Primary KeyCUSTOMER_ID NUMBER Foreign key to CUSTOMER_ID column of the CUSTOMER tableORDER_DT DATE ORDER_AMT NUMBER (7,2) SHIP_METHOD VARCHAR2 (5)

You need to create a report that displays customers without a sales order.

A: SELECT c.customer_name FROM customers c WHERE c.customer_id not in (SELECT s.customer_id FROM sales_order s);

SELECT id_number, name, department_id, SUM(salary)

FROM employees

WHERE salary > 25000

GROUP BY department_id, id_number, name

ORDER BY hire_date;

Why will this statement cause an error?

The HAVING clause is missing.

The WHERE clause contains a syntax error.

The SALARY column is NOT included in the GROUP BY clause.

The HIRE_DATE column is NOT included in the GROUP BY clause. (*)

OBSERVATIE

ORDER BY cu GROUP BY

SELECT employee_id,department_id, SUM(salary)

FROM employees

WHERE salary>10000

GROUP BY employee_id,department_id,hire_date

ORDER BY hire_date;

TRY IT / SOLVE IT

day 3 - caiet de informatică și tic · exista in jur de 14 tipuri de date varchar2 –lungine...

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