v 1.0 oe nik 2013 1 php+sql 10 (sql 3) group by, having multi-table queries subqueries examples

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PHP+SQL10

(SQL 3)

Group By, HavingMulti-table queriesSubqueriesExamples

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SELECTDisplayed order of suffixes

1. INTO2. FROM3. WHERE4. GROUP BY5. HAVING6. UNION/MINUS7. INTERSECT8. ORDER BY

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PHP+SQL10

(SQL 3)


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Grouping/Aggregate functions • SUM - Sum• AVG - Average• MIN - Minimum• MAX - Maximum• COUNT - Number of non null values (records)• GROUP_CONCAT - Concatenated list of elements• STDDEV - Standard deviation• VARIANCE - Variance

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Non-grouping usage• select avg(sal) as Average from emp;• select min(sal) from emp;• select min(sal) from emp where sal>2000;• select avg(distinct sal) as Average from emp;• select count(sal) from emp;• select count(comm) from emp where sal>2000;• select comm from emp where sal>2000;• select count(*) from emp where sal>2000;• select avg(comm) from emp; NULL values are not

included!

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Grouping• select distinct deptno from emp;• select avg(sal) from emp where deptno=10;• select avg(sal) from emp where deptno=20;• select avg(sal) from emp where deptno=30; select deptno, avg(sal) from emp group by deptno;

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Grouping

IN THE SELECTION LIST (FIELD LIST) ONLY THE GROUPED FIELD(s) AND THE

GROUPING FUNCTION(s) ARE ALLOWED!

(YES, IN MYSQL AS WELL!!!)(ONLY_FULL_GROUP_BY)

• select deptno, avg(sal) as Average, min(sal) as Minimum, count(*) as Num from emp group by deptno;

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Grouping and suffixes

• select mgr, avg(sal) from emp group by mgr;• select ifnull(mgr, "none") as boss, lpad(avg(sal), 15, '#')

as "Averagesal" from emp group by mgr;• HAVING vs. WHERE• select mgr, avg(sal) from emp where ename like '%E%'

group by mgr;• select mgr, avg(sal) from emp where ename like '%E%'

group by mgr having avg(sal)>1300;• select mgr, avg(sal) as average from emp where ename

like '%E%' group by mgr having avg(sal)>1300 order by average desc;

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More complex grouping queries• select min(max(sal)), max(max(sal)),

round(avg(max(sal))) from emp group by deptno; -- In Oracle this works, in MySQL „Invalid use of group function”

• select min(sal+ nvl(comm,0)), mod(empno,3) from emp group by mod(empno,3) having min(sal+nvl(comm,0)) > 800;

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• select distinct job, substr(job, 2, 1) from emp;• select avg(sal) as average, substr(job, 2, 1) from emp

group by substr(job, 2, 1);

• select ename, sal, round(sal/1000) from emp;• select round(sal/1000) as SalCat, count(sal) as Num

from emp group by round(sal/1000);

More complex grouping queries

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• select ename, round(datediff(curdate(), hiredate)/365.25) as diff from emp;

• select count(*), round(datediff(curdate(), hiredate)/365.25) as diff from emp group by round(datediff(curdate(), hiredate)/365.25);


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• select distinct depno, job from emp;• select deptno, job, avg(sal), min(sal), max(sal) from

emp group by deptno, job order by deptno, job;

Oracle-specific „extras” (not required):– GROUP BY GROUPING SETS– GROUP BY CUBE– GROUP BY ROLLUP


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PHP+SQL10

(SQL 3)


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SET OPERATORS• [query] [set operator] [query]• Important: two queries with same number of columns!• Operators: UNION, INTERSECT, MINUS / EXCEPT• E.g.: select min(sal) as MIN_AVG_MAX from emp UNION

select avg(sal) from emp UNION select max(sal) from emp;

• MySQL S***CKS: http://www.bitbybit.dk/carsten/blog/?p=71

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Table structure diagram

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Preparations• We already have a department without a worker

• create table emp1 as select * from emp;• create table dept1 as select * from dept;• update emp1 set deptno=50 where ename='WARD'; The new tables will be inconsistent, but we need this to

test everything• update emp1 set deptno=NULL where ename='WARD'; Now we have a department without worker (40) and a

worker without department (WARD)

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Querying multiple tables• select * from emp1, dept1;• select * from emp1, emp1;• select * from emp1 a, emp1 b;• select a.empno, a.ename, b.empno, b.ename from emp1

a, emp1 b;• select a.ename, a.deptno, b.deptno, b.dname from

emp1 a,dept1 b;• „Cross Join” Cartesian product EMP1 DEPT1

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"MANUAL JOIN"• select a.ename, a.deptno, b.deptno, b.dname from

emp1 a, dept1 b where a.deptno=b.deptno;• Where is Ward???• select a.empno, a.ename, a.mgr, b.empno, b.ename,

b.mgr from emp1 a, emp1 b where a.mgr=b.empno;• Where is King???

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JOIN• Goes into the FROM part of a query• SELECT * FROM table1, table2

SELECT * FROM table1 JOIN_EXPRESSION table2

JOIN_CONDITION• Joining tables = connecting the foreign keys to the

primary keys• Joining indexed fields is fast, joining non-indexed (non-

key) fields is very slow good-to-avoid, cannot-always-avoid

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NATURAL / INNER JOIN• Same result as with "MANUAL JOIN" By default, it

only works with the same field names! (…)• select * from emp1 natural inner join dept1; -- Not in

MySQL!• select * from emp1 natural join dept1;• select * from emp1 inner join dept1 using (deptno);• select * from emp1 inner join dept1 on

(emp1.deptno=dept1.deptno);

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LEFT JOIN … ON …• It must be used if we want to display the records where

there is no matching primary key record• The nonexistent records' fields will be filled with NULL

values use IFNULL if needed

• select a.ename, a.deptno, b.deptno, b.dname from emp1 a LEFT JOIN dept1 b ON a.deptno=b.deptno;

• select a.empno, a.ename, a.mgr, b.empno, b.ename, b.mgr from emp1 a LEFT JOIN emp1 b ON a.mgr=b.empno;

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LEFT/RIGHT JOIN … ON …

• select a.ename, a.deptno, b.deptno, b.dname from emp1 a RIGHT JOIN dept1 b ON a.deptno=b.deptno;

• select a.ename, a.deptno, b.deptno, b.dname from dept1 b LEFT JOIN emp1 a ON a.deptno=b.deptno;

The LEFT JOIN is more used

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JOIN / ORACLE

• SELECT * FROM emp1 d, emp1 fWHERE d.mgr=f.empno (+);

SELECT * FROMemp1 d LEFT JOIN emp1 f ON (d.mgr=f.empno)

• SELECT * FROM emp1 d, emp1 fWHERE d.mgr(+)=f.empno;

SELECT * FROMemp1 d RIGHT JOIN emp1 f ON (d.mgr=f.empno)

• SELECT level, empno, ename, mgr FROM emp START WITH empno=7839 CONNECT BY mgr=prior empno ORDER BY level desc;

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FULL JOIN

• LEFT JOIN + RIGHT JOIN• select a.empno, a.ename, a.deptno, b. deptno,

b.dname from emp1 a FULL JOIN dept1 b ON (emp1.deptno=dept1.deptno);

• [The standard names this type as UNION JOIN; unavailable in MySQL…]

Use the "MANUAL" JOIN and the LEFT JOIN

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JOIN MISTAKE???• SELECT ename, sal, dname FROM emp, dept WHERE

emp.deptno=dept.deptno;

• SELECT ename, sal, dname FROM emp, dept WHERE emp.deptno=dept.deptno AND ename like '%E%';

• SELECT ename, sal, dname FROM emp, dept WHERE emp.deptno=dept.deptno AND ename like '%E%' OR sal<3000;

• SELECT ename, sal, dname FROM emp, dept WHERE emp.deptno=dept.deptno AND (ename like '%E%' OR sal<3000);

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JOINING TABLES

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JOINING TABLES

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JOINING TABLES

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JOINING TABLES

• SELECT * FROMemp1 worker, emp1 boss,dept1 work_dept, dept1 boss_dept

WHEREworker.mgr=boss.empno ANDworker.deptno=work_dept.deptno ANDboss.deptno=boss_dept.deptno;

• Good, if we don't want to see the unpaired records

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JOINING TABLES

• SELECT * FROMemp1 worker LEFT JOIN emp1 boss

ON (worker.mgr=boss.empno),dept1 work_dept, dept1 boss_dept

WHEREworker.deptno=work_dept.deptno ANDboss.deptno=boss_dept.deptno;

• NOT CHANGES ANYTHING

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JOINING TABLES

• SELECT * FROMemp1 worker

LEFT JOIN emp1 boss ON(worker.mgr=boss.empno)LEFT JOIN dept boss_dept ON(boss.deptno=boss_dept.deptno)LEFT JOIN dept work_dept ON (worker.deptno=work_dept.deptno);

• PERFECT

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JOINING TABLES• After the join is made in the FROM/WHERE, the query

can be built up just like as if it was a single-table query. Every suffix can be used, the result-table can be used the same way as with a single-table query

• PRACTICE and STRUCTURED QUERIES are required!!!

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EXAMPLE

Display every boss' name and the average salary of those sub-workers who earn more than 1000 USD. Only display those records where this average is smaller than 5000 USD.

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EXAMPLE

SELECT avg(a.sal) as AVERAGE, b.ename as BOSSFROM emp a, emp bWHERE (a.mgr=b.empno) AND (a.sal>1000)GROUP BY b.enameHAVING avg(a.sal)<5000ORDER BY AVERAGE desc;

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EXAMPLE

SELECT avg(a.sal) as AVERAGE, b.ename as BOSSFROM emp a

LEFT JOIN emp b ON a.mgr=b.empnoWHERE a.sal>1000GROUP BY b.enameHAVING avg(a.sal)<5000ORDER BY AVERAGE desc;

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PHP+SQL10

(SQL 3)


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SUB-QUERIES• Basic principle: in some parts of the main query (field

list, WHERE suffix, FROM suffix) there is another query instead of a simple field name/expression/constant/ table

sub-query

• MySQL only supports it with the newer (>4.1) versions (with some angry limitations…)

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Sub-queries in the field list• You can put a constant in the field list you can put any

sub-query there that returns with exactly one value [more than one row / too many values / operand should contain 1 column(s)]

• SELECT sal,(select max(sal) from emp) as MAX,(select max(sal) from emp)-sal as DELTA FROM empORDER BY DELTA desc;

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Sub-queries in the FROM (inline view)• You can put any table in the FROM every query

returns a table almost any sub-query can be used in the FROM, that we can use as a table in that query

• The sub-query must have an alias and must be written between parentheses

• select worker.salary from (select sal as salary from emp) worker order by salary desc;

• select * from (select a.empno, a.ename, a.mgr, b.empno, b.ename from emp a, emp b where a.mgr=b.empno) order by ename asc; -- Field names must be unique – use an alias!!!

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Sub-queries in the FROMSELECT bosses.work_name as Worker,

bosses.boss_name as BossFROM ( select a.empno as work_id,

a.ename as work_name, a.mgr as work_bossid, b.empno as boss_id, b.ename as boss_name

from emp a, emp b where a.mgr=b.empno) bossesORDER BY Worker asc;

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Sub-queries in the FROMSELECT emp.deptno, min, enameFROM ( Select deptno, min(sal) as min From emp Group by deptno) minimums, empWHERE emp.sal=minimums.min and

emp.deptno=minimums.deptno;

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Sub-query in the WHERE• Use constant values in the WHERE use any sub-query

that returns with exactly one value [more than one row / too many values / operand should contain 1 column(s)]

• ALWAYS VERY SLOW SUBQUERY IN THE FROM IS BETTER

• select ename, sal from emp where sal>(select avg(sal) from emp);

• select ename, sal from emp where sal=(select min(sal) from emp);

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Sub-query in the WHERE

• Use lists in the WHERE use any sub-query that returns with exactly one column

• Operators: [NOT] IN, ANY, ALL• Példa:

select sal, mod(round(sal/1000), 2) from emp;

select sal from emp where mod(round(sal/1000), 2)=0;

select sal from emp where deptno = 10;

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• select ename, sal from emp where sal IN (select sal from emp where deptno = 10);

• select ename, sal from emp where sal NOT IN (select sal from emp where deptno = 10);

Correlated sub-query We’ll talk about optimization later Optimization is not important this semester

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• select ename, sal from emp where sal> ANY (select sal from emp where deptno = 10);

• select ename, sal from emp where sal>(select min(sal) from emp where deptno = 10);

Same results Oracle does an internal sort with the ANY!!!

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• select ename, sal from emp where sal> ALL (select sal from emp where deptno = 10);

• select ename, sal from emp where sal> ALL (select sal from emp where deptno = 30);

• select ename, sal from emp where sal>(select max(sal) from emp where deptno = 30);

Same results No automatic sort Always use the ORDER BY, if required

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PHP+SQL10

(SQL 3)


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Zero• List the name, job and income for the workers who work

in Dallas or Chicago.

• Looks easy… Let’s use operator IN

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Solution using the INSELECT

ename, job, sal+nvl(comm, 0)FROM emp, deptWHERE

emp.deptno=dept.deptno ANDupper(dept.loc) IN ('DALLAS', 'CHICAGO');

SELECT ename, job, sal+nvl(comm, 0)

FROM empWHERE deptno IN (

SELECT deptno FROM deptWHERE upper(loc) IN ('DALLAS', 'CHICAGO')

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BAD Solution using manual joinSELECT

ename, job, sal+nvl(comm, 0)FROM

emp, deptWHERE

emp.deptno=dept.deptno ANDupper(dept.loc)='DALLAS' ORupper(dept.loc)='CHICAGO';

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Solution using manual joinSELECT

ename, job, sal+nvl(comm, 0)FROM

emp, deptWHERE

emp.deptno=dept.deptno AND(upper(dept.loc)='DALLAS' ORupper(dept.loc)='CHICAGO');

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• List the name of the workers, their salary, the name of their boss, the boss' departments' location and the sum salary of that place.

First

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• Sum salary of a location join EMP and DEPT, use sum()

• Worker name and income EMP D• Boss name and income join EMP D and EMP F, use

EMP F• Place of the boss' department join EMP F and DEPT

CHOOSING TABLES AND FIELDS

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SALARIES IN A DEPARTMENT• A simple multi-table GROUP BY• EMP_1 <deptno, deptno> DEPT_1, and then sum(sal)

grouped by the LOC field

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SALARIES

SELECT sum(emp.sal) as dept_sal,dept.loc as dept_loc

FROM emp, deptWHERE emp.deptno=dept.deptnoGROUP BY dept.loc;

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• Workers' name and salary EMP_2• Boss' name and salary join EMP_2 and EMP_3, use

fields in EMP_3• Boss' department's name Join EMP_3 and DEPT_2

EMP_2 <empno, mgr> EMP_3 EMP_3 <deptno, deptno> DEPT_2

WORKERS-BOSSES-BOSS' DATA

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SELECT work.ename as work_name,work.sal as work_sal,boss.ename as boss_name,boss.sal as boss_sal,dept.loc as boss_loc

FROM emp work, emp boss, deptWHERE work.mgr=boss.empno and

boss.deptno=dept.deptno;


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RESULT

• WORKERS-BOSSES-BOSS' DATA data1 • SUM-SALARIES data2• data1 <boss_loc, dept_loc> data2

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SELECT * from(SELECT work.ename as work_name,

work.sal as work_sal,boss.ename as boss_name,boss.sal as boss_sal,dept.loc as boss_loc

FROM emp work, emp boss, deptWHERE work.mgr=boss.empno and

boss.deptno=dept.deptno) data1,(SELECT sum(emp.sal) as dept_sal,

dept.loc as dept_locFROM emp, deptWHERE emp.deptno=dept.deptnoGROUP BY dept.loc) data2

WHERE data1.boss_loc=data2.dept_locORDER BY boss_name, work_name;

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Second

List the number of employees for each boss, along with their departments, average income, and the difference between that average and the boss' income.

Now do we need the department of the boss (A), or the workers (B)???

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• Number of employees, their average income GROUP BY EMP.MGR

• Department Joining EMP and DEPT

CHOOSING FIELDS AND TABLES

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GROUP BY EMP.MGR• We can create a table that looks like boss_id –

avg.income – num.of.workers, that can be joined to the other tables

• How to join: that depends if we interpret the exercise as "A" or "B"

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SELECT avg(sal+nvl(comm, 0)) as avg_work_sal,count(*) as work_num,mgr as boss_id

FROM empGROUP BY mgr

Inline view: Data After this: joining tables

Worker's number and average income

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Joining tables (A)

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Joining tables (B)

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Joining tables (A)

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SELECT emp.ename as boss_name,emp.sal+nvl(emp.comm, 0) as boss_sal,data.avg_work_sal as boss_worker_sal,data.work_num as boss_worker_num,dept.loc as boss_loc,emp.sal+nvl(emp.comm, 0) - data.avg_work_sal as sal_delta

FROM( SELECT avg(sal+nvl(comm, 0)) as avg_work_sal,

count(*) as work_num,mgr as boss_id

FROM empGROUP BY mgr) data, emp, deptWHERE data.boss_id=emp.empno and emp.deptno=dept.deptno; 68OE NIK 2013

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Joining tables (B)

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SELECT d.ename as work_name,f.ename as boss_name,f.sal+nvl(f.comm, 0) as boss_sal,data.avg_work_sal as boss_worker_sal,data.work_num as boss_worker_num,dept.loc as worker_loc,f.sal+nvl(f.comm, 0) -data.avg_work_sal as

sal_delta FROM(SELECT avg(sal+nvl(comm, 0)) as avg_work_sal,

count(*) as work_num,mgr as boss_id

FROM empGROUP BY mgr) data, emp d, emp f, deptWHERE data.boss_id=f.empno and

f.empno=d.mgr and d.deptno=dept.deptno ;

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ThirdList the jobs according to the departments, and the

number of workers, the minimum and maximum salaries according to that job. Order the list by the department (ascending) and job (descending).

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JOB-DATA DATA1SELECT count(*) as Work_Num,

min(sal+nvl(comm, 0)) as Minimum,max(sal+nvl(comm, 0)) as Maximum,avg(sal+nvl(comm, 0)) as Average,job

FROM empGROUP BY job;

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DEPARTMENTS AND JOBS DATA2SELECT distinct loc, jobFROM emp, deptWHERE emp.deptno=dept.deptnoORDER BY loc;

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RESULT (A)SELECT * FROM(SELECT distinct dept.deptno, loc, job

FROM emp, deptWHERE emp.deptno=dept.deptnoORDER BY loc) data2,

(SELECT count(*) as Work_Num,min(sal+nvl(comm, 0)) as Minimum,max(sal+nvl(comm, 0)) as Maximum,avg(sal+nvl(comm, 0)) as Average,jobFROM empGROUP BY job) data1

WHERE data1.job=data2.jobORDER BY data2.deptno asc, data1.job desc;

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RESULT (B)

SELECT count(*) as Work_Num,min(sal+nvl(comm, 0)) as Minimum,max(sal+nvl(comm, 0)) as Maximum,avg(sal+nvl(comm, 0)) as Average,job, deptnoFROM empGROUP BY job, deptnoORDER BY deptno asc, job desc;

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Fourth• Display every worker’s name, their department’s name,

their boss’ name. The list should include the income parameter for both the boss and the worker.

• Income parameter: (income) – (the average income of those who entered the company the same year)

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v 1.0 oe nik 2013 1 php+sql 10 (sql 3) group by, having multi-table queries subqueries examples

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