Nested SQL Dr. Zhen Jiang

Transcription

1 Nested SQL Dr. Zhen Jiang Sometimes we need the result of one query prior to the querying of another.consider the following: List the name and age of all employees who are younger than the average age for all the employees in the company. The following WILL NOT WORK- because the where clause can t be applied to each row. No such a value of avg(age exists in each row for the Boolean check. SELECT enum,lname, age From emp where age >avg(age ; To answer this query we need to first find the average age as a separate query and then use this result to find all the employees as required. Consider the following query that simply finds the average age for all employees and then nest this query inside our original query as SELECT enum,lname, age From emp where age > (SELECT avg(age From emp ; Consider the following request: List the employee number, name and age, for all employees who work in departments with more than 5 people. This will not work - Why? SELECT enum, Lname, age having count(enum>5; The solution is expect to find a list of departments with more than 5 employees first and then use this result to check if each employee is in one of those departments. SELECT enum, Lname, age wherednum IN ( Select Dnum having count(enum>5

2 Exercises: write SQL queries for each of the following. QN1 List the name, number and age of all employees make less than the average salary. 1. First write a query to simply find the average salary of all employees. 2. Next write a nested query where we look at each employee and see if that employee makes less than the value calculated by the above sub query. Select Lname, Enum, age where salary < ( SELECT avg(salary FROM emp QN2 Find all the employees (enum, lname, dnum who work in departments whose average salary is >51, First write a sub query that lists all departments whose average salary is >51, Next write a nested query which list all employees who are in any of the departments listed in the above sub query. select enum, lname, dnum where dnum in ( select dnum having avg(salary>51000 * why use <in>? QN3 For all employees in D25 who make more than the average salary in the department, List enum, lname and salary. 1. First write a sub query that lists the average salary for those in d25 (Does not require grouping 2. Next write a nested query which lists all employees who are in d25 AND make more than the average salary calculated in the above sub query.

3 select enum, lname, salary where dnum like 'd25' and salary > ( select avg(salary where dnum like 'd25' * why two dnum like d25? QN4 Find all employees (enum, lname, dnum who work in departments that have no senior person (age>= First write a sub query that lists the department that has any person older than Next write a nested query which lists all employees who are NOT in department calculated in the above sub query. select enum, lname, dnum where dnum not in ( select dnum where age > 55 * Why not use <>, and why use not in? QN5 Find the department(s which contains the most junior workers (age <= First write a sub query that lists the count of juniors in each department. 2. Second write a nested query to get the maximum in such counting, which must have been prepared in a temporary table in advance in the above sub query. 3. Third, write another (higher level nested query to select the departments while having its counting of juniors matches the above maximum value.

4 select dnum where age <=30 having count(id = ( select max(c from( select count(id as C where age <=30 * how the tie problem can be solved here in case we have 2 departments reaching the maximum as the result? So far, we handle all distinct record, i.e., there is no duplicate in headcount (counting of id. Consider a query similar to QN5 in the following table: TC Num A B S X PA Easy X20 X22 X25 X27 X33 X38 X49 X55 X58 X62 X88 X PA Hard D 8 50 NJ Impossible 2 30 DE Hard 3 80 NJ Easy 4 20 DE Easy 9 40 NJ Hard 2 60 DE Easy 7 18 NJ Easy 4 20 PA Impossible 4 70 PA Hard 3 40 DE Easy 9 36 DE Hard

5 QN6 Find the S which contains the most D records. If you modified the code of QN5 by the replacement of Departmentà S and Junior workers count(id à count(d : select S having count(d = ( select max(c from( select count(d as C * The result is DE, not PA and NJ! Here is the correct answer: 1. First write a sub query that lists the distinct D records in each S. 2. Second write a nested query to get the count of distinct D in each S (group by. 3. Third, write another (higher level nested query to select the maximum in such counting, which must have been prepared in a temporary table in advance in the above sub query. 4. At the end, write the highest level nested query to select the S from distinct record table with the corresponding counting of D matches the above maximum value. Select S from ( Select S, D, D having count(d = ( Select max(c from ( select count (D as C from ( Select S, D, D