I am using SQL Developer and I have this table structure:
Name Null Type
-------------- -------- ------------
EMPLOYEE_ID NOT NULL NUMBER(6)
FIRST_NAME VARCHAR2(20)
LAST_NAME NOT NULL VARCHAR2(25)
DEPARTMENT_ID NUMBER(4)
SALARY NUMBER(8,2)
What I want to do is print out the department_id which has AT LEAST 2 employees with salaries greater than 10000.
I though this would be the query
select department_id
from employees
having count(select * from employees where salary > 10000) > 2
group by department_id;
but, from what I found out, you can't put a SELECT statement inside COUNT so now I an stuck and I don't know how else am I supposed to do this query. Any suggestion is welcome.
UPDATE: Please note that I want AT LEAST 2 employees to have salary > 10000, not all of them
SELECT Department_Id,
COUNT(*)
FROM Employee
WHERE Salary > 10000
GROUP BY Department_Id
HAVING COUNT(*) > 1
SQL Fiddle example.
SELECT department_id FROM employees
WHERE salary > 10000 GROUP BY department_id HAVING COUNT(*) >= 2;
Related
I have a practice that I should find the employees who earn more than average salary and works in the departments with employees whose last name contains the letter u
the select statement I have used was
SELECT employee_id,
last_name,
salary
FROM employees
WHERE salary > (SELECT AVG(salary)
FROM employees )
AND department_id IN(SELECT department_id
FROM employees
WHERE LOWER(last_name) LIKE '%u%')
Could anyone check this statement is suitable or not ?
thank you
That looks fine to me, assuming you mean the average salary across all departments in the database, and all employees (active or not) across all of time.
I would think you might be more interested in all active employees in this current financial year, for example.
You haven't provided the schema, so be careful to check for conditions like:
inactive departments
inactive / terminated employees
period you are interested in for comparing the salary
Your queries looks like it will work. You can rewrite it to remove all the sub-queries (that will require additional table/index scans) and just use analytic queries:
SELECT employee_id,
last_name,
salary
FROM (
SELECT employee_id,
last_name,
salary,
AVG( salary ) OVER () AS avg_salary,
COUNT( CASE WHEN LOWER( last_name ) LIKE '%u%' THEN 1 END )
OVER ( PARTITION BY department_id ) AS num_last_name_with_u
FROM employees
)
WHERE salary > avg_salary
AND num_last_name_with_u > 0;
db<>fiddle
My first Question are you getting the expected result ?
Let me break down your Query
SELECT department_id FROM employees WHERE LOWER(last_name)
Here you are selecting the department so it retrieve the department id, what is the need of selecting department Id when all you need employee_id with last name contains u so change it to employee_id instead of department_id
select avg(salary) over (partition by department_id order by employee_id)
So using partition by you must get the avg salary per department
SELECT employee_id,last_name,salary
FROM
employees
WHERE salary>(SELECT AVG(salary) OVER (PARTITION BY department_id)
FROM
employees )
AND employee_id IN
( SELECT employee_id
FROM
employees
WHERE LOWER(last_name) LIKE '%u%')
Let me know if you have any issues running it, any corrections to Query is appreciated
Can anyone explain why this query:
SELECT employee_id, last_name, salary
FROM employees
WHERE department_id IN (SELECT department_id
FROM employees
WHERE last_name LIKE '%u%'
)
AND salary > (SELECT AVG(salary)
FROM employees);
returns way less rows than this nested one:
SELECT employee_id, last_name, salary
FROM employees
WHERE department_id IN (SELECT department_id
FROM employees
WHERE last_name LIKE '%u%'
AND salary > (SELECT AVG(salary)
FROM employees);
)
The first returns all employees who meet the following conditions:
The employee is in a department has a "u" employee.
The employee has a salary larger than the average.
The second returns all employees who meet these conditions:
The employee is in a department that has a "u" employee who has a salary larger than the average.
The two are very different conditions. I wouldn't expect them to return the same result set.
Also, whenever you have more than one table in a query, you should use table aliases that are abbreviations of the table name and you should qualify all column names.
I have this table:
Name Null? Type
-------------------------- -------- ------------
EMPLOYEENO NOT NULL NUMBER(4)
ENAME VARCHAR2(15)
JOB VARCHAR2(15)
MGR NUMBER(4)
HIREDATE DATE
SAL NUMBER
COMM NUMBER
DEPTNO NUMBER(2).
I want to get the department with minimum employees who have a given job (for example all the employees with 'Analyst' job).
Can you please help me with the query?
Here the key is to get the count of Employee doing particular job in each department. In below query, this is achieved by subquery. Then, we want to get the Department with minimum no. of employee doing that job so we ordered the records returned by subquery in ascending and then select the first result by using rownum = 1
SELECT DEPTNO from (
SELECT COUNT(*) AS NO_OF_EMP , DEPTNO
FROM EMPLOYEE EMP
WHERE EMP.JOBNAME = 'Analyst'
GROUP BY DEPTNO
ORDER BY NO_OF_EMP ASC )
where ROWNUM = 1;
Given that the minimum number of such employees could be 0, you need to be a bit clever about how you do this:
select d.*
from (select deptno,
sum(case when jobname = 'Analyst' then 1 else 0 end) as numAnalysts
from employees
group by deptno
order by numAnalysts asc
) d
where rownum = 1;
i want to display department_id's along with count,and count should be more than 5, and i want to have employees who are not hired in January.
i tried the below query
SELECT * FROM EMPLOYEES
WHERE DEPARTMENT_ID IN
(
SELECT DEPARTMENT_ID
FROM EMPLOYEES
GROUP BY DEPARTMENT_ID
HAVING COUNT(*)>5
)
AND HIRE_DATE NOT LIKE '%JAN%';
but here I didnt get count.I want count Also.
SELECT department_ID, count(employee_id) as '# of Employees' FROM EMPLOYEES
WHERE DEPARTMENT_ID IN
(
SELECT DEPARTMENT_ID
FROM EMPLOYEES
GROUP BY DEPARTMENT_ID
HAVING COUNT(*)>5
)
AND HIRE_DATE NOT LIKE '%JAN%'
group by department_ID;
This query returns the department_id and because I group by department_id, the count of employees that belong to each department will be returned
Output will look something like this
Department_Id | # of Employees
1 7
2 6
4 9
If you want the dept id and count of employees (where employee hire date is not in Jan) then something like the following should work. I say "something like the following" because I suspect the WHERE hire_date NOT LIKE '%JAN%' could be improved, but it would just depend on the format of that column.
SELECT
DEPARTMENT_ID,
COUNT(*)
FROM EMPLOYEES
WHERE HIRE_DATE NOT LIKE '%JAN%'
GROUP BY DEPARTMENT_ID
HAVING COUNT(*)>5;
If you also want to list the individual employees along with these departments, then something like this might work:
SELECT a.*, b.count(*)
FROM EMPLOYEES AS a
INNER JOIN (
SELECT
DEPARTMENT_ID,
COUNT(*)
FROM EMPLOYEES
WHERE HIRE_DATE NOT LIKE '%JAN%'
GROUP BY DEPARTMENT_ID
HAVING COUNT(*)>5) AS b
ON a.department_id = b.department_id
WHERE a.HIRE_DATE NOT LIKE '%JAN%';
Again, though, I think you can leverage your schema to improve the where clause on HIRE_DATE. A like/not-like clause is generally going to be pretty slow.
Select the count from your inner query and join to it:
SELECT E.*, DEPT_COUNT
FROM EMPLOYEES E
JOIN (
SELECT DEPARTMENT_ID, COUNT(*) DEPT_COUNT
FROM EMPLOYEES
GROUP BY DEPARTMENT_ID
HAVING COUNT(*) > 5
) DC ON E.DEPARTMENT_ID = DC.DEPARTMENT_ID
AND HIRE_DATE NOT LIKE '%JAN%'
I am using oracle's SQL Developer. To begin with, I have this table:
Name Null Type
-------------- -------- ------------
EMPLOYEE_ID NOT NULL NUMBER(6)
FIRST_NAME VARCHAR2(20)
LAST_NAME NOT NULL VARCHAR2(25)
DEPARTMENT_ID NUMBER(4)
I would like to retrieve(select) the ID of the department in which are the most employees.
I managed through a statement to retrieve all the numbers of the employees in every department:
select count(employee_id), department_id
from employees
group by department_id;
It gives me something like:
count(employee_id) | department_id
---------------------|------------------
6 100
16 30
1 12
What I would like to do is ONLY to retrieve the department_id 30, which (in this case) has the most employees.
A typical way to do this in Oracle:
select department_id
from (select count(employee_id), department_id
from employees
group by department_id
order by count(employee_id) desc
) t
where rownum = 1;
If you have potential duplicates and want all the department ids, then a join to the max or analytic function is a better approach. For example:
select department_id
from (select count(employee_id), department_id,
rank() over (order by count(employee_id) desc) as seqnum
from employees
group by department_id
) t
where seqnum = 1;