sql query to count number of employees in each department

For each department whose average employee salasy is more than $30,000, retrive the department name and the number of employees working for that department. For example, we can use the AVG function to calculate the average salary of employees in each department, and apply the AVG function one more time to calculate the average … In this example you will be grouping by deptno. For this SQL server Derived Table example, we are using Inner Join to get a column from both Employees and Department tables. The SUM() function returns the total sum of a numeric column. Type ----- ----- ----- EMPLOYEE_ID NOT NULL NUMBER(6) FIRST_NAME VARCHAR2(20) LAST_NAME NOT NULL VARCHAR2(25) EMAIL NOT NULL VARCHAR2(25) PHONE_NUMBER VARCHAR2(20) HIRE_DATE NOT NULL DATE JOB_ID NOT NULL VARCHAR2(10) SALARY NUMBER… use Company; select Dname, COUNT (*) Number_Of_Employee: from DEPARTMENT, EMPLOYEE: where Dno = Dnumber: group by Dname: having AVG … The COUNT() function returns the number of rows that matches a specified criterion. We can apply AVG function multiple times in a single SQL statement to calculate the average value of a set of average values. List the employees whose salary is higher than the average salary with self join Write a SQL query that will return all the unique department numbers that are represented in the EMPLOYEE table, ordered by department number. Something like this should do SQL> select * 2 from scott.emp e 3 where not exists 4 ( select null 5 from scott.emp 6 where deptno = e.deptno 7 and sal > e.sal ); EMPNO ENAME JOB MGR HIREDATE SAL COMM … 1. Twitter Interview Online Test SQL A company uses 2 data tables, Employee and Department, to store data about its employees and departments. SQL AVG with a subquery. Out should be like this : - EmplD NoofEmployees 1234 10 2562 25 Where 10 & 25 is actually the total number of employees working under the manager of EmplD 1234 and 2562 … Which query will return the minimum salary in each department? Write a query to print the respective Department Name and number of employees for all departments in the Department table (even unstaffed ones). Include the deptno column in your SQL query. Write a SQL query to return the employee number, name and salary of all ‘CLERK’ who work in Department 10. Display job ID, number of employees, sum of salary of each … Need to Pass employee Id and get total number of employees working under the manager whose employee id is passed in SQL. Answer: SELECT DEPARTMENT_ID, COUNT (EMPLOYEE_ID) FROM EMPLOYEES GROUP BY DEPARTMENT_ID; Result: Posted by Prosen Ghosh at 08:51. Create a SQL query that will use the count function to return the number of employees in each department in the emp table. COUNT() Syntax. Display the employees who are working in “Oxford” (should use sub query) SELECT EMPLOYEES.EMPLOYEE_ID,EMPLOYEES.FIRST_NAME,EMPLOYEES.LAST_N AME,LOCATIONS.CITY FROM EMPLOYEES . 2.6.6. Employees Per Department. List department who has more than 4 employee born after 1960: 2.6.8. SQL> DESC employees Name Null? The SQL COUNT(), AVG() and SUM() Functions. SELECT COUNT(column_name) ... --57. In what city do we have the most employees? Display department number and its employee count if it has more than 4 employees: 2.6.7. The AVG() function returns the average value of a numeric column. Apologies, I misread the question as "show all employees and show the highest salary per dept". Copy and paste your code into your assignment and include a screenshot of your result set in your … 25. The following Query count the total number of employees present in each department, their sales amount, and total annual salaries. Write an SQL query to print the name of employees having the highest salary in each department. Display depart numbers and total number of employees working in each department SQL Query Solution. Could you please help me to get the highest salary for each department. If multiple employees having the highest salary in the same department then I need the get oldest employee details(who joined first in …

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