StringPattern-Sorting-Grouping_Solution_Script.sql

--Query 1.1--

SELECT F_NAME , L_NAME
FROM EMPLOYEES
WHERE ADDRESS LIKE '%Elgin,IL%';

--Query 1.2--

SELECT F_NAME , L_NAME
FROM EMPLOYEES
WHERE B_DATE LIKE '197%';

--Query 1.3--

SELECT *
FROM EMPLOYEES
WHERE (SALARY BETWEEN 60000 AND 70000) AND DEP_ID = 5;

--Query 2.1--

SELECT F_NAME, L_NAME, DEP_ID 
FROM EMPLOYEES
ORDER BY DEP_ID;

--Query 2.2--

SELECT F_NAME, L_NAME, DEP_ID 
FROM EMPLOYEES
ORDER BY DEP_ID DESC, L_NAME DESC;

--Optional Query 2.3--

SELECT D.DEP_NAME , E.F_NAME, E.L_NAME
FROM EMPLOYEES as E, DEPARTMENTS as D
WHERE E.DEP_ID = D.DEPT_ID_DEP
ORDER BY D.DEP_NAME, E.L_NAME DESC;

--Query 3.1--

SELECT DEP_ID, COUNT(*)
FROM EMPLOYEES
GROUP BY DEP_ID;

--Query 3.2--

SELECT DEP_ID, COUNT(*), AVG(SALARY)
FROM EMPLOYEES
GROUP BY DEP_ID;

--Query 3.3--

SELECT DEP_ID, COUNT(*) AS "NUM_EMPLOYEES", AVG(SALARY) AS "AVG_SALARY"
FROM EMPLOYEES
GROUP BY DEP_ID;

--Query 3.4--

SELECT DEP_ID, COUNT(*) AS "NUM_EMPLOYEES", AVG(SALARY) AS "AVG_SALARY"
FROM EMPLOYEES
GROUP BY DEP_ID
ORDER BY AVG_SALARY;

--Query 3.5--

SELECT DEP_ID, COUNT(*) AS "NUM_EMPLOYEES", AVG(SALARY) AS "AVG_SALARY"
FROM EMPLOYEES
GROUP BY DEP_ID
HAVING count(*) < 4
ORDER BY AVG_SALARY;