Choose the letter of the correct answer based on the table EMPLOYEES as shown be
ID: 3902341 • Letter: C
Question
Choose the letter of the correct answer based on the table EMPLOYEES as shown below.
Table 1.0 EMPLOYEES
Select one:
a. SELECT MIN(LASTNAME||FIRSTNAME) AS NAME, DEPARTMENT_ID
FROM EMPLOYEES
GROUP BY DEPARTMENT_ID
WHERE DEPARTMENT_ID IN(90,50)
HAVING MIN(DEPARTMENT_ID)<=90;
b. SELECT MIN(LASTNAME||FIRSTNAME) AS NAME, DEPARTMENT_ID
FROM EMPLOYEES
WHERE DEPARTMENT_ID NOT IN(90,50)
GROUP BY DEPARTMENT_ID
HAVING(DEPARTMENT_ID)<=90;
c. SELECT MIN(LASTNAME||FIRSTNAME) AS NAME, DEPARTMENT_ID
FROM EMPLOYEES
WHERE DEPARTMENT_ID IN(90,50)
HAVING MIN(DEPARTMENT_ID)<=90;
d. SELECT MIN(LASTNAME||FIRSTNAME) AS NAME, DEPARTMENT_ID
FROM EMPLOYEES
WHERE DEPARTMENT_ID IN(90,50)
GROUP BY DEPARTMENT_ID
HAVING MIN(DEPARTMENT_ID)<=90;
2.
Select one:
a. SELECT JOB_ID, COUNTDISTINCT(JOB_ID)"NO. OF JOB_ID", SUM(SALARY), AVG(SALARY)
FROM EMPLOYEES
GROUP BY JOB_ID;
b. SELECT JOB_ID, COUNT(JOB_ID)"NO. OF JOB_ID", SUM(SALARY), AVG(SALARY)
FROM EMPLOYEES
GROUP BY JOB_ID;
c. SELECT JOB_ID, COUNT(JOB_ID)"NO. OF JOB_ID", SUM(SALARY), AVG(SALARY)
FROM EMPLOYEES;
d. SELECT JOB_ID, COUNT*(JOB_ID)"NO. OF JOB_ID", SUM(SALARY), AVG(SALARY)
FROM EMPLOYEES
GROUP BY JOB_ID;
a. SELECT MAX(FIRSTNAME), MANAGER_ID, COUNT(SALARY), AVG(SALARY)
FROM EMPLOYEES
WHERE SALARY BETWEEN 6000 AND 10000
GROUP BY MANAGER_ID
HAVING (SALARY) >= 5000;
b. SELECT MAX(FIRSTNAME), MANAGER_ID, COUNT(SALARY), AVG(SALARY)
FROM EMPLOYEES
GROUP BY MANAGER_ID
WHERE SALARY BETWEEN 6000 AND 10000
HAVING MIN(SALARY) >= 5000;
c. SELECT MAX(FIRSTNAME), MANAGER_ID, COUNT(SALARY), AVG(SALARY)
FROM EMPLOYEES
WHERE SALARY BETWEEN 10000 AND 6000
GROUP BY MANAGER_ID
HAVING MIN(SALARY) >= 5000;
d. SELECT MAX(FIRSTNAME), MANAGER_ID, COUNT(SALARY), AVG(SALARY)
FROM EMPLOYEES
WHERE SALARY BETWEEN 6000 AND 10000
HAVING MIN(SALARY) >= 5000
GROUP BY MANAGER_ID;
EMPLOYEE ID FIRSTNAME LASTNAME JOB ID SALARY MANAGER ID DEPARTMENT ID 24000 17000 17000 90 90 90 100 STEVEN 101 NENA 102 LEX 03 ALEXANDER HUNOLD 104 BRUCE 107 DIANA 124 KEVIN 141 TRINA 142 CURTIS 43 RANDALLMATOS 144 PETER 149 ELENI 174 ELLEN 176 JONATHAN TAYLOR 78 KIMBERLEY GRANT 200 JENNIFER WHALEN AD ASST 201 MICHAELHARTSTEIN MK MAN 202 PAT 205 SHELLEYHIGGINS AC MGR 206 WILLIAM GIETZ KING KOCHAR E HAAN AD VP AD PRES AD VP 100 100 101 102 103 100 124 124 124 124 100 149 149 149 101 100 201 101 205 IT PROG IT PROG 6000 4200 5800 3500 3100 2600 2500 10500 11000 8600 7000 4400 13000 6000 12000 8300 ERNST LORENT2 IT PROG MOURGOS ST MAN RAJS DAVIES 50 ST CLERK ST CLERK ST CLERK ST CLERK 50 50 50 80 50 80 VARGAS LOTKEY SA MAN ABEL SA REP SA REP SA REP 10 20 20 110 110 FAY MK REP AC ACCOUNTExplanation / Answer
1.
Order of syntax: WEHRE GROUP_BY HAVING
d. SELECT MIN(LASTNAME||FIRSTNAME) AS NAME, DEPARTMENT_ID
FROM EMPLOYEES
WHERE DEPARTMENT_ID IN(90,50)
GROUP BY DEPARTMENT_ID
HAVING MIN(DEPARTMENT_ID)<=90;
2.
Results are grouped by job_id
To get the count, COUNT is used
b. SELECT JOB_ID, COUNT(JOB_ID)"NO. OF JOB_ID", SUM(SALARY), AVG(SALARY)
FROM EMPLOYEES
GROUP BY JOB_ID;
3.
Order of syntax: WHERE GROUP_BY HAVING
a. SELECT MAX(FIRSTNAME), MANAGER_ID, COUNT(SALARY), AVG(SALARY)
FROM EMPLOYEES
WHERE SALARY BETWEEN 6000 AND 10000
GROUP BY MANAGER_ID
HAVING (SALARY) >= 5000;