SUBQUERY & JOIN

SUBQUERY & JOIN

Professional Development

11 Qs

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SUBQUERY & JOIN

SUBQUERY & JOIN

Assessment

Quiz

Other

Professional Development

Practice Problem

Easy

Created by

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11 questions

Show all answers

1.

REORDER QUESTION

1 min • 1 pt

Cari Gaji Rata-Rata Per Departemen

SELECT d.dept_name, AVG(s.salary) AS avg_salary

FROM departments d JOIN dept_emp de

JOIN salaries s ON de.emp_no = s.emp_no

GROUP BY d.dept_name;

ON d.dept_no = de.dept_no

Answer explanation

SELECT d.dept_name, AVG(s.salary) AS avg_salary

FROM departments d

JOIN dept_emp de ON d.dept_no = de.dept_no

JOIN salaries s ON de.emp_no = s.emp_no

GROUP BY d.dept_name;

2.

REORDER QUESTION

1 min • 1 pt

Temukan Karyawan dengan Jabatan Tertentu di Departemen

de.dept_no = d.dept_no JOIN titles t

ON e.emp_no = de.emp_no JOIN departments d ON

SELECT e.first_name, d.dept_name, t.title FROM employees e JOIN dept_emp de

ON e.emp_no = t.emp_no

WHERE t.title = 'Senior Engineer';

Answer explanation

SELECT e.first_name, d.dept_name, t.title

FROM employees e

JOIN dept_emp de ON e.emp_no = de.emp_no

JOIN departments d ON de.dept_no = d.dept_no

JOIN titles t ON e.emp_no = t.emp_no

WHERE t.title = 'Senior Engineer';

3.

REORDER QUESTION

1 min • 1 pt

Hitung Total Karyawan di Setiap Departemen

FROM employees e

COUNT(*) AS total_employees

JOIN dept_emp de ON e.emp_no = de.emp_no

SELECT d.dept_name,

JOIN departments d ON de.dept_no = d.dept_no GROUP BY d.dept_name;

Answer explanation

SELECT d.dept_name, COUNT(*) AS total_employees

FROM employees e

JOIN dept_emp de ON e.emp_no = de.emp_no

JOIN departments d ON de.dept_no = d.dept_no

GROUP BY d.dept_name;

4.

REORDER QUESTION

1 min • 1 pt

Cari Karyawan yang Pernah Bekerja di Beberapa Departemen

GROUP BY e.emp_no

FROM employees e

JOIN dept_emp de ON e.emp_no = de.emp_no

HAVING COUNT(de.dept_no) > 1;

SELECT e.first_name, e.last_name

Answer explanation

SELECT e.first_name, e.last_name

FROM employees e

JOIN dept_emp de ON e.emp_no = de.emp_no

GROUP BY e.emp_no

HAVING COUNT(de.dept_no) > 1;

5.

REORDER QUESTION

1 min • 1 pt

Temukan Manager dengan Gaji Tertinggi di Departemennya

JOIN salaries s ON t.emp_no = s.emp_no

FROM departments d JOIN dept_emp de ON d.dept_no = de.dept_no JOIN titles t ON de.emp_no = t.emp_no

ORDER BY MAX(s.salary) DESC;

SELECT d.dept_name, MAX(s.salary) AS highest_salary

WHERE t.title = 'Manager' GROUP BY d.dept_name

Answer explanation

SELECT d.dept_name, MAX(s.salary) AS highest_salary

FROM departments d

JOIN dept_emp de ON d.dept_no = de.dept_no

JOIN titles t ON de.emp_no = t.emp_no

JOIN salaries s ON t.emp_no = s.emp_no

WHERE t.title = 'Manager'

GROUP BY d.dept_name

ORDER BY MAX(s.salary) DESC;

6.

REORDER QUESTION

1 min • 1 pt

Cari Karyawan yang Gajinya di Atas Rata-Rata Departemennya

SELECT e.first_name, e.last_name, s.salary FROM employees e

JOIN departments d ON de.dept_no = d.dept_no WHERE d.dept_name = d.dept_name );

SELECT AVG(s.salary) FROM salaries s JOIN dept_emp de ON s.emp_no = de.emp_no

WHERE s.salary > (

JOIN dept_emp de ON e.emp_no = de.emp_no JOIN departments d ON de.dept_no = d.dept_no JOIN salaries s ON e.emp_no = s.emp_no

Answer explanation

SELECT e.first_name, e.last_name, s.salary

FROM employees e

JOIN dept_emp de ON e.emp_no = de.emp_no

JOIN departments d ON de.dept_no = d.dept_no

JOIN salaries s ON e.emp_no = s.emp_no

WHERE s.salary > (

SELECT AVG(s.salary)

FROM salaries s

JOIN dept_emp de ON s.emp_no = de.emp_no

JOIN departments d ON de.dept_no = d.dept_no

WHERE d.dept_name = d.dept_name

);

7.

FILL IN THE BLANK QUESTION

1 min • 1 pt

Temukan nama karyawan yang gajinya lebih tinggi dari rata-rata gaji seluruh karyawan.

SELECT e.first_name, e.last_name, s.salary

FROM employees e

JOIN salaries s ON e.emp_no = s.emp_no

WHERE s.salary > ( _______ );

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