
SUBQUERY & JOIN
Quiz
•
Other
•
Professional Development
•
Practice Problem
•
Easy
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11 questions
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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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