Hey team!
I can’t download the solution posted in the SQL Subqueries nested in SELECT and FROM – solution 2 part of the sql course. I tried using chrome, safari and mozilla but the download didn’t work. Highly appreciate if you can help me on this.
Hi Paul!
Thanks for reaching out!
In similar situations, please try using another browser to download the necessary course resources – this sometimes helps.
Otherwise, please find the code pasted below. Thank you.
Hope this helps.
Best,
Martin
INSERT INTO emp_manager
SELECT
u.*
FROM
(SELECT
a.*
FROM
(SELECT
e.emp_no AS employee_ID,
MIN(de.dept_no) AS department_code,
(SELECT
emp_no
FROM
dept_manager
WHERE
emp_no = 110022) AS manager_ID
FROM
employees e
JOIN dept_emp de ON e.emp_no = de.emp_no
WHERE
e.emp_no <= 10020
GROUP BY e.emp_no
ORDER BY e.emp_no) AS a UNION SELECT
b.*
FROM
(SELECT
e.emp_no AS employee_ID,
MIN(de.dept_no) AS department_code,
(SELECT
emp_no
FROM
dept_manager
WHERE
emp_no = 110039) AS manager_ID
FROM
employees e
JOIN dept_emp de ON e.emp_no = de.emp_no
WHERE
e.emp_no > 10020
GROUP BY e.emp_no
ORDER BY e.emp_no
LIMIT 20) AS b UNION SELECT
c.*
FROM
(SELECT
e.emp_no AS employee_ID,
MIN(de.dept_no) AS department_code,
(SELECT
emp_no
FROM
dept_manager
WHERE
emp_no = 110039) AS manager_ID
FROM
employees e
JOIN dept_emp de ON e.emp_no = de.emp_no
WHERE
e.emp_no = 110022
GROUP BY e.emp_no) AS c UNION SELECT
d.*
FROM
(SELECT
e.emp_no AS employee_ID,
MIN(de.dept_no) AS department_code,
(SELECT
emp_no
FROM
dept_manager
WHERE
emp_no = 110022) AS manager_ID
FROM
employees e
JOIN dept_emp de ON e.emp_no = de.emp_no
WHERE
e.emp_no = 110039
GROUP BY e.emp_no) AS d) as u;