Ages of people above average MySQL ages WHERE, HAVING How?

Asked Jul 13 '21 at 20:52

Active Jul 13 '21 at 21:01

Viewed 21 times

Question: Show which people are alive, their ages to people above the average age of living people.

SELECT nome, TIMESTAMPDIFF (YEAR, data_nasc,CURDATE()) AS "Idade"
FROM pessoas
WHERE data_morte IS NULL AND TIMESTAMPDIFF (YEAR, data_nasc,CURDATE()) > AVG(TIMESTAMPDIFF (YEAR, data_nasc,CURDATE()));

When I put it in Mysql, this error appears, I've tried in several ways and I couldn't solve it, can you help me please? ERROR 1111 (HY000): Invalid use of group function

edited Jul 13 '21 at 21:01

asked Jul 13 '21 at 20:52

Bruno

I put it the correct way, thanks for letting me know – Bruno Jul 13 '21 at 20:56
Does this answer your question? [ERROR 1111 (HY000): Invalid use of group function](https://stackoverflow.com/questions/28171977/error-1111-hy000-invalid-use-of-group-function) – B001ᛦ Jul 13 '21 at 20:57
You can't use aggregate functions in the `WHERE` clause. Use `HAVING` instead. – Barmar Jul 13 '21 at 20:57
How Could a Use? Could you show me? I don't know "HAVING" – Bruno Jul 13 '21 at 20:58
Why my question is closed? It's different! – Bruno Jul 13 '21 at 21:04

Ages of people above average MySQL ages WHERE, HAVING How?

0 Answers0