Mysql query does not work, displays an empty field

Asked Feb 18 '20 at 14:40

Active Feb 18 '20 at 16:16

Viewed 30 times

The first table with posts (posts with columns: id, post_id, post_name, post_date)

The second table with a view counter (post_views with columns: id, post_id, post_views).

$count_boards = $db-> super_query ("SELECT * FROM posts AS p
JOIN post_views AS pv ON p.post_id = pv.post_id
WHERE p.post_date> = (CURDATE () - INTERVAL 3 DAY) ";

$count_boards = $count_boards['post_views'];
echo $count_boards;

When outputting, it doesn’t output anything, that is, it is empty ... What could be the problem?

edited Feb 18 '20 at 16:16

Dan

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asked Feb 18 '20 at 14:40

keaks88

What is "`super_query`"? Also, I hope the spaces between $ and the var names are just a copy fault and not in your code. – N3x Feb 18 '20 at 14:42
Did you check your error logs? – Andrea Golin Feb 18 '20 at 14:45
1

Please remove the spaces. – Dharman Feb 18 '20 at 14:45
`$count_boards = $db->super_query("SELECT * FROM posts AS p JOIN post_views AS pv ON p.post_id = pv.post_id WHERE p.post_date> = (CURDATE () - INTERVAL 3 DAY) "; $count_boards = $count_boards['post_views']; echo $count_boards;` – keaks88 Feb 18 '20 at 14:47
@keaks88 just edit your question... – N3x Feb 18 '20 at 14:48
Try `echo` $count_boards before the line that says, `$count_boards = $count_boards['post_views'];` – Dan Feb 18 '20 at 15:04
super_query = mysql_query – keaks88 Feb 18 '20 at 15:10

Mysql query does not work, displays an empty field

0 Answers0