num_rows()?

[sabret00the]

why won't this work?

PHP Code:

    $totalquery = $DB_site->query_first("SELECT* FROM grps WHERE title LIKE '%" . addslashes($search) . "%' ORdescription LIKE '%" . addslashes($search) . "%'");
    $totalresults = number_format($DB_site->num_rows($totalquery)); 


the error i'm getting is

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in \includes\db_mysql.php on line 312

[cinq]

PHP Code:

$totalquery = $DB_site->query("SELECT * FROM grps WHERE title LIKE '%".addslashes($search)."%' OR description LIKE '%".addslashes($search)."%'");

$totalresults = number_format($DB_site->num_rows($totalquery)); 


?

[noppid]

Using query fiirst returns the first row, not the identifier.

Edited:

[filburt1]

You should use COUNT(), not SELECT *.

[sql]
SELECT COUNT(unique id column name) AS totalcount FROM grps WHERE title LIKE "expression%
[/sql]
It will return a single result row with a column name of totalcount containing the number of matching rows.

Don't forget TABLE_PREFIX.

[deathemperor]

it's because he wrote "SELECT*" instead of "SELECT *"

[sabret00the]

Originally Posted by filburt1

You should use COUNT(), not SELECT *.

[sql]
SELECT COUNT(unique id column name) AS totalcount FROM grps WHERE title LIKE "expression%
[/sql]
It will return a single result row with a column name of totalcount containing the number of matching rows.

Don't forget TABLE_PREFIX.

thanks that fixed it perfectly

Originally Posted by noppid

Using query fiirst returns the first row, not the identifier.

Edited:

seems i should've used query, didn't think of that.