[SOLVED] MySql LIMIT problem

[Msword]

new problem, this is giving me syntax error. The variables are right, $page == 0. $by == pid. i want it to display 1-25 and i just gives error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/.marja/msword/draftx.net/test/displaytable.php on line 54

 

Code:

if ($sort == "asc")
    {
    $result = mysql_query("SELECT pid, name,location, quest, membonly FROM pdatabase LIMIT '$page', 25
    ORDER BY " . $by);
    }
    ELSE
    {
    $result = mysql_query("SELECT pid, name,location, quest, membonly FROM pdatabase LIMIT '$page', 25
    ORDER BY " . $by . " DESC");
    }

technicly the error is when i'm displaying, but i know that works

Code:

while($row = mysql_fetch_array($result))

[teng84]

first remove the ' ' in this '$page', this is supposedly an INT

 

and try to put a die mysql_error to see the error but i think removing this '' will do

[matto]

When using limit you should use it like this:

 

SELECT pid, name, location, quest, membonly FROM pdatabase
order by pid
limit $page, 25

 

Bascially the limit should be the last part of the SQL statement....

 

 

[teng84]

yahhhh 8) 8)

[Msword]

Seems to have worked, thanks again guys