2 replies to this topic

[Dorian]

I am currently programming the v2 of my site
but I came across an annoying problem :
I keep on getting the same error message

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in... line 56

with this code :

CODE

54. $sql = "SELECT * from DP_news WHERE passage=1184408630";
55. $result = mysql_query($sql);
56. print_r(mysql_fetch_array($result));

one hour I'm trying to resolve the problem... what's wrong?
I really don't understand, I even copy-pasted an exemple on the w3school website and it still won't work !
any suggestion?

[Saffith]

The query's failing for some reason...
Try this:

CODE

$sql = "SELECT * from DP_news WHERE passage=1184408630";
$result = mysql_query($sql) or die(mysql_error());
print_r(mysql_fetch_array($result));

That way, it should tell you why it's failing.

[Dorian]

that doesn't let me know more or less ...

EDIT: this is strange, the homepage was working yesterday but now there are tons of mysql resource errors...
something must be wrong about the database :/

EDIT2 : i copied the table content to another, made the codes changes and the index page was working again.
then I noticed that I didn't selected my database (it was selected but only in the if's first branch) on the "else" branch.
after some code cleaning it's working

thanks anyway

Edited by Dorian, 26 July 2007 - 12:25 PM.