This is crazy..

[Cody]

Hi!

I am using this code, which should and does work when i run a manual query from mysql-front interface but when i call it from php i get:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in ..

What the hell is wrong? As i see, theres nothing wrong with it !!! It's driving me crazy !!!

Here's the code:

$News_query = mysql_query("SELECT * FROM News WHERE ID='$NID'");
$News = mysql_fetch_array($News_query);

Thank you 🙂

[Drakla]

1) where is $NID coming from? that's one factor in things going wrong.

2) less likely, because of a change in environment capitalisation of table names might matter on one system, and not on another.

Put after your query

or die(mysql_error())'

[Cody]

oh! I am using two MySQL servers and on the one I was testing now I forgot to create the table. lol.

Thank you Drakla! 🙂