Re: [PHP] MySQL error... but it works!!??

Hi, So far the best way to troubleshoot mysql+php for me has been to add an "or die" after the sql exec. $crdate = date("Y-m-d"); $result = mysql_query("SELECT * FROM sites WHERE creation_date = '$crdate' AND status = 'T'") or die ('CAN NOT EXEC SQL'); Or you can do $sql = "SELECT * FROM sit

Fw: [PHP] MySQL error... but it works!!??

- Original Message - From: "Bas van Rooijen" <[EMAIL PROTECTED]> To: "Adam Baratz" <[EMAIL PROTECTED]> Sent: Saturday, January 05, 2002 4:39 PM Subject: Re: [PHP] MySQL error... but it works!!?? > > maybe try: > > if (!$result) > { >

Re: [PHP] MySQL error... but it works!!??

> Does anyone have any idea why this is giving me a "Warning: Supplied > argument is not a valid MySQL result resource in..." error? You'll have to give more code, which line you get that error from, etc., to give us a better idea why it's "not working." However, a quick tip: > $crdate = date("

[PHP] MySQL error... but it works!!??

Does anyone have any idea why this is giving me a "Warning: Supplied argument is not a valid MySQL result resource in..." error? The funny thing is that apart from that error message it works perfectly!! --- $crdate = date("Y-m-d"); $result = mysql_query("SELECT * FROM s