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[mySQL] Error I cannot Debug





NewGuyinTown
OK guys...

I got an error message
Quote:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource


and my code, from the beginning to where it crashes, is
Code:

<?php
$dbh=mysql_connect ("localhost", "$user", "$pass") or die ('I cannot connect to the database because: ' . mysql_error());
mysql_select_db ("$db");
$result = mysql_query("SELECT * FROM $tab");
//$num = mysql_num_rows($result);
while($row = mysql_fetch_array($result)){?>


I executed the script select from $tab in mySQL and it works, but when I used php to execute it gives me an error. I spent hours trying to debug this already.

I also tried num_rows and that does work. I have 2 rows of data in the table.

Can someone explain to me what's wrong.

Thank you.
Rhysige
Code:

<?php
$dbh = mysql_connect ("localhost", "$user", "$pass") or die ('I cannot connect to the database because: ' . mysql_error());
mysql_select_db ("$db");
$result = mysql_query("SELECT * FROM $tab");
//$num = mysql_num_rows($result);
while($row = mysql_fetch_array($result)){
?>


Ok lets look into a few possibles.

1/ mysql_select_db ("$db");
The specified variable when connecting to the DB was $dbh
2/ it will return that error if whats going into mysql_fetch_array() is not a resource AKA has no data being retrieved. Where is $tab sepecified?
n0obie4life
mysql_select_db( $dbname, $dbh );

name of db, database link (the connect code).

And are you giving the full code?

I see missing parts here and there.

What's $tab? IF it's not a var, there's something wrong.
NewGuyinTown
I found out what's wrong... It was at the selecting part. I didn't give the any users the selecting privileges... Confused I was looking at my script for more than 2 hours, changing it, testing it... There wasn't any problem with my script... Ugh.

Yeah... it wasn't full the code... I copied from the mysql section to the error.

Thank you for trying to help.
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