[SOLVED] No error but still get mysql_fetch_array(): supplied argument is not a valid MyS

[chamster]

HI. This one's got me stumped.

 

My query works, returns a record, does what it's supposed to do, but I still get mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/.sites/95/site29/web/data_recovery/admin/dre_fns.php on line 354.

 

I echo the query, run it via PhpMyadmin, and it returns a record and no error.

 

I use $result = mysql_query($query) or die(mysql_error()); and again, no error.

 

Here's the code:

 

*************************

 

$query = "SELECT *, CONCAT(tbl_customers.first_name, ' ', tbl_customers.last_name) as 'full_name' FROM tbl_cases LEFT JOIN tbl_customers ON tbl_customers.customer_id = tbl_cases.customer_id WHERE case_number='$case_number';";

echo"$query<br><br>";

$result = mysql_query($query) or die(mysql_error());

while ($row = mysql_fetch_array($result))

{

extract($row);

do some stuff;

}

 

**************************

 

It does the stuff but still gives the warning. Anybody have an ideas?

 

Thanks a million!

 

Bob

[PFMaBiSmAd]

extract($row);
do some stuff; 

The part of your code that you omitted is overwriting $result. Either there is a column named result or you are setting $result to something else inside of the loop. Post your code to get the quickest solution.

[chamster]

You are correct! I was setting $result to something else inside the loop.

 

All other discussions I found regarding this error/warning concluded that 99.9% of the time the error was due to a bad query. I guess this is one of the other 0.01%.

 

Thank you!

[PFMaBiSmAd]

Knowing that your code die()'ed when the query failed, eliminated a failed query as the reason. Leaving only the possibility that $result was being overwritten at some point after the query had been successfully executed.