[SOLVED] Going nuts...fresh set of eyes anyone?

[Pandolfo]

Hey everyone,

 

I have some code which is really giving me some grief. The script always dies at this exact point, giving me the same errors over and over. Shown first is the offending snippet, followed by the errors.

$result=mysql_query("select currentcar, currenthome from progress where userid = $userid;");
while($result=mysql_fetch_array($result))
        {
		$carpurchased = $result["currentcar"]; 
		$homepurchased = $result["currenthome"];
	}

 

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/.joliet/blah/blah.org/main/updatehomequestions.php on line 20

 

Warning: Cannot modify header information - headers already sent by (output started at /home/.joliet/blah/blah.org/main/updatehomequestions.php:7) in /home/.joliet/blah/blah.org/main/updatehomequestions.php on line 38

 

Now, i know that the most common cause of the first error is a typo in the database connecting code. I've checked that. There are no typos. I can hit the database perfectly if i hard code and then echo the $userid. Neither field is empty either. Is there a mistake in the snippet? I know that sometimes if you work on something too long you can miss the obvious. Any help or suggestions would be a godsend!

 

Thanks!

[tibberous]

Might be that your query is failing, then your sending the junk query into mysql_fetch_array.

 

Right after your query line, echo mysql_error(); Might be something simple, like $userid not getting set.

[akitchin]

two things.  first off, you have an erroneous semi colon at the end of the query, which is unnecessary when passing the query to mysql_query():

 

$result=mysql_query("select currentcar, currenthome from progress where userid = $userid");

 

second, this code will only work for the first row.  in the while() loop, you're overwriting the $result variable (which contains the resource ID of the query's results) with the first row; when it runs the second time, you're going to get an invalid resource error as well.  change $result to anything else in the while:

 

while($anything_else=mysql_fetch_array($result))

[Pandolfo]

Perfect! Awesome! Kick-&#*!

 

Thank you so much! You rock!