Unexplainable php WARNING

Shuriken1

If am receiving the following warning:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /usr/local/psa/home/vhosts/shurikenet.34sp.com/httpdocs/ec2/includes/members_online.php on line 28

on the following bit of code:

$sql = "SELECT * FROM ".$LOGIN_TABLE." WHERE logged_in = '1' AND expiry_time < time() ";
$result = mysql_query($sql, $db);
//$myrow = mysql_fetch_array($result);
while ($myrow = mysql_fetch_array($result)) { <<<<THIS IS LINE 28
$sql = "UPDATE ".$LOGIN_TABLE." SET logged_in = '0' AND expiry_time = '' LIMIT 1";
$result = mysql_query($sql);
}

Any ideas? Syntax Error?

Shuriken1.

rinjani

Probably the double quotes in your sql.

The warning means your result from the query failed.

Try checking the errors e.g.

$err = mysql_error();

echo $err;

to see what mysql thinks about your query.

rinjani

Shuriken1

Could it be because it is an included file, thats the only difference I can see in this code and every while i use in SQL.

Shuriken1.

Weedpacket

What happens when you follow rinjani's advice?

superwormy

time () is not a MySQL function, so why are you using it?

If you meant to use the PHP time () function, you can't use it within a string. Use something like:

$string = " first part " . time () . " second part ";

And ALWAYS use:

mysql_query (" my query here ") or die (mysql_error ()); when testing!

Shuriken1

Do you have to escape the sql string to use a variable, like so:

$sql = "SELECT * FROM ".$LOGIN_TABLE." bhah blah";

or can i do this?:

$sql = "SELECT * FROM $LOGIN_TABLE blah blah";

the error i am now getting is:

You have an error in your SQL syntax near '() ' at line 1

which is this:

$sql = "UPDATE ".$LOGIN_TABLE." SET logged_in = 0 AND expiry_time = 0 LIMIT 1";

Rinjani; double quotes? You mean after expiry time, well if you did, you can see that it wasn't that 🙂.

Thank you everyone for your replies btw, much appreciated 🙂

Shuriken1.

Tracer

Shouldnt you select a RECORD to set new values for?

I.E. WHERE ID = Something

Tracer

Additionally...If you should try running your query in PHPMyAdmin......It will tell you where in your query the mistake is 🙂

rinjani

I misread the SQL before, sorry, the double quotes issue was wrong, I didn't notice the concat period.

Try:

$sql = "UPDATE ".$LOGIN_TABLE." SET logged_in = '0' AND expiry_time = '0' LIMIT 1";

where you have single quotes around the numbers, that seems to be a common issue with my version of MySQL.

rinjani

Shuriken1

I've got it, it was a bad use of the time function, I had it in an SQL statement higher up and it is fine now.
Thanks for your help everybody, greatly appreciated.

Shuriken1.