Web Hosting Talk : Web Hosting Main Forums : Programming Discussion : C'mon man... mysql_fetch_array!!!

[seodevhead]

I am getting this error when trying to count number of records in table 'users':

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

with this code:

PHP Code:

$query = "SELECT COUNT(*) FROM users";
    $result = mysql_query($query);
    $row = mysql_fetch_array ($result, MYSQL_NUM);
    $num_users = $row[0];
    echo "There are currently " . $num_users . " registered!"; 


What the hell is causing the error??? My code looks spot on!??! Any help???

[seodevhead]

HAHA... nevermind...

I forgot I had a msyql_close(); above it. Sorry

[Burhan]

Always use mysql_error():

PHP Code:

$query = "SELECT COUNT(*) as `num_rows` FROM users";
$result = mysql_query($query);
if (!$result)
{
   die($query."<br />".mysql_error());
}
echo "Number of users is ".mysql_result($result,0); 


[ochiba]

Originally posted by fyrestrtr
Always use mysql_error():

...in dev only, right?

In production, might not that cause a security issue? By sending bad vars which will purposely cause an error, script-monsters can find out more information about your DB structure and have a better chance of figuring out what they can break.

I dunno, just my opinion ^_^;; but then again, I'm paranoid.

[jimlundeen]

There is a happy medium, which is to look at the response from the mysql server, and if it's an error, then gracefully generate your own error message which contains no "helpful" information that bad people might use against you.