Weird Stuff with fetch array

[cqball2000]

I'm usually good at debugging my scripts. I've also used the code below many times with no problems. However, I'm now working with php4 and receiving the following errors.

Code:

$sql = "select * from sponsors_login where username = '$spuser' and password = '$sppass'";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result))
{

When I run the script I get:

Warning: Supplied argument is not a valid MySQL result resource in /libs/modules/file.php on line 19...which is >> while($row = mysql_fetch_array($result)).

Is it the mysql fetch array call or what? The code looks fine, right? Any suggestions?

Thanks! 🙂

[plutarck]

If I ever get that error again in my life it will be too soon.

But I'm about to write another script, so it will be too soon 😉

Feeling like an expert on that frickin' error, it means that your mysql_query returned false.

First, right after your mysql_query, echo out $result. If it doesn't say something like "Resource id 2", there is something wrong in your query.

Before you assign your query to $sql, echo out the variables that you are going to make part of $sql (sepperate all the echo's with a "<BR>" so you can easily see what variable is what).

Then after that assign the variables to $sql as you are doing, then echo out $sql before the mysql_query call.

If that STILL doesn't give you the information you need, use this right after your query call:

printf ("MySQL Error: %s\n", mysql_error ());

Beyond that ensure that your table name and column names are valid.

If none of this gives you something interesting to chew on, use your computer as a kufu (meditation pillow) and become a luddite 😉

[[deleted]]

In short, always (and I do mean always) use error checking like this:

$sql = "select * from sponsors_login where username = '$spuser' and password = '$sppass'";

if ($result = mysql_query($sql))
{

query worked, fetch results and work with them

while($row = mysql_fetch_array($result))
{

do the work

};
}
else
{
#query failed, print debug info
echo "SQL ERROR: ".mysql_error();
echo "<br>";
echo $sql;
};