Problem with script

[Chris_Evans]

Hi - I've got a small :rolleyes: problem with a script I've written. Nothing comes from the variables so I get no results... any ideas?

<?php
include("includes/confstring.php");

$studentfind = mysql_query("SELECT fname,lname,email,courseid,chapter FROM student")
	or die(mysql_error());
While($row = mysql_fetch_array($studentfind))
	{
	$x = 1;
	$fname.$x = $row['fname'];
	$lname.$x = $row['lname'];
	$email.$x = $row['email'];
	$courseid.$x = $row['courseid'];
	$chapter.$x = $row['chapter'];

	$retrievetest = mysql_query("SELECT test FROM test WHERE courseid='$courseid.$x' AND chapter='$chapter.$x'")
		or die(mysql_error());
	While($roger = mysql_fetch_array($retrievetest))
		{
		echo 'Added the following test to email for $fname$x $lname$x:';
		echo $roger['test'];
		}
	mysql_free_result($retrievetest);

$x++;
}
mysql_free_result($studentfind);
?>

It appears to be this bit that is the problem:

	$x = 1;
	$fname.$x = $row['fname'];
	$lname.$x = $row['lname'];
	$email.$x = $row['email'];
	$courseid.$x = $row['courseid'];
	$chapter.$x = $row['chapter'];

Since I have an $x there it appears to be replacing the $fname with $x.. but I really needed to have $fname called $fname1 ... $fname2 etc which is why I added .$x on the end.... I'm using the incorrect syntax, any help would be appreciated.

Chris

[ir4z0r]

at first: put your $x = 1; above the while loop, cause now it's always
initialised with 1

the correct syntax would be:

${fname . $x} =

otherwise the $fname (which is empty) would be replaced with nothing

[rupertbj]

if

$x = 1;

then all the extensions added will just be 1 so you will only get one set out. There is no auto-incrementing / counting.

[Chris_Evans]

Thanks for the above...

rupertbj, the syntax is correct, but the while loop does an $x++ at the end.. I don't need to add a new number for each variable, just each loop.

Thanks,

Chris

[rupertbj]

thanks - just read your extract only - not your longer code!