Why doesnt $_GET ['$example'] Work?

[ljking20003]

hi
dont really know how to explain it so i'll just show the code

what i need it to do is
$_get information from the url that is stored in a variable
for example
abc=1,bcd=2

(load abc and bcd in a while loop from database)
set $name=$_get[$variable]
display the name then wipe before movin on to the next variable in the database


this is where the information is sent

while($row = mysql_fetch_array($result)) {
$ddriver=$row['driver'];
echo"<br> Enter $ddriver 's Hours For $date <input type='text' name='$ddriver' value='0' size='5'>";
};
echo"<br><input type='submit' value='Generate Preview Profit Loss Sheet For $date'></form>";

and this is the section where its supposed pick it up

$query="SELECT * FROM `drlist`";
$result=mysql_query($query);
while($row = mysql_fetch_array($result)) {
$dname=$row['dname'];
$dhourrate=$row['dhourrate'];
$ddriver=$_GET [$dname];
echo "$dname,$ddriver, test ,$dhourrate";
if ($ddriver<>'') {
$driverwork=$ddriver-0.75;
$drivercost=$driverwork*$dhourrate;
															
echo " Driver $dname, Hours After Break = $driverwork, Driver Cost @ $dhourrate per hour = 				
$drivercost";}; 
$ddriver='';
};
															
 
};

i know this method isnt very secure but at the moment i just need this working can anyone help

[dlite922]

$_GET [$dname]; <--- Why is there a space between $_GET and [$dname] ?

Also, you didn't tell us where the code breaks, What's the error you get if any. Is error_reporting turned on?



Dan

[ljking20003]

sorry that was just a typo there is no space in the code as for code breaks, i dont get any errors with reporting on, the rest of the page loads fine ive used echos to display all the variables after loading however the $_GET[$dname]; is the only one it doesnt retrieve is there a different way of formating the [$dname]; or something due to it being a variable not a set name?

[Markus]

Check the validity of your claim that $_GET[$dname] exists.

if (!isset($_GET[$dname])) {
     printf("%s does not exist in the URL (%s",
             $dname, $_SERVER['QUERY_STRING']);
}

If you receive the printed error, then you know you are doing something wrong.

Also post the results of that test.

Mark.