Parameter through browser

[mikerh9]

I have link that prints to the image(see below).

File test.php take image_name as parameter. In the browser the parameter has the value.
But when I print the variable it is empty. Any ideas why image_name is empty?

$thumb = "<A href=test.php?image_name=$image_small>". "<img src=images/"."$image_small"." border=1>"."</A>";

Thanks

[LordShryku]

How are you printing the variable? Could it be a register_globals problem?

[Gary_King]

Print out $_REQUEST[image_name] instead.

[mikerh9]

Originally posted by LordShryku
How are you printing the variable? Could it be a register_globals problem?

I tried to print

$_REQUEST[image_name];
echo $image_name

Outputs nothing.

[LordShryku]

try $_GET['image_name']

[mikerh9]

I printer phpinfo() and the register_globals is OFF. Can this be a problem? PHP version is 4.3.2
register_globals Off Off

Should I change the register_globals to on ?

Thanks

[LordShryku]

No, register_globals off, it should work either way. You may be overwritting it somewhere? Hard to say without the code.

[mikerh9]

Originally posted by LordShryku
No, register_globals off, it should work either way. You may be overwritting it somewhere? Hard to say without the code.

<?php

$host ="localhost";
$user = linuxuser";
$pass ="qwerty";

dbConnect($host, $user, $pass);
$sql_images = mysql_query("SELECT image_small from ALBUM.album");
$mogrify ="/usr/bin/mogrify";
?>

<table align="center" border=0 cellpadding=2 cellspacing =5>
<?php
$count = 0;

while(list($image_small) = mysql_fetch_array($sql_images)){
if($count == 0)
echo "<tr><td>";
if($count != 0)
echo "<td>";
$count = $count + 1;
cropimage("images/$image_small", 150, 100, $mogrify);
//$thumb = "<A href=image_medium/$image_small border=1>". "<img src=images/"."$image_small"." border=1>"."</A>";
//$thumb = "<img src=images/"."$image_small"." border=1>";
$thumb = "<A href=test.php?image_name=$image_small>". "<img src=images/"."$image_small"." border=1>"."</A>";

list($width, $heigth) = getimagesize("images/$image_small");
echo $thumb. "<br>";
echo "<b><font size='2'> Image Size:</b> ".$width." x ";
echo $heigth. "</font>";

?> <td>
<?php
if($count / 4 == 1) {
$count =0;
echo "</tr>";
}

}?>
</table>

<?php
database_close($host, $user, $pass);

function database_close($host, $user, $pass) {
mysql_close(mysql_connect($host, $user, $pass));
}

function dbConnect($host,$user,$pass) {
$link = mysql_connect($host, $user, $pass) or die("Could not connect: ".mysql_error());
mysql_select_db("mysql");

}

function cropimage($picture,$fixedwidth,$fixedheight,$mogrify) {
// GET IMG
$img = imagecreatefromjpeg($picture);
$width= imagesx($img);
$height= imagesy($img);
// CROP WIDTH
if($width!=$fixedwidth){
$ratio =$fixedwidth/$width;
$NewHeight=round($height$ratio);
$NewWidth=round($width$ratio);
exec( $mogrify." -resize ".$NewWidth."x".$NewHeight."! $picture");
exec( $mogrify." -crop ".$fixedwidth."x".$fixedheight."+0+0 $picture");
// REFRESH
$img = imagecreatefromjpeg($picture);
$width= imagesx($img);
$height= imagesy($img);
}
// CROP HEIGHT
if($height!=$fixedheight){
$ratio =$fixedheight/$height;
$NewHeight=round($height$ratio);
$NewWidth=round($width$ratio);
exec( $mogrify." -resize ".$NewWidth."x".$NewHeight."! $picture");
exec( $mogrify." -crop ".$fixedwidth."x".$fixedheight."+0+0 $picture");
}

ImageDestroy($img);
}

?>

I looked on the code again, can't find my mistake. Thanks

[Gary_King]

Originally posted by mikerh9
I tried to print

$_REQUEST[image_name];
echo $image_name

Outputs nothing.

You do NOT print it like that.

You do this:

print $_REQUEST[image_name];