Using selecting data from database in PHP Dynamic image - php

Question

I have been working on a dynamic image script.The main reason I was searching for it is I want to show user info from a database.Here's my problem:
<?php
header("Content-type:image/png");
$array=array("I am a monument to all your sins", "Currently making pizza","Best before 12/7/09", "Farming Onions");
function imagettftext_cr(&$im, $size, $angle, $x, $y, $color, $fontfile, $text)
{
// retrieve boundingbox
$bbox = imagettfbbox($size, $angle, $fontfile, $text);
// calculate deviation
$dx = ($bbox[2]-$bbox[0])/2.0 - ($bbox[2]-$bbox[4])/2.0; // deviation left-right
$dy = ($bbox[3]-$bbox[1])/2.0 + ($bbox[7]-$bbox[1])/2.0; // deviation top-bottom
// new pivotpoint
$px = $x-$dx;
$py = $y-$dy;
return imagettftext($im, $size, $angle, $px, $y, $color, $fontfile, $text);
}
$image = imagecreate(500,90);
$black = imagecolorallocate($image,0,0,0);
$grey_shade = imagecolorallocate($image,40,40,40);
$white = imagecolorallocate($image,255,255,255);
$text = $array[rand(0,sizeof($array)-1)];
$otherFont = 'open.ttf';
$font = 'open.ttf';
$name = "erlis";
$name = substr($name, 0, 25);
//BG text for Name
while($i<10){
imagettftext_cr($image,rand(2,40),rand(0,50),rand(10,500),rand(0,200),$grey_shade,$font,$name);
$i++;
}
//BG text for saying
while($i<10){
imagettftext_cr($image,rand(0,40),rand(90,180),rand(100,500),rand(200,500),$grey_shade,$otherFont,$text);
$i++;
}
// Main Text
imagettftext_cr($image,35,0,250,46,$white,$font,$name);
imagettftext_cr($image,10,0,250,76,$white,$otherFont,$text);
imagepng($image);
?>
This part works perfectly with simple text and no mysql queryes. But when I try to initiate a mysql query..let's say $query = mysql_query("SELECT * FROM serverplayers WHERE id=1"); it just breaks the image.It doesn't matter what kind of form I use but it just brakes my image.How can I implement some "fetch from the database" code? Maybe $_GET[''] or something like that.
I need the $name="erlis"; to be changed to a users info.

Answer 1

This seems to be a matter of how to debug that beast.
Build yourself a URL that opens the image AND test it in any browser. Something like
http :// yourserver / images?id=1 (adapt this to your needs please - was just an example)
Add your mysql sequence to the code
Uncomment your mimetype header modification like this:
// header("Content-type:image/png");
run the link from step 1 in the browser -> the php/mysql debugging potentially hints you on the error
if you cannot see anything on the browser try 'view source code' from the browser's menu