Echo does not show up a picture - php

Question

Okay i have a site wich makes signatures. But because some header problems i had to put it in another file and use echo. this is the code:
<h1>Create your European Trucking signatures</h1>
<?php
echo '<img src="SigGen.php?player=$Playername&score=$Score&money=$Money" />';
?>
</div>
But it shows a broken image thingy. But when i go to the SigGen.php itself it does show up like here:
http://european-trucking.com/SigGen.php?player_name=Thimo
This is the code of SigGen:
<?
/*
***Made by: Nodroz***
*** Enjoy your signatures! ***
*/
$username="10528_Thimo"; //Your MySQL Username.
$password="*********"; // Your MySQL Pass.
$database="10528_Thimo"; // Your MySQL database.
$host="95.211.***.***"; // Your MySQL host. This is "localhost" or the IP specified by your hosting company.
$player_name=$_GET['player_name']; // This gets the player his name from the previous page.
/* Next, we will make a connection to the mysql.
If it can't connect, it'll print on the screen: Unable to select database. Be sure the databasename exists and online is. */
mysql_connect($host,$username,$password); // Connection to the database.
#mysql_select_db($database) or die( "Unable to select database. Be sure the databasename exists and online is."); //Selection of the database. If it can't read the database, it'll give an error.
/* To protect MySQL injection. */
$player_name = stripslashes($player_name);
$player_name = mysql_real_escape_string($player_name);
$query="SELECT * FROM Users WHERE Name='$player_name'"; // Gets all the information about the player.
$result=mysql_query($query);
$i=mysql_num_rows($result); // Here we are counting how many rows this result gives us.
/* We will now put the player's information into variables so we can use them more easily. */
/* DON'T FORGET: The names should be exact the same as in your mysql db.*/
if ($i == 1) // If the user has been correct, then it'll give us 1 row. If its 1 row, then it'll proceed with the code.
{
$Playername=mysql_result($result,0,"Name"); // Gets the username of the player and put it in the variable $Playername.
$Money=mysql_result($result,0,"Money"); // Gets the money of the player and put it in the variable $Money.
$Score=mysql_result($result,0,"Score"); // Gets the score of the player and put it in the variable $Score.
// Creating of the .png image.
header('Content-Type: image/png;'); // Don't touch this. We use this to tell the script we're working with an image.
$im = #imagecreatefrompng('PlayerSig.png') or die("Cannot select the correct image. Please contact the webmaster."); // Don't forget to put your picture there. Eg: playersig.png
$text_color = imagecolorallocate($im, 000,000,000); // RED, GREEN, BLUE --> Go to www.colorpicker.com, select a nice color, copy the R/G/B letters provided by colorpicker and put them here.
$text_username = "$Playername"; // This gets the information: player name to be showed in the picture.
$text_score = "$Score"; // Same as above but with score.
$text_money = "$Money"; // Same as above but with money.
$font = 'myfont.ttf'; //Upload your custom font to the directory where this file is placed. Then change the name here.
/* USAGE OF THE imagettftext: First ($im) shouldn't be changed. (16) is the text-size. (0) is the angle of your text. Change it, and you'll see what's going on. (20) is de X-coordinate of the text.
(36) is the Y-coordinate of the text. */
imagettftext($im, 16, 0, 5, 20, $text_color, $font, $text_username); // Prints the username in the picture.
imagettftext($im, 16, 0, 5, 40, $text_color, $font, "Score:");
imagettftext($im, 16, 0, 72, 40, $text_color, $font, $text_score); // Prints the score in the picture.
imagettftext($im, 16, 0, 5, 60, $text_color, $font, "Money:");
imagettftext($im, 16, 0, 72, 60, $text_color, $font, $text_money); // Prints the money in the picture.
imagepng($im);
imagedestroy($im);
} else echo('Username is not in our database. Please try again.'); // If the username doesn't exist (so the row is 0) then it'll give en error.
mysql_close();
?>
Anyone knows why it doesnt show up as an image using echo?

Answer 1

i think the problem is that you cant use variable names in string like this :
THIS WONT WORK
echo 'Hi $userName';
THIS WILL WORK
echo "Hi $userName";
so your line :
echo '<img src="SigGen.php?player=$Playername&score=$Score&money=$Money" />';
needs to be changed to :
echo "<img src='SigGen.php?player=$Playername&score=$Score&money=$Money' />";

Answer 2

try this...
<?php echo '<img src="SigGen.php?player='.$Playername.'&score='.$Score.'&money='.$Money.'" />'; ?>
use the concatenation features of the echo to properly append the variables to the image link