[Resolved] PHP problem (GD, HTML, PHP)

[Bah__Name]

I've got this function, well its not a function, its a file,
DrawPiccy.php

It has to be a file, not a function because it outputs an image (At least, thats my thinking. It works, so I'm happy. I'd rather use a function, but I'll worry about that later).

Heres the code

print "<img src='DrawPiccy.php?Background=$Name'border=0>";

and in DrawPiccy.php the parameter is picked up thus:

$_SESSION['BackgroundName'] = $_GET['BackGround'];

(Yes its a session variable, don't worry).

Here's the problem - I now want to pass a second parameter in the img src line

print "<img src='DrawPiccy.php?Background=$Name'border=0>";

I just don't know the syntax. Theres a lot of apostrophes in there, and quotes. I've tried separating variable names with ampersands,

print "<img src='DrawPiccy.php?Background=$Name&mode=$mode'border=0>";

and tried with &amp too,

print "<img src='DrawPiccy.php?Background=$Name&ampmode=$mode'border=0>";

neither worked, the parameter never arrives.

I've also tried passing an array, that completely didn't work either :/

Frankly, I'd rather it was a function, but I am sticking with it as a php file because of said image creation issues.

I wish I had a good manual.

I'd be grateful if you could help.

[dnast]

You didn't put a semi-colon on & but the previous one you showed should have worked.

Check if $mode has a value, or just look at the status bar when you hover over the link and see if it's displayed correctly. It could also have something to do with there being no space between the src value and "border".

print "<img src='DrawPiccy.php?Background=$Name&amp;amp;mode=$mode' border=0>";

^ Should work fine, though.

[Bah__Name]

Yeay!

I didn't have a semi-colon. Now it works.

May small tribes in Borneo forever worship all of our source code..