[Cocoadialog-users] Help with incorporating with Platypus/PHP/Cocoadialog

[Andy H <an...@me...>]

Hello all. First off, I've just started to use Cocoadialog today. I'm  
trying to incorporate my PHP script that has args into a GUI using  
Platypus and Cocoadialog. So, my problem is that cocoadialog is  
reporting the wrong args to the PHP script. My bash shell script looks  
like this:

#!/bin/bash

CD="CocoaDialog.app/Contents/MacOS/CocoaDialog"
un=`$1/Contents/Resources/$CD inputbox --title "Username" --no-newline  
--width 400 --height 250 --informative-text "Enter your username here"  
--button1 "Enter"`
pass=`$1/Contents/Resources/$CD inputbox --width 400 --no-newline -- 
height 250 --no-show --informative-text "Enter your password" -- 
button1 "Enter"`
coins=`$1/Contents/Resources/$CD inputbox --width 400 --height 250 -- 
informative-text "Enter the amount of coins you wish to have here  
(must be between 1 and 1,000,000. the higher you go, the better chance  
you have of being banned)" --button1 "Enter"`
php $1/Contents/Resources/files/mm.php $un $pass $coins

My PHP script's variables look like this:
$Username = $argv[1];
$Password = $argv[2];
$coins = $argv[3];

So, when I run the script, and tell it to echo the args I gave it from  
bash (echo "$un" echo "$pass" etc etc..) and my username is foo,  
password is fooie, and coins is 10 it returns:
1
foo
1
fooie
1
10

I read up on this, and it seems that it echos what button was pressed  
then in a new line it says the text that was entered. Is there a way  
that I can remove the number that says what button was pressed and  
just make it say the text entered in three lines?

Thanks,
Andy