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

[Bill L. <wl...@sw...>]

Mark, you example is simple and clean.  The CocoaDialog developers  
have examples of using Bash and Perl, but I think this would be a good  
addition.

To clean up, I played with this and found a whole script of:

#!/usr/bin/php
<?php
$return_values=`/Applications/CocoaDialog.app/Contents/MacOS/ 
CocoaDialog \
     standard-inputbox --text "hello world" --string-output`;
list($button, $value)=explode("\n", $return_values);
echo "User typed \"$value\" and pressed the \"$button\" button.\n";
?>

Andy, I completely agree with Mark, skip Bash and write PHP scripts  
directly.

Bill Larson

On Dec 17, 2008, at 6:16 AM, Mark A. Stratman wrote:

> Hello Andy,
>
> My recommendation is to just simply not use bash. Since your main
> program is in PHP anyway, just run CocoaDialog from PHP (within your
> 'mm.php' program).
>
> PHP also uses backticks for capturing program output, if i recall
> correctly.  So just put a dollar sign in front of those variables,  
> add a
> semicolon to the end of each statement, and you should be good to go.
>
> Here's a simple inputbox example for php:
> $return_values = `/path/to/CocoaDialog standard-inputbox --text "hello
> world" --string-output`;
> list($button, $value) = explode("\n", $return_values);
> echo "User typed $value and pressed the $button button.\n";
>
> hope this helps.
> - mark
>
> Andy H wrote:
>> 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
>>
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