Re: [PHP] Re: Opening a Dynamic Pop Up Window

2003-05-30 Thread Kevin Stone 
In 4.06 you have to use $HTTP_GET_VARS  in place of $_GET.  It's the same
thing, just a different name.

Ditto with $_SERVER, $_POST and $_SESSION

- Kevin


- Original Message -
From: "Jeffrey L. Fitzgerald" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, May 29, 2003 4:05 PM
Subject: Re: [PHP] Re: Opening a Dynamic Pop Up Window


>
>  Actually, I just found out we run PHP 4.0.6... Will the query
> string/GET process still work? Right now it's not and I thought I saw in
> the docs that the POST/GET isn't in pre 4.1...
>
>
> >Just add some Javascript to the links around your thumbnails, like this:
> >
> >
> >
> >Then you need to create a script called show_photo.php that takes the
"img"
> >var passed by the Javascript above and loads that image onto the page.
> >
> >$image = $_GET['img'];
> >
> >echo "";
> >
> >Hope that helps.
> >
> >Monty
>
>
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>



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Re: [PHP] Re: Opening a Dynamic Pop Up Window

2003-05-30 Thread Jeffrey L. Fitzgerald 
Actually, I just found out we run PHP 4.0.6... Will the query 
string/GET process still work? Right now it's not and I thought I saw in 
the docs that the POST/GET isn't in pre 4.1...


Just add some Javascript to the links around your thumbnails, like this:



Then you need to create a script called show_photo.php that takes the "img"
var passed by the Javascript above and loads that image onto the page.
$image = $_GET['img'];

echo "";

Hope that helps.

Monty


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Re: [PHP] Re: Opening a Dynamic Pop Up Window

2003-05-30 Thread Jeffrey L. Fitzgerald 
First thanks for your help!..   I have the variable passing OK to 
this html page below. But the filename is not making it to the  
tag. I think the syntax is off??  I'm thinking maybe the img src tage needs 
to be broken up to allow the $image var to process?





Test image



$image = $_GET['img'];
echo "";
?>



Just add some Javascript to the links around your thumbnails, like this:



Then you need to create a script called show_photo.php that takes the "img"
var passed by the Javascript above and loads that image onto the page.
$image = $_GET['img'];

echo "";

Hope that helps.

Monty


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[PHP] Re: Opening a Dynamic Pop Up Window

2003-05-30 Thread Monty 
This should be fairly easy. It sounds like the script you are using is
opening only the image in a new window, and not a page that loads the image
in a new window.

Just add some Javascript to the links around your thumbnails, like this:



Then you need to create a script called show_photo.php that takes the "img"
var passed by the Javascript above and loads that image onto the page.

$image = $_GET['img'];

echo "";

Hope that helps.

Monty

> From: [EMAIL PROTECTED] (Jeffrey L. Fitzgerald)
> Newsgroups: php.general
> Date: Thu, 29 May 2003 14:28:11 -0400
> To: [EMAIL PROTECTED]
> Cc: [EMAIL PROTECTED]
> Subject: Opening a Dynamic Pop Up Window
> 
> 
> Help! :-)
> 
> Have a need for a pop up window to show images when the thumbnail is
> clicked. But the script I found only shows the image and no background,
> etc. I want to be able to have the pop up open and then send it a php
> created dynamic page with the proper html and image.
> 
> In perl I know how to do this, by calling a perl script, sending the
> filename variable and returning with a generated page. But I am not as
> familiar with PHP. Here is the script I found:
> 
> http://www.ciarasprettygoodpictures.com/gallery.php
> 


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