Re: [PHP] Re: Opening a Dynamic Pop Up Window
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 img src
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?
!DOCTYPE HTML PUBLIC -//W3C//DTD HTML 4.01 Transitional//EN
html
head
titleTest image/title
/head
body bgcolor=#00 text=#ff
?
$image = $_GET['img'];
echo img src='$image' border=0;
?
/body
/html
Just add some Javascript to the links around your thumbnails, like this:
a href=javascript:window.open('show_photo.php?img=image.jpg');
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 = $_GET['img'];
echo img src='$image' border=0;
Hope that helps.
Monty
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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:
a href=javascript:window.open('show_photo.php?img=image.jpg');
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 = $_GET['img'];
echo img src='$image' border=0;
Hope that helps.
Monty
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
