Pass if via the URL
Link
On the display page, access the data you passed by...
$_GET['picture_data']
Do a google search for "get query".
Ron Clark wrote:
Hello all,
I have a php script that reads all the files in a directory (pictures), the
outputs each picture as a link to be displayed in t
You just need to add a question mark, the name of the variable, an equal
sign and its value (properly encoded):
echo '';
In the receiving script, you can read the value using $_GET['myvalue'];
Hope this helps!
Marco
--
php|architect - The magazine for PHP Professionals
The first
Hello all,
I have a php script that reads all the files in a directory (pictures), the
outputs each picture as a link to be displayed in the main frame of the html
page. All of that works fine. I want to be able to "pretty up" the picture
out put by placing the picture into a html table so it is c
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