Passing image name

David_Bethel

This is tall a very confusing language...

I can get this into the address bar:

port.php?imagename=test.jpg

port.php:

<html>

<body>

<?php $in=$_GET['imagename'];?>

</body>
</html>

I looked into the image function in php. It looks like you need a degree just to display1 image. Add libraries, compile this etc..

I must be missing something very basic about php and all of the stuff on passing variables.

I need to add a lot of html to the final result.

Any help would be greatly appriciated. -David

jpmoriarty

to get it into the address bar you'd need to do a link to it on the previous page:

assuming your file name was stored in a variable called $name you'd do:
echo "<a href='port.php?imagename=$name'>";

Your problem on the port page is that your image link isnt within the PHP tags, and therefore will just be outputted to the browser as <img src=$in> without inserting your variable name.

You'd need to do this:

or alternatively

<html>

<body>

<?php

$in = $imagename;
echo "<img src='".$in."'>";

</body>

</html>

That should do it for you...

David_Bethel

John

That did exactly what I needed! I must have a thick head when it comes to php. The basics just aren't sinking in...

Thank you. -David