Quesion about - isset($_GET['image']) !

[ayachuca]

Hello,

 

I'm having one problem. Since I'm not PHP developer, I don't know how to solve it.

 

I'm having two pages.

 

Page1 have two images that are linked to:

 

...page2.php?image=1

...page2.php?image=2

 

On click on one of those links, on Page2, I need to change one image with another.

So, for example, when I click on link "...page2.php?=image1", it should open me Page2 with new image loaded in some place.

Same, when I click on "...page2.php?=image2", it opens me another image on Page2.

 

Since I need to update website and there was code only for one image:

 

<img src="pic/<?php echo isset($_GET['image']) ? 'promospace2.gif' : 'middle_highlights.gif';?>" height="740" width="450"/>

 

Can anyone help me and tell me how to make this code for 2 images/links?

I'm guessing that there should be an "if and then/else" loop, but I'm dumb enough, so anyone please ?

[JonnoTheDev]

<?php
// array of images
$images[1] = "promospace2.gif";
$images[2] = "anotherimage.gif";

if(is_numeric($_GET['image']) && array_key_exists($_GET['image'],$images)) {
$image = $images[$_GET['image']];
}
else {
// image does not exist
$image = "middle_highlights.gif";
}
// display image
print "<img src=\"pic/".$image."\" />";
?>