Image Linking (Is this possible?)

[miles_rich]

Why wont this work? is it possible to do it like this?

Page 1:

<?php

echo "<img src='main.php?img=1'>";

?>

Page 2:

<?php

$img = $_REQUEST['img'];

switch($img)
{
	case "1":
		echo "<img src='test.jpg' width='200' height='20' alt='Picture - 1'>";
		break;

	case "2":
		echo "<img src='test.jpg' width='200' height='20' alt='Picture - 2'>";
		break;

	case "3":
		echo "<img src='test.jpg' width='200' height='20' alt='Picture - 3'>";
		break;
}

?>

Should this not choose image three and display it??

[andymoo]

calling an image tag inside an image tag will fail, you need to get the file and output it with the right mimi type in main.php

[mattsinclair]

It outputs

<img src='<img src='test.jpg'>'>

andymoo is right, you'll have to open the file and dump it... or create the initial <img tag dynamically.

you might be able to use a header() redirect to add the same functionality.

in both cases you lose the ability to predefine the width/height/alttext

[devinemke]

you cannot specify a PHP file as the source in your img tag. why not just do this:

page.php?img=3

echo '<img src="image' . $_GET['img'] . '.jpg" width="200" height="20" alt="Picture - ' . $_GET['img'] . '">';

this will display image3.jpg. simply pass the appropraite number on the URL depending on what image you want

[andymoo]

gd library will let you get the image size in a function funnily enough called getimagesize

list($width, $height, $type, $attr) = getimagesize("$image");

then call the image like

echo "<img src=\"main.php?pic=1\" $attr alt=\"blah\">";

hth

[mattsinclair]

you cannot specify a PHP file as the source in your img tag

Yes you can... http://www.mattsinclair.com/pics.php?id=18

gd library will let you get the image size in a function funnily enough called getimagesize

That's true, but since he's having the main.php file generate the image name AFTER the referring page is generated, getimagesize() wouldn't be avaliable to use on the referring page.