asked on

Load different images different page

I am using this code below so when the URL is domain.com/registration/ load the following images for my image slideshow and everything else use the other set.

For some reason it is just loading the first image 4.jpg.

Any ideas?

No PHP errors too.

Thanks,

Ryan

<?php
		if($_GET )switch(basename($_SERVER['REQUEST_URI'])){
		
			case "registration":
				echo '	<img src="bloginfo('.template_url.')/images/middle/4.jpg" alt="middle image" width="998" height="177" />
				<img src="bloginfo('.template_url.')/images/middle/5.jpg" alt="middle image" width="998" height="177" /> 
				<img src="bloginfo('.template_url.')/images/middle/6.jpg" alt="middle image" width="998" height="177" />';
			break;	
			
			default:
				echo '	<img src="bloginfo('.template_url.')/images/middle/1.jpg" alt="middle image" width="998" height="177" />
				<img src="bloginfo('.template_url.')/images/middle/2.jpg" alt="middle image" width="998" height="177" /> 
				<img src="bloginfo('.template_url.')/images/middle/3.jpg" alt="middle image" width="998" height="177" />';
			break;
		
		}
	?>

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kevin_u

I'd suggest doing an echo basename($_SERVER['REQUEST_URI']); in order to examine what you actually receive. It may be /registration, or /registration/.

catonthecouchproductions

ASKER

I did this: echo basename($_SERVER['REQUEST_URI']);

And it gave me: "registration" no slashes.

Ryan

profya

Try this:

<?php
                if($_GET )switch(preg_match('/registration\/$/', $_SERVER['REQUEST_URI'])){
                
                        case 1:
                                echo '  <img src="bloginfo('.template_url.')/images/middle/4.jpg" alt="middle image" width="998" height="177" />
                                <img src="bloginfo('.template_url.')/images/middle/5.jpg" alt="middle image" width="998" height="177" /> 
                                <img src="bloginfo('.template_url.')/images/middle/6.jpg" alt="middle image" width="998" height="177" />';
                        break;  
                        
                        default:
                                echo '  <img src="bloginfo('.template_url.')/images/middle/1.jpg" alt="middle image" width="998" height="177" />
                                <img src="bloginfo('.template_url.')/images/middle/2.jpg" alt="middle image" width="998" height="177" /> 
                                <img src="bloginfo('.template_url.')/images/middle/3.jpg" alt="middle image" width="998" height="177" />';
                        break;
                
                }
        ?>

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catonthecouchproductions

ASKER

Hey, thanks!

This is the source after load:

<div id="middle_image" class="grid_12">
</div>

Using that code above, not sure why its not printing anything out.

Ryan

ASKER CERTIFIED SOLUTION

kevin_u

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profya

In addition to what said above by kevin_u, may be there is a fatal syntax error, please add this line to the first line of your php script:
error_reporting(E_ALL);

And let's know what's up.