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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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catonthecouchproductions
Asked:
catonthecouchproductions
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1 Solution
 
kevin_uCommented:
I'd suggest doing an echo basename($_SERVER['REQUEST_URI']); in order to examine what you actually receive. It may be /registration, or /registration/.
0
 
catonthecouchproductionsAuthor Commented:
I did this:       echo basename($_SERVER['REQUEST_URI']);

And it gave me: "registration" no slashes.

Ryan
0
 
profyaCommented:
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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catonthecouchproductionsAuthor Commented:
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
0
 
kevin_uCommented:
If its doing nothing, it means that the if ($_GET) is failing.  Are there query name/value paires on the request?
0
 
profyaCommented:
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.
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