Solved

Load different images different page

Posted on 2009-07-15
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Last Modified: 2012-05-07
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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6 Comments
 
LVL 12

Expert Comment

by:kevin_u
ID: 24866610
I'd suggest doing an echo basename($_SERVER['REQUEST_URI']); in order to examine what you actually receive. It may be /registration, or /registration/.
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Author Comment

by:catonthecouchproductions
ID: 24872327
I did this:       echo basename($_SERVER['REQUEST_URI']);

And it gave me: "registration" no slashes.

Ryan
0
 
LVL 14

Expert Comment

by:profya
ID: 24872375
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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LVL 1

Author Comment

by:catonthecouchproductions
ID: 24874379
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
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LVL 12

Accepted Solution

by:
kevin_u earned 500 total points
ID: 24875306
If its doing nothing, it means that the if ($_GET) is failing.  Are there query name/value paires on the request?
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LVL 14

Expert Comment

by:profya
ID: 24885390
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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