display image depending on url

[dazbot]

Hi everyone, pleased to be here.

 

I am a complete newby to php and Im trying to learn how to do something, here is what im trying to achieve.

 

something like:

 

if url="page1.php" then print("<img src="image1.gif" alt="image1">");

if url="page2.php" then print("<img src="image2.gif" alt="image2">");

if url="page3.php" then print("<img src="image3.gif" alt="image3">");

 

I know its wrong but you get the idea. Ive searched around google etc for some kind of example but I can't find anything, can anyone help me go in the right direction?

 

Cheers

 

Dazbot

[xyn]

$url = explode("/", $_SERVER["SCRIPT_NAME"]);
if($url == "page1.php")
{
    echo("<img src=image1.jpg>");
}
elseif($url == "page2.php")
{
       echo("<img src=image2.jpg>");
}

[ToonMariner]

<?php
switch($url)
{
case 'page1.php':
 echo '<img src="image1.gif" alt="image1">';
 break;
case 'page2.php':
 echo '<img src="image1.gif" alt="image1">';
 break;
case 'page3.php':
 echo '<img src="image1.gif" alt="image1">';
 break;
default:
 echo '<img src="imagedefault.gif" alt="image1">';
 break;
}
?>

 

 

 

[dazbot]

<?php
switch($url)
{
case 'page1.php':
  echo '<img src="image1.gif" alt="image1">';
  break;
case 'page2.php':
  echo '<img src="image1.gif" alt="image1">';
  break;
case 'page3.php':
  echo '<img src="image1.gif" alt="image1">';
  break;
default:
  echo '<img src="imagedefault.gif" alt="image1">';
  break;
}
?>

 

 

 

 

thank you both for such a quick response, heres what I have so far with your suggestion ToonMariner

 

<?php
switch($url)
{
case 'index.php?department=44':
  echo '<img src="imgs/inc/1000mile.gif" alt="1000 mile">';
  break;
case 'index.php?department=45':
  echo '<img src="imgs/inc/muller.gif" alt="Muller">';
  break;
case 'index.php?department=46':
  echo '<img src="imgs/inc/noene.gif" alt="Noene">';
  break;
default:
  echo '<img src="imgs/inc/default.gif" alt="blah">';
  break;
}
?>

 

but it displays the default image all the time. I have tried using full urls and just pagename but having no luck.

 

thanks in advance

 

 

 

 

[dazbot]

Just playing around with with your post as well xyn which is:

 

<?php

$url = explode("/", $_SERVER["SCRIPT_NAME"]);

if($url == "index.php?department=44")
{
    echo("<img src="imgs/inc/1000mile.gif" alt="1000 mile">");
}
elseif($url == "index.php?department=45")
{
    echo("<img src="imgs/inc/muller.gif" alt="Muller">");
}



?>

but that gives me the error:

 

Parse error: parse error in /web/inc/department.inc.php on line 121

 

thanks in advance

 

 

[bronzemonkey]

On the pages you are trying to get to show these images, you will want the value of $url to be the same as the output that you get from this when it is included on your page (remove it when you've got the output):

 

<?php var_dump($url); ?>

 

And in xyn's post he didn't escape the " " within " ". ToonMariner used " " within ' ' so no escape is required. This is the correction to xyn's code:

 

<?php

$url = explode("/", $_SERVER["SCRIPT_NAME"]);

if($url == "index.php?department=44")
{
    echo("<img src=\"imgs/inc/1000mile.gif\" alt=\"1000 mile\">");
}
elseif($url == "index.php?department=45")
{
    echo("<img src=\"imgs/inc/muller.gif\" alt=\"Muller\">");
}

?>

 

Hopefully that helps

[xyn]

@ Bronzemonkey;

Thanks for correcting the code heh, though i never test my coding.

 

however i noticed a problem i forgot. using explode() creates an array

of which we need to ammend like so:

<?php

$url = explode("/", $_SERVER["SCRIPT_NAME"]);

if($url[1] == "index.php?department=44")
{
    echo("<img src=\"imgs/inc/1000mile.gif\" alt=\"1000 mile\">");
}
elseif($url[1]== "index.php?department=45")
{
    echo("<img src=\"imgs/inc/muller.gif\" alt=\"Muller\">");
}

?>

[dazbot]

you guys are great! thanks for all the responses but alas it still isn't working! When I tried :

 

<?php var_dump($url); ?>

 

it returns NULL ???????

 

so I searched around and found this:

 

$uri = 
    'http' . (!empty($_SERVER['HTTPS']) ? 's' : null) . '://' .
    $_SERVER['HTTP_HOST'] .
    ($_SERVER['SERVER_PORT'] != 80 ? ":{$_SERVER['SERVER_PORT']}" : null) .
    $_SERVER['REQUEST_URI'];
    
echo $uri;

 

which gave me this address:

 

http://www.domain.com/shop/index.php?department=44

 

 

however the altered code displays nothing (but doesn't give a parse error now!), I checked the page source for code but nothings there.

 

 

 

<?php

$url = explode("/", $_SERVER["SCRIPT_NAME"]);

if($url[1] == "http://www.domain.com/shop/index.php?department=44")
{
    echo("<img src=\"imgs/inc/1000mile.gif\" alt=\"1000 mile\">");
}
elseif($url[1]== "http://www.domain.com/shop/index.php?department=45")
{
    echo("<img src=\"imgs/inc/muller.gif\" alt=\"Muller\">");
}

?>

 

 

 

[bronzemonkey]

Try this:

 

<?php

$url = $_SERVER['PHP_SELF'];

if($url == "/index.php?department=44")
{
    echo("<img src=\"imgs/inc/1000mile.gif\" alt=\"1000 mile\">");
}
elseif($url == "/index.php?department=45")
{
    echo("<img src=\"imgs/inc/muller.gif\" alt=\"Muller\">");
}

?>

 

If it doesn't work, use var_dump again to see what value of url is required. If that still doesn't work, then we've already reached the limits of my PHP 

[ToonMariner]

OMG what are you all doing????????

 

leading the poor lad astray switch is the order of the day when haveing MANY possible values - far better than lots of if statements!!!

 

OK I was thinking your would not actually use the scriptname but a url var in my first post but as you made no mention I decided to leave to what you posted...

 

Try this.

<?php

$dept = isset($_GET['department']) ? $_GET['department'] : NULL;

switch($dept)
{
case 44:
  echo '<img src="imgs/inc/1000mile.gif" alt="1000 mile">';
  break;
case 45:
  echo '<img src="imgs/inc/muller.gif" alt="Muller">';
  break;
case 46:
  echo '<img src="imgs/inc/noene.gif" alt="Noene">';
  break;
default:
  echo '<img src="imgs/inc/default.gif" alt="blah">';
  break;
}
?>