$_SERVER['REQUST_URI'] unique animation

Question

I'm attempting to make a different animation depending on from where the user navigated from, but this isn't working. Thanks!

    <?php 
    $ref =$_SERVER['REQUEST_URI'];      
?>

<script type="text/javascript">
    if(<?php $ref; ?>) == 'http://livvedesign.com/content/philosophy.php/'){
        $('#wrapper1').css('left','-150vw');
        $('#wrapper1').delay(500).animate({left:'0vw', opacity:1}, 800);
    };
    if(<?php $ref;?> == 'http://livvedesign.com/content/meet.php/'){
        $('#wrapper1').css('left','150vw');
        $('#wrapper1').delay(500).animate({left:'0vw', opacity:1}, 800);
    };
    if(<?php $ref;?> == 'http://livvedesign.com/content/contact2.php/'){
        $('#wrapper1').css('top','250vh');
        $('#wrapper1').delay(500).animate({top:'0vh', opacity:1}, 800);
    };
     if(<?php echo $ref;?> == 'http://livvedesign.com/content/projects.php'){
        $('#wrapper1').css('top','250vh');
        $('#wrapper1').delay(500).animate({top:'0vh', opacity:1}, 800);
    };
    if(<?php $ref ?> == "" || <?php $ref ?> == 'index.php') {$('#wrapper1').animate({opacity:1}, 700);};
        </script>

you cant use it like this, you need to use `echo` like this `if() ` — dreamweiver, Jul 23 '13 at 05:58
Why would you throw all that javascript to the client? IMHO there are 2 better options (1) It seems the javascript could just be fixes, and inspect `window.location.href`, so zero PHP needed (2) If you _do_ want to use PHP, just use a switch in PHP itself (omit the protocol & domain for matching), and only echo out the javascript portion relevant for _that_ page. — Wrikken, Jul 23 '13 at 06:06

Kara · Answer 1 · 2013-07-23T06:01:09.057

3

If you want to print the $ref variable you should do:

<?php echo $ref;?>

or the shortcut syntax if you have short open tags enabled:

<?=$ref?>

You will also need to put the string in quotes so each if statement should look something like:

if('<?=$ref?>' == 'http://livvedesign.com/content/philosophy.php/'){

You also have extra brackets in some lines like this:

if(<?php $ref; ?>) == 'http://livvedesign.com/content/philosophy.php/'){
                 ^ //This should not be here

edited Jul 23 '13 at 06:01

answered Jul 23 '13 at 05:56

Kara

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I made the changes but the animation still isn't working. – cgauss Jul 23 '13 at 07:31

score 0 · Answer 2 · answered Jul 23 '13 at 05:58

Try to print_r/var_dump $_SERVER. You would see that $_SERVER['REQUEST_URI'] doesn't give you the data you expect. I guess you really meant to use $_SERVER['HTTP_REFERER'], and as a comment say you are missing some echoes.

Also when echoing $ref you should make sure to escape the data. It is easily altered by a potential attacker.

score 0 · Answer 3 · answered Jul 23 '13 at 06:00

0

The REQUEST_URI which was given in order to access the page not referer

$ref = $_SERVER['REQUEST_URI']; 

echo $ref;

// Output
/content/philosophy.php/

Use referer:

$ref = $_SERVER['HTTP_REFERER'];

answered Jul 23 '13 at 06:00

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$_SERVER['REQUST_URI'] unique animation

3 Answers3