Check if URL = Desired URL (PHP)

Question

I'm trying to check if the current URL equals the desired URL for making the link class active (Bootstrap).

Example

<?php
    If ($currenturl = "show.php?name=123") {
      ?><li class="active"><a href="show.php?name="123"">123</a></li><?php
    }else{
        ?><li class=""><a href="show.php?name="123"">123</a></li><?php
    }
?>

Current URL -> $currenturl = $_SERVER['REQUEST_URI']

Now, I have this if I'm on show.php?name=456

see

basename($_SERVER['PHP_SELF']); – I wrestled a bear once. Dec 08 '14 at 00:57 — I wrestled a bear once., Dec 08 '14 at 00:57

Kevin · Accepted Answer · 2014-12-08T00:51:11.533

Make sure you have the correct comparison inside the if. You're currently using the assignment operator:

Note the difference:

if($currenturl = "show.php?name=123") // assignment =
  // changes $currenturl to "show.php?name=123"
  // and then tests if("show.php?name=123") equivalent to if(true)
  // see also http://stackoverflow.com/questions/5940626/evaluation-of-assignment-in-php

if($currenturl == "show.php?name=123") // comparison ==

Second: You have set $urlpage but comparing $currenturl. Use $urlpage

Code:

<?php $urlpage = $_SERVER['REQUEST_URI']; ?>
<?php if ($urlpage == "/show.php?nom=123") { ?>
    <li class="active"><a href="/show.php?nom=123">123</a></li>
<?php } else { ?>
    <li class=""><a href="/show.php?nom=123">123</a></li>
<?php } ?>

An alternative using a ternary operator:

<li <?php echo ($urlpage == "/show.php?nom=123") ? 'class="active"' : ''; ?>><a href="/show.php?nom=123">123</a></li>

Applying to all pages:

<?php $pages = array('123', '456', '789'); ?>
<ul>
    <li <?php echo (!isset($_GET['nom']) ? 'class="active"' : ''); ?>><a href="show.php">Home</a></li>
    <?php foreach($pages as $page): ?>
        <?php if (isset($_GET['nom']) && $_GET['nom'] == $page) { ?>
            <li class="active"><a href="/show.php?nom=<?php echo $page; ?>"><?php echo $page; ?></a></li>
        <?php } else { ?>
            <li class=""><a href="/show.php?nom=<?php echo $page; ?>"><?php echo $page; ?></a></li>
        <?php } ?>
    <?php endforeach; ?>
</ul>

@dotpush its okay to add supplemental info, no need to remove the ternary way as this was an alternative and part of the answer also — Kevin, Dec 08 '14 at 00:33
It's not working... It's just not actived! Also, when I click on the link, it's goes to `show.php?name=` but with no 123 or other. — Félix Desjardins, Dec 08 '14 at 00:35
@Ghost: Oups, sorry. I didn't even saw for the ternary way part removal. — dotpush, Dec 08 '14 at 00:43
Tanks a lot! Just change `/show.php?nom=` to `show.php?nom=`. — Félix Desjardins, Dec 08 '14 at 01:07

score 0 · Answer 2 · answered Dec 08 '14 at 00:23

0

An alternate way is to use the $_GET

if(isset($_GET['name']))
 if($_GET['name'] == '123') 
  ...

and so forth

answered Dec 08 '14 at 00:23

DdD

453
4
19

Yes, i change to `If ($_GET['name'] == '123')` but I get an error : `Notice: Undefined index: name in C:\xampp\htdocs\cadeau noel\cadeau.php on line 40`. Line 40 -> `If ($_GET['name'] == '123') {` – Félix Desjardins Dec 08 '14 at 00:38
What does URL look like? (In your browser) – DdD Dec 08 '14 at 00:39
Also thats why I suggested to put `if(isset($_GET['name']))` if your URL does not have the ?name=123 or name=124, name=125 etc at the end of the URL, it will just give a warning. So either earlier in the code add `if(!isset($_GET['name'])) $_GET['name'] = '';` – DdD Dec 08 '14 at 00:42
Tank you, but I used the Ghost answer, it was better. – Félix Desjardins Dec 08 '14 at 01:08

Check if URL = Desired URL (PHP)

2 Answers2