CODEWITHSUNDEEP
phpcss
jenny
jenny

Reputation: 229

How to set current page "active" in php

Hi I have a menu on my site on each page, I want to put it in it's own menu.php file but i'm not sure how to set the class="active" for whatever page i'm on. Here is my code: please help me

menu.php:

<li class=" has-sub">
    <a class="" href="javascript:;"><i class=" icon-time"></i> Zeiten<span class="arrow"></span></a>
    <ul class="sub">
       <li><a class="" href="offnungszeiten.php">Öffnungszeiten</a></li>
       <li><a class="" href="sauna.php">Sauna</a></li>
       <li><a class="" href="frauensauna.php">Frauensauna</a></li>
       <li class=""><a class="" href="custom.php">Beauty Lounge</a></li>
       <li><a class="" href="feiertage.php">Feiertage</a></li>
    </ul>
</li>

Upvotes: 21

Views: 95876

Answers (17)

Sakauat Hossen
Sakauat Hossen

Reputation: 1

For making a dynamic active menu link I follow this method. first, In the menu link, I always use the full address:

//HTML CODE

<ul class="menu">
  <li>
    <a href="http://localhost/index.php">Home</a>
  </li>
  <li>
    <a href="http://localhost/about.php">About us</a>
  </li>
  <li>
    <a href="http://localhost/contact.php">Contact</a>
  </li>
</ul>

//Javacript Code
  const menus = document.querySelectorAll('.menu li a');
menus.forEach((menu) => {
  const currentLocation = window.location.href;
  if (currentLocation === window.origin) {
    menus[0].classList.add('active');
  } else if (menu.href === currentLocation) {
    menu.classList.add('active');
  } else {
    return;
  }
});

and then I will use vanilla javascript code to do the rest

Upvotes: 0

user14181536
user14181536

Reputation: 45

I have a similar issue with my web app menu. I also have sub menus which do not appear as top level menu buttons.

My solution is as follows:

a) Partial php file with menu html and a little php function at the top that checks GET variables against the menu buttons. I have two GET variables to check: the page and (if necessary) the menu_button.

b) Adding any new php page with a href links to either menu pages or sub menu pages.

The variable "menu_button" is optional and can be used to link to submenu php files.

Of course the security concerning GET variables should be considered.

From my point of view, this solution has less effort than having to maintain an array of pages or links somewhere. You just use a get variable "menu_button" where you pass the top level menu button that should be marked visually in any link which targets your php file.

Code examples:

Partial menu.php (has to be included in every php file):

<?php
  function active($page_link){
    $menu_button = $_GET("menu_button") ?: $_GET("page");  // sets the menu button either to the given top level menu or it defaults to the page itself
    if($menu_button === $page_link) return "active";
  }
?>

<div>
<a href="?page=one" class="<?= active('one') ?>"Link one</a>
<a href="?page=two" class="<?= active('two') ?>">Link two</a>
</div>

Any php file with links to sub menu file:

<div>
 <a href="?page=one">Link one</a>
 <a href="?page=three&menu_button=two">Link to sub menu page "three" of menu "two"</a>
</div>

Works for me. Hope someone else can use this.

Upvotes: 0

Bandula Dharmadasa
Bandula Dharmadasa

Reputation: 867

Assume you have a navbar with the following items:

<ul>
   <li id="menu-item-home"><a href="index.php">HOME</a></li>
   <li id="menu-item-services"><a href="/services.php">SERVICES</a></li>
   <li id="menu-item-about-us"><a href="/about-us.php">ABOUT US</a></li>
   <li id="menu-item-contact"><a href="/contact.php">CONTACT</a></li>      
</ul>

Then, declare a javascript variable in each page as below:

<script>
    <?php echo("var active = 'menu-item-home';"); ?>
</script>

The variable "active" is assigned with the corresponding item of each page.

Now, you can use this variable to highlight the active menu item as below.

$(window).ready(function(){$("#" + active).addClass("active");});

Upvotes: 0

Steve Shead
Steve Shead

Reputation: 131

Better late than never - I like to keep it simple, to be honest, especially if there's a ton of scripting and PHP going on.

I place this code on the top of each page to identify the page:

<?php
  $current_page = 'home';
  include 'header.php';
?>

Then your menu/navigation (mine is bootstrap 4) looks like this:

<ul class="navbar-nav mx-auto">
  <li class="nav-item <?php if ($current_page=="home") {echo "active"; }?>">
      <a class="nav-link" href="<?php echo SITEURL;?>/">Home</a>
  </li>
  <li class="nav-item <?php if ($current_page=="about") {echo "active"; }?>">
      <a href="<?php echo SITEURL;?>/about" class="nav-link ">About</a>
  </li>
  <li class="nav-item <?php if ($current_page=="store") {echo "active"; }?>">
      <a href="<?php echo SITEURL;?>/store" class="nav-link">Store</a>
  </li>
  <li class="nav-item <?php if ($current_page=="news") {echo "active"; }?>">
        <a href="<?php echo SITEURL;?>/news" class="nav-link">News</a>
  </li>
  <li class="nav-item <?php if ($current_page=="contact") {echo "active"; }?>">
      <a href="<?php echo SITEURL;?>/content/contact-us" class="nav-link ">Contact</a>
  </li>
</ul>

I'm not saying this is the optimal method, but it works for me and it's simple to implement.

Upvotes: 1

Remzi Cavdar
Remzi Cavdar

Reputation: 187

I have some simple example, see below:

<?php
function active($currect_page) {
    $url = $_SERVER['REQUEST_URI'];

    if($currect_page == $url){
        echo 'active';
    }
}
?>

<ul class="navbar-nav mr-auto">
  <li class="nav-item <?php active('/');?>">
    <a class="nav-link" href="/">Home</a>
  </li>
  <li class="nav-item <?php active('/other');?>">
    <a class="nav-link" href="/other">Other page</a>
  </li>
</ul>

Upvotes: 0

mrmut
mrmut

Reputation: 494

This worked for me:

    function active_page($script){

    $actual = basename($_SERVER['PHP_SELF']);

  if($script == $actual){
      return 'active-page'; //class name in css
  }
}

Upvotes: 0

asprin
asprin

Reputation: 9823

Put all the below code in menu.php and everything will be taken care of.

// function to get the current page name
function PageName() {
  return substr($_SERVER["SCRIPT_NAME"],strrpos($_SERVER["SCRIPT_NAME"],"/")+1);
}

$current_page = PageName();

Use the above to get the current page name then put this in your menu 

<li><a class="<?php echo $current_page == 'offnungszeiten.php' ? 'active':NULL ?>" href="offnungszeiten.php">Öffnungszeiten</a></li>
<li><a class="<?php echo $current_page == 'sauna.php' ? 'active':NULL ?>" href="sauna.php">Sauna</a></li>
<li><a class="<?php echo $current_page == 'frauensauna.php' ? 'active':NULL ?>" href="frauensauna.php">Frauensauna</a></li>
<li><a class="<?php echo $current_page == 'custom.php' ? 'active':NULL ?>" href="custom.php">Beauty Lounge</a></li>
<li><a class="<?php echo $current_page == 'feiertage.php' ? 'active':NULL ?>" href="feiertage.php">Feiertage</a></li>

where active is the name of the class which will highlight your menu item

Upvotes: 10

iraqi_love4ever
iraqi_love4ever

Reputation: 583

You can use

<?php
function active($current_page){
    $page = $_GET['p'];
    if(isset($page) && $page == $current_page){
        echo 'active'; //this is class name in css
    }
}
?>
<ul>
    <li><a class="<?php active('page1');?>" href="?p=page1">page1</a></li>
    <li><a class="<?php active('page2');?>" href="?p=page2">page2</a></li>
    <li><a class="<?php active('page3');?>" href="?p=page3">page3</a></li>
    <li><a class="<?php active('page4');?>" href="?p=page4">page4</a></li>
</ul>

Upvotes: -1

iraqi_love4ever
iraqi_love4ever

Reputation: 583

this method gets the current page using php which will pass a word in this case active and places it inside the class parameter to set the page active.

<?php
function active($currect_page){
  $url_array =  explode('/', $_SERVER['REQUEST_URI']) ;
  $url = end($url_array);  
  if($currect_page == $url){
      echo 'active'; //class name in css 
  } 
}
?>
<ul>
    <li><a class="<?php active('page1.php');?>" href="http://localhost/page1.php">page1</a></li>
    <li><a class="<?php active('page2.php');?>" href="http://localhost/page2.php">page2</a></li>
    <li><a class="<?php active('page3.php');?>" href="http://localhost/page3.php">page3</a></li>
    <li><a class="<?php active('page4.php');?>" href="http://localhost/page4.php">page4</a></li>
</ul>

Upvotes: 35

jimmy
jimmy

Reputation: 411

A bit late on the ball, but I just had to solve this myself and ended up using this Javascript method, with a small modification. This has the advantage on not requiring many changes to the current code, just run the script and voila.

window.onload = activateCurrentLink;

function activateCurrentLink(){
    var a = document.getElementsByTagName("A");
    for(var i=0;i<a.length;i++)
        if(a[i].href == window.location.href.split("#")[0])
            a[i].className = 'activelink';
}

Upvotes: 1

Alex
Alex

Reputation: 1030

I solved this using jQuery/javascript by running the code below each time my any page is loaded: 

 $(document).ready(function () {    
    //Get CurrentUrl variable by combining origin with pathname, this ensures that any url appendings (e.g. ?RecordId=100) are removed from the URL
    var CurrentUrl = window.location.origin+window.location.pathname;
    //Check which menu item is 'active' and adjust apply 'active' class so the item gets highlighted in the menu
    //Loop over each <a> element of the NavMenu container
    $('#NavMenu a').each(function(Key,Value)
        {
            //Check if the current url
            if(Value['href'] === CurrentUrl)
            {
                //We have a match, add the 'active' class to the parent item (li element).
                $(Value).parent().addClass('active');
            }
        });
 });

This implementation assumes your menu has the 'NavMenu' ID, and uses http://hostname/scriptname.php href attributes like so:

 <ul id="NavMenu">
   <li><a href="http://localhost/index.php">Home</a></li>
   <li><a href="http://localhost/smartphone.php">Smartphone</a></li>
   <li><a href="http://localhost/tablet.php">Tablet</a></li>
   <li><a href="http://localhost/about.php" class="active">About Us</a></li>
   <li><a href="http://localhost/contact.php">Contact Us</a></li>
</ul>

Read the javascript comments to see what's going on. If you prefer to use a different href layout (like in your original example), you have to play with the CurrentUrl variable a bit to get it to use the same layout as your href attributes.

For me this was the easiest solution since I had an existing sites with a big menu and many pages, and wanted to avoid having to modify all pages. This allows me to throw in a piece javascript code in the header file (which was a central file already) which solves the problem for all existing pages.

Upvotes: 1

CodyBugstein
CodyBugstein

Reputation: 23312

Simplere solution:

Borrowing the code from asprin above;

Create a new file menu.php where you will store the one and only copy of the menu. In this file, you will create a function addMenu($pageName) that take a parameter as the page name and returns a string consisting of the menu after having added the current tag.

In your HTML code, you would include(menu.php) and then call the function addMenu with the current page name. So your code will look like this:

menu.php

     <?php

        function addMenu($pageName){
            $menu = 

                '<ul>
                    <li><a href="Öffnungszeiten.php"' . ($pageName == "Öffnungszeiten" ? "class=\"current\"" : "") . '><span>Öffnungszeiten</span></a></li>
                    <li><a href="sauna.php"' . ($pageName == "Öffnungszeiten" ? "class=\"current\"" : "") . '><span>Sauna</span></a></li>
                    <li><a href="frauensauna.php"' . ($pageName == "Frauensauna" ? "class=\"current\"" : "") . '><span>Frauensauna</span></a></li>
                    <li><a href="custom.php" ' . ($pageName == "lounge" ? "class=\"current\"" : "") . '><span>Beauty Lounge</span></a></li>
                    <li><a href="Feiertage.php"' . ($pageName == "feiertage" ? "class=\"current\"" : "") . '><span>Feiertage</span></a></li>                                                                                
                </ul>';
          return $menu;
    }
        ?>

And in your HTML, say this:

<div id="menu">
            <?php
                include('menu.php');
                echo addMenu("index");
                echo $hello;     
            ?>
        </div>

Upvotes: 0

Angela Taylor
Angela Taylor

Reputation: 151

I have done it with php in this way, 

function createTopNav($active)
{
    $pages = array(
        array(
            'name'=>'Home',
            'link'=>'index'
        ),
        array(
            'name'=>'Smartphone',
            'link'=>'smartphone'
        ),
        array(
            'name'=>'Tablet',
            'link'=>'tablet'
        ),
        array(
            'name'=>'About Us',
            'link'=>'about'
        ),
        array(
            'name'=>'Contact Us',
            'link'=>'contact'
        )
    );
    $res = "<ul>";
    $activePage = "";
    foreach($pages as $key=>$val)
    {
        if($val['link']==$active)
        {
            $res.= "<li><a href='".$val['link']."'  class='active'  >".$val['name']."</a></li>";
        }
        else
        {
            $res.= "<li><a href='".$val['link']."'>".$val['name']."</a></li>";          
        }
    }
    $res.="</ul>";
    return $res;
}

And then to call this function

echo createTopNav("about");

and the output will be like this 

<ul>
   <li><a href="index">Home</a></li>
   <li><a href="smartphone">Smartphone</a></li>
   <li><a href="tablet">Tablet</a></li>
   <li><a href="about" class="active">About Us</a></li>
   <li><a href="contact">Contact Us</a></li>
</ul>

Upvotes: 1

package
package

Reputation: 4801

It would be easier if you would build an array of pages in your script and passed it to the view file along with the currently active page:

//index.php or controller

$pages = array();
$pages["offnungszeiten.php"] = "Öffnungszeiten";
$pages["sauna.php"] = "Sauna";
$pages["frauensauna.php"] = "Frauensauna";
$pages["custom.php"] = "Beauty Lounge";
$pages["feiertage.php"] = "Feiertage";

$activePage = "offnungszeiten.php";


//menu.php
<?php foreach($pages as $url=>$title):?>
  <li>
       <a <?php if($url === $activePage):?>class="active"<?php endif;?> href="<?php echo $url;?>">
         <?php echo $title;?>
      </a>
  </li>

<?php endforeach;?>

With a templating engine like Smarty your menu.php would look even nicer:

//menu.php
{foreach $pages as $url=>$title}
   <li>
       <a {if $url === $activePage}class="active"{/if} href="{$url}">
         {$title}
      </a>
   </li>
{/foreach}

Upvotes: 18

Netorica
Netorica

Reputation: 19327

there is two things you can do.

first you can read the current filename of the php file you request by using $_SERVER['PHP_SELF'] or $_SERVER['REQUEST_URI'] or any other $_SERVER global variables that you can use to read your current page and compare it with the link's url, something like this

  <a href="offnungszeiten.php" <?php if($_SERVER['PHP_SELF']=='offnungszeiten.php'){ ?>class="activatepage" <?php } ?> >
       Öffnungszeiten
    </a>

the second one is to create a variable that you can read globally that would store the current name of the current page, like this

<?php
$cur_page ="offnungszeiten"
?>

<a href="offnungszeiten.php" <?php if($cur_page=='offnungszeiten'){ ?>class="activatepage" <?php } ?> >
 Öffnungszeiten
</a>

Upvotes: 2

Ajeet Singh
Ajeet Singh

Reputation: 32

Send page name in query string and check it on every page by getting the variable.

Upvotes: 0

chandresh_cool
chandresh_cool

Reputation: 11830

Create a variable in each of your php file like :

$activePage = "sauna"; (different for each page)

then check that variable in your html page like this

<?php if ($activePage =="sauna") {?>
 class="active" <?php } ?>

Upvotes: 13

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