用于检测当前页面的代码

Question

Is there a php or javascript code that can detect the current user's page and then add <a class="active"> to an item in a ul (my menu). I include my menu in my pages with PHP include so making change is easy; I only have to edit it once. But, with that method, I can't individually set each page to have a class="active". How can I do this?

Answer 1

If I'm understanding your question correctly, what I usually do is set a variable before I include the header, like

$current = "home";

And then in the header I'd have an if statement in each link

<a href="/home" <?php if ( $current == "home" ) { echo "class='active'" } ?>>Home</a>

Could be ways to improve it, but it's simple if your menu isn't too big.