带有if语句的变量/函数

Question

I want a menu showing links/buttons depending on the permissions of the account. And I don't want to copy/paste this code to every page. I use an if-statement to check if the link/button will show for the user who is logged in at the moment. How can I like create a variable or function to just print the menu that is on another PHP file?

Like this menu:

<div id="menu">
    <a href="#">Link 1</a>
    <a href="#">Link 2</a>
    <a href="#">Link 3</a>
    <a href="#">Link 4</a>';

    if(condition) {
        echo '
        <a href="#">Link 5</a>';
    }

</div>

Answer 1

Store your menu HTML in a file, and include it in every other page in which you want your menu to appear.

menu.php:

<div id="menu">
    <a href="#">Link 1</a>
    <a href="#">Link 2</a>
    <a href="#">Link 3</a>
    <a href="#">Link 4</a>';

    <?php
    if(condition) {
        echo '
        <a href="#">Link 5</a>';
    }
    ?>

</div>

foo.php:

<?php

include 'menu.php';

# code ...

That way, the contents of menu.php will get evaluated and be put into your foo.php file as if it were copied and pasted.