How to display different links for logged in and logged out users?

Question

When a user is not logged in I am trying to show

Support || Log In

When they are logged out it should say

Support || Log Out 
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Answer 1

Add this to functions.php in your theme or create a plugin. Change 'menu' to the menu location eg. 'primary-menu' in your theme. Change 'logged-in' to the name of the logged in menu and logged out respectively.

<?php
  function my_wp_nav_menu_args( $args = '' ) {

    if( is_user_logged_in() ) { 
        $args['menu'] = 'logged-in';
    } else { 
        $args['menu'] = 'logged-out';
    } 
        return $args;
    }
    add_filter( 'wp_nav_menu_args', 'my_wp_nav_menu_args' );
?>

referenced from: WPBeginner

Answer 2

Apart from coding you can always use a plugin called Nav Menu Roles

Answer 3

Not sure about the result of your function, but to start you were not escaping properly. Secondly, why not just use one li to house the correct link as follows:

<div class="fr">             
    <ul class="rss">                  
        <li><a href="http://example.com/go/wp-login.php">Support</a></li>                 
        <li><?php if (is_user_logged_in() ) { 
                echo " <a href=\"" . wp_logout_url() . "\" title=\"Logout\">Logout</a>";
            }else{ 
                echo " <a href=\"http://example.com/" title=\"Login\">Member Login</a>";
            } ?>
        </li>             
    </ul>         
</div> 

Answer 4

Another way to display content for users with a shortcode. Post this into functions.php

// When member is logged in [memberin]

add_shortcode( 'memberin', 'member_check_shortcode' ); 
function member_check_shortcode( $atts, $content = null ) {
function member_check_shortcode( $atts, $content = null ) {
  if ( is_user_logged_in() && !is_null( $content ) && !is_feed() )
return $content;
return '';
}

// When member is logged out [memberout]

add_shortcode( 'memberout', 'member_check_shortcode_out' );
function member_check_shortcode_out( $atts, $content = null ) {
  if (!is_user_logged_in() && !is_null( $content ) && !is_feed() )
return $content;
return '';
}

Answer 5

Have you tried changing ?php else if (!is_user_logged_in() ) to just ?php if (!is_user_logged_in() ) ?

Answer 6

What about the following?

<div class="fr">
    <ul class="rss">
        <li><a href="http://example.com/wp-login.php">Support</a></li>
        <?php if (is_user_logged_in() ) { echo " <li><a href=" . wp_logout_url() . " title=\"Logout\">Logout</a></li>";}?>
        <?php else if (!is_user_logged_in() ) { echo " <li><a href="http://example.com/wp-login.php" title=\"Logout\">Member Login</a></li>";}?>
    </ul>
</div>

I moved the php if statements so you don't get empty li elements. You should still do the escaping that others have noticed.