Help with my functions.php coding to create a logout link

[pmcg911]

Hi all

 

I currently have this code in my functions.php to control my footer menu:

add_filter('omega_footer_insert', 'my_omega_footer_insert', 11);

function my_omega_footer_insert( $settings ) {
    $year = date( 'Y' );
    $siteurl = get_bloginfo( 'url' );
    $sitename = get_bloginfo( 'name' );
    return "<a href='http://bibliophone.com/about/'>About</a> | <a href='http://bibliophone.com/terms-of-use/'>Terms of Use</a> | <a href='http://bibliophone.com/privacy-policy/'>Privacy Policy</a><br>© <a href='$siteurl'>$sitename</a> $year</p>";
}

I want to add a logout link to this, which displays ONLY when a user is registered and logged in. I have this code:

if ( is_user_logged_in() ) {
echo ' | <a href="<?php echo wp_logout_url( get_permalink() ); ?>" title="Logout">Logout</a>';
}

..but I'm not having mu luck with it because I don't know for certain that it's correct, and secondly I'm not sure where it should be placed within the first bit of code above. Can anyone help please?

 

Thank you!

 

Paul

[Ch0cu3r]

That functions returns the footer menu as a string, so you cant use echo. You need to concatenate the logout link if the user is logged in. Something like this maybe

function my_omega_footer_insert( $settings ) {
    $year = date( 'Y' );
    $siteurl = get_bloginfo( 'url' );
    $sitename = get_bloginfo( 'name' );

    $footer = '';

    // prepend logout link to footer
    if(is_logged_in())
        $footer .= '<a href="' . wp_logout_url(get_permalink()) . '" title="Logout">Logout</a> | ';

    $footer .= "<a href='http://bibliophone.com/about/'>About</a> | <a href='http://bibliophone.com/terms-of-use/'>Terms of Use</a> | <a href='http://bibliophone.com/privacy-policy/'>Privacy Policy</a><br>© <a href='$siteurl'>$sitename</a> $year</p>";

   return $footer;
}