Question

Sorry i am a newb in PHP, i am trying the following but it isnt working.

else{
            $output = '<form name="loginform" id="loginform" action="%s" method="post">
            <div class="title">test</div>
            <div class="simplemodal-login-fields">
            You are logged in
            </form/></div></div>';
            return $output;
            echo (wp_logout_url( get_permalink() ));
            }

It only shows the $output data what i am trying to do is show the wordpress logout link
<?php echo wp_logout_url( $redirect ); ?>

Here is the reference: http://codex.wordpress.org/Function_Reference/wp_logout_url

Any advice is appreciated

Edit:

I just tried the following without success

else {
                $html = '
                <form name="loginform" id="loginform" action="%s" method="post">
                    <div class="title">test</div>
                        <div class="simplemodal-login-fields">
                        You are logged in
                        </div>
                    </div>
                </form/>
                ';
                $html .= '<a href="' . wp_logout_url( get_permalink() ) . '">Logout</a>';
                return $html;
            }

Answer 1

The issue is that i was not displaying the url inside the div elements

the following is the right code

$html = '<form name="loginform" id="loginform" action="%s" method="post">
                <div class="title">test</div>
                    <div class="simplemodal-login-fields">
                    You are logged in
                    <a href="' . wp_logout_url( get_permalink() ) . '">Logout</a>
                    </div>
                </div>
            </form/>
            ';
            return $html;