2014-02-22 51 views
0

我想在這裏做的是顯示我的登錄信息,並在顯示別的用戶名和密碼的幫助

實例一旦登錄就在這裏

<?php 
    $user = JFactory::getUser(); 
    $status = $user->guest; 

    if($status == 1){ 
    echo ' 
    <form id="login-form" class="form-inline" method="post" action="/home"> 
    <div class="topUser"> 
     Name: 
     <input id="modlgn-username" class="TopinputReg" type="text" placeholder="User Name" tabindex="0" name="username" /> 
     </div> 
     <div class="toppass">Password: 
     <input id="modlgn-passwd" class="TopinputReg" type="password" placeholder="Password" tabindex="0" name="password" pwfprops="," /> 
     </div> 
     <div class="topsignin"> 
     <input class="Signin" name="Submit" tabindex="0" type="submit" value="" /> 
     </div> 
     <div class="topregister"> 
     <a href="http://gamers-league.com/forums/user/register"><input name="button" type="submit" class="Register" id="button" value="" /></a> 
       </div> 
    <input type="hidden" value="com_users" name="option"> 
    <input type="hidden" value="user.login" name="task"> 
    <input type="hidden" value="aW5kZXgucGhwP0l0ZW1pZD0xMzM=" name="return"> 
    <input type="hidden" value="1" name="39ea5446d876078c6f6221d396ef5bd4"> 
       </form> 
    '; 
    } 
    else 
    { 
    echo ' 
    <div class="topUser">Welcome Back!</div> 
     <div class="toppass">Enjoy you stay.</div> 
     <form id="login-form" class="form-vertical" method="post" action="/log-out"> 
     <div class="topsignin"> 
     <input class="Signout" type="submit" value="" name="Submit" /> 
     <input type="hidden" value="com_users" name="option"> 
       </div> 
    <input type="hidden" value="com_users" name="option"> 
    <input type="hidden" value="user.logout" name="task"> 
    <input type="hidden" value="aW5kZXgucGhwP0l0ZW1pZD0xMDE=" name="return"> 
    <input type="hidden" value="1" name="994c61f8ab4ccf23fe9dae6546a87089"> 

     </form> 
     <div class="topregister"> 
     <a href="http://gamers-league.com/forums/user"><input name="button" type="submit" class="account" id="button" value="" /></a> 
     </div> 
    '; 
    } 
    ?> 

沒有,我知道我在做什麼錯只是沒有線索如何解決它。兩個

需要有一個不同的生成的數字每次但我怎麼做到這一點。

,因爲這個數字是相同的,每次當我登錄再出來,然後嘗試重新登錄,它會給我「無效令牌」

當我進入登錄模板它給了我 本作登陸

<input type="hidden" name="return" value="<?php echo base64_encode($this->params->get('login_redirect_url', $this->form->getValue('return'))); ?>" /> 
<?php echo JHtml::_('form.token'); ?> 

這對於註銷

<input type="hidden" name="return" value="<?php echo base64_encode($this->params->get('logout_redirect_url', $this->form->getValue('return'))); ?>" /> 
<?php echo JHtml::_('form.token'); ?> 

現在,如果我替換已有的預生成的號碼與PHP的塞入隱藏字段上面的f當我嘗試刷新我的頁面時,它完全變爲空白。

我對此感到不知所措,請參閱我在php中不怎麼高級。任何幫助任何人都不能提供將大量有用的..

+0

究竟是你想做些什麼?如果你不熟悉PHP,那麼你最不想做的事情就是玩弄密碼。你只是想在用戶登錄時向用戶顯示一些東西? – Elin

+0

即時通訊設法顯示登錄和註冊,如果登錄有一個選項註銷或帳戶。 – user3238054

回答

0

爲什麼不嘗試這樣,

<?php 
    $user = JFactory::getUser(); 
    $status = $user->guest; 

    if($status == 1): 
?> 

<form id="login-form" class="form-inline" method="post" action="/home"> 
    <div class="topUser"> 
     Name: 
     <input id="modlgn-username" class="TopinputReg" type="text" placeholder="User Name" tabindex="0" name="username" /> 
     </div> 
     <div class="toppass">Password: 
     <input id="modlgn-passwd" class="TopinputReg" type="password" placeholder="Password" tabindex="0" name="password" pwfprops="," /> 
     </div> 
     <div class="topsignin"> 
     <input class="Signin" name="Submit" tabindex="0" type="submit" value="" /> 
     </div> 
     <div class="topregister"> 
     <a href="http://gamers-league.com/forums/user/register"><input name="button" type="submit" class="Register" id="button" value="" /></a> 
       </div> 
    <input type="hidden" value="com_users" name="option"> 
    <input type="hidden" value="user.login" name="task"> 
    <input type="hidden" name="return" value="<?php echo base64_encode($this->params->get('login_redirect_url', $this->form->getValue('return'))); ?>" /> 
<?php echo JHtml::_('form.token'); ?> 
       </form> 

<?php else: 
?> 

<div class="topUser">Welcome Back!</div> 
     <div class="toppass">Enjoy you stay.</div> 
     <form id="login-form" class="form-vertical" method="post" action="/log-out"> 
     <div class="topsignin"> 
     <input class="Signout" type="submit" value="" name="Submit" /> 
     <input type="hidden" value="com_users" name="option"> 
       </div> 
    <input type="hidden" value="com_users" name="option"> 
    <input type="hidden" value="user.logout" name="task"> 
    <input type="hidden" name="return" value="<?php echo base64_encode($this->params->get('logout_redirect_url', $this->form->getValue('return'))); ?>" /> 
<?php echo JHtml::_('form.token'); ?> 

     </form> 
     <div class="topregister"> 
     <a href="http://gamers-league.com/forums/user"><input name="button" type="submit" class="account" id="button" value="" /></a> 
     </div> 
<?php endif;?> 
+0

因爲它的原因,當我添加,http://gamers-league.com/home – user3238054

+0

檢查我的編輯註銷url正確放置,也是什麼您的要求? –

+0

好吧,我用你所做的更新了它,並且它在php之後仍然空白了剩下的內容。我想要做的是讓人們可以登錄,如果他們登錄,它會顯示註銷和帳戶。這是它應該看起來只有工作。鏈接的一個只是CSS而不是一個真正的登錄http://gamers-league.com/wild-star-home – user3238054