Button Bug（CodeIgniter＆Twitter Bootstrap）

Question

CodeIgniter中的編碼跟蹤/取消關注系統（PHP）& Twitter-Bootstrap。我也有路由活動的URLS。按照VIEW中的按鈕代碼。

<!-- Follow Button Start --> 
<?php $is_logged_in = $this->session->userdata('is_logged_in'); ?> 
<?php if(!(empty($is_logged_in)) && $sID != $vID && !in_array($sID, $following)): ?> 
    <button class="btn" type="submit" onClick="location.href='<?php echo site_url("follow/$vUsername"); ?>'">Follow <?php echo $vUsername; ?> </button> 
<?php elseif (in_array($sID, $following)):?> 
    <button class="btn" type="submit" onClick="location.href='<?php echo site_url("unfollow/$vUsername"); ?>'">UnFollow <?php echo $vUsername; ?> </button> 
<?php else: ?> 
    <button class="btn disabled" type="submit">Follow <?php echo $vUsername; ?> </button> 
<?php endif; ?> 
<!-- Follow Button End --> 

即使用戶沒有比得過按鈕下面顯示UnFollow $vUsername

Answer 1

我不能在這裏充分測試，但這樣會使我更有意義：

<?php if((!empty($is_logged_in)) && ($sID != $vID) && (!in_array($sID, $following))): ?> 

編輯：

那麼，在這種情況下，只有簡單的一步一步的調試將爲您節省：

<?php 
$is_logged_in = $this->session->userdata('is_logged_in'); 

if(!empty($is_logged_in)) 
{ 
    echo '$is_logged_in is not empty'; 
} 

echo '$sID is: '.$sID.' and it should not be equal to $vID'. $vID; 

if($sID != $vID){ 
    echo 'and indeed it is not'; 
} 
else 
{ 
    echo 'but it is. Here is the problem'; 
} 

echo 'This is the $following array:'; 
print_r($following); 

if(!in_array($sID, $following)) 
{ 
    echo '$sID is not in the $following array'; 
} 
else 
{ 
    echo '$sID IS in the $following array'; 
} 

?>