嘗試用預先準備好的語句來捕獲

Question

我對這個查詢很困難，並且建議我學習嘗試並捕獲塊，以便更容易地找出錯誤。所以這是我的第一次嘗試。

我得到了我的@ stmt2沒有被定義爲我的print_r($stmt2)行的錯誤。

Notice: Undefined variable: stmt2 

這是我在try try上的嘗試。我爲這個錯誤出現了什麼問題嗎？

try { 
$con = mysqli_connect("localhost", "", "", ""); 
if (mysqli_connect_errno()) { 
    throw new Exception("Connect failed: %s\n", mysqli_connect_error()); 
    exit(); 
} 
$cid = $_GET['cid']; 
$tid = $_GET['tid']; 
$userid = (isset($_SESSION['user']) ? $_SESSION['user'] : ""); 

//Prepare 
if($stmt = mysqli_prepare($con, "SELECT * FROM forum_topics WHERE `category_id`=? AND `id`=? LIMIT 1")) { 

    mysqli_stmt_bind_param($stmt, "ii", $cid, $tid); 
    mysqli_stmt_execute($stmt); 

     /* fetch value */ 
    mysqli_stmt_fetch($stmt); 

    if (!$stmt) { 
     throw new Exception($con->error); 
    } 
} 
$stmt->store_result(); 
$numrows = $stmt->num_rows; 
if($numrows == 1){ 
    echo "<table width='100%'>"; 
    if ($_SESSION['user']) { 
     echo "<tr><td colspan='2'><input type='submit' value='Add Reply' onclick=\"window.location = 
    'forum_post_reply.php?cid=".$cid."$tid=".$tid."'\"> <hr />"; 
    } else { 
     echo "<tr><td colspan='2'><p>Please log in to add your reply</p><hr /></td></tr>"; 
    } 

/*} 
catch (Exception $e) 
{ 
    //print_r($e); 
    echo "There has been an error with the $stmt block."; 
} 

print_r($stmt); 

try {*/ 
    foreach($stmt as $row) { 

     //Prepared SELECT stmt to get forum posts 
     if($stmt2 = mysqli_prepare($con, "SELECT * FROM forum_posts WHERE `category_id`=? AND `topic_id`=?")) { 
     //var_dump($stmt2); 

      mysqli_stmt_bind_param($stmt2, "ii", $cid, $tid); 
      mysqli_stmt_execute($stmt2); 
     } 
      while (mysqli_stmt_fetch($stmt)) { 
       echo "<tr><td valign='top' style='border: 1px solid #000000;'> 
       <div style='min-height: 125px;'>".$row['topic_title']."<br /> 
       by ".$row2['post_creator']." - " .$row2['post_date']. "<hr />" . $row2['post_content'] ."</div></td> 
       <td width='200' valign='top' align='center' style='border: 1px solid #000000;'>User Info Here!</td></tr> 
       <tr><td colspan='2'><hr /></td></tr>"; 
      } 
    }  
    } else { 
     echo "<p>This topic does not exist.</p>"; 
     } 
} 
catch (Exception $e) 
{ 
    //print_r($e); 
    echo "There has been an error with the foreach $stmt2 block."; 
} 
print_r($stmt2); 

Answer 1

如果你想跟蹤所有在你的代碼中發生的錯誤，你可以使用嘗試，趕上語句來處理他們的正確方法。在try塊內部，您可以放置​​所有代碼或其中的一部分。如果失敗了，將會產生例外。

該異常使用下面的語句也可以自己拋出：

throw new Exception("I think something's failing"); 

只是後嘗試塊，你必須聲明捕捉塊，將處理例外，使它提供給你。在以下示例中，我將打印錯誤消息。

<?php 

try { 
    /* all your code */ 
} catch (Exception $e) { 
    /* if something fails inside try block will be catched here */ 
    print $e->getMessage(); 
} 

你可以找到更多關於這個在這裏： http://php.net/manual/en/language.exceptions.php