试着用一个准备好的陈述来抓住

Question

我在这个问题上经历了一段非常艰难的时光，有人建议我学习如何尝试捕捉块，以便更容易地找出错误所在。这是我第一次尝试。

我的@stmt2没有为我的print\r（$stmt2）行定义，这是一个错误。

Notice: Undefined variable: stmt2 

这是我尝试接球的尝试。我有没有做错什么，导致这个错误出现？

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

如果您想跟踪代码中发生的所有错误，您可以使用try，cat语句以正确的方式处理它们。在try块中，你可以把你所有的代码或其中的一部分。如果某个东西失败了，它将生成一个异常。

您也可以使用下一条语句引发该异常：

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

在try block之后，您必须声明将处理异常的catkblock，使其对您可用。在下面的示例中，我将打印错误消息。

<?php

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

您可以在此处找到更多关于此的信息：http://php.net/manual/en/language.exceptions.php