Question

此代码已经工作了一段时间，但突然之间没有任何改变。代码的第一部分是检查用户的凭据是否正确。如果是，则将表单中输入的数据提交给mysql，并发送电子邮件通知。这是代码：

 <?php
    if(isset($_POST['insert'])){
    $hostname = "localhost";
    $username = "root";
    $password = "";
    $databaseName = "change_management";
    $connect = mysqli_connect($hostname, $username, $password, $databaseName);

    $user = $_POST['user'];
    $passcode = $_POST['passcode'];

    $sql = "SELECT * FROM admin WHERE username = '" . $user . "' and passcode ='".md5($passcode)."'";
    $result = mysqli_query($connect,$sql);
    $row = mysqli_fetch_array($result,MYSQLI_ASSOC);
    $count = mysqli_num_rows($result);
    if($count ==0) {
    echo "
<font color=red><strong>Authentication failed - Invalid credentials</font></strong><br><br>";
    }
else {
    $change_id = $_POST['change_id'];
    $approval_disposition = $_POST['approval_disposition'];
    $approval_reason = $_POST['approval_reason'];
    $approval_impact = $_POST['approval_impact'];
    $user = $_POST['user'];
    $approval_date = date('Y/m/d');
    $approval_expiry_date = date('Y/m/d', strtotime('+2 weeks'));

    $query = "UPDATE change_request_tbl SET approval_disposition='$approval_disposition', approval_reason='$approval_reason', approval_impact='$approval_impact', approval_name='$user', approval_date='$approval_date', approval_expiry_date='$approval_expiry_date' WHERE change_id='$change_id'";
    $query = "SELECT `change_requestor` FROM `change_request_tbl` WHERE `change_id` = $change_id";
    $result1 = mysqli_query($connect,$query);
    while ($row = mysqli_fetch_array($result1))

    $to = "******, ****, ****, *****, $row[change_requestor]";
    $subject = "New Change Acceptance Board (CAB) Approval";
    $message = " 
    <html>
    <head>
    <style>
    body {font-family: Calibri;}
    body {font-style: italic;}
    .table {
    background-color: white;
    }
    </style>
    <title>Change Acceptance Board Approval</title>
    </head>
    <body>
    <p>A new Change Acceptance Board (CAB) Approval has been submitted:</p>
    <table border=4 bordercolor=black class=table>
    <tr>
    <th>Change ID:</th>
    <th>Disposition:</th>
    <th>Reason for Rejection or Exceptions:</th>
    <th>Impact (By CAB):</th>
    <th>Approved/Rejected by:</th>
    <th>Date Approved:</th>
    <th>Approval Expiry Date*:</th>
    </tr>
    <tr>
    <td>$change_id</td>
    <td>$approval_disposition</td>
    <td>$approval_reason</td>
    <td>$approval_impact</td>
    <td>$user</td>
    <td>$approval_date</td>
    <td>$approval_expiry_date</td>
    </tr>
    </table>
    <p>*This change must be completed by the Expiry Date. If this is not achieved a second CAB Request must be raised referencing this Change ID.</p>
    </body>
    </html>";
    $headers = "MIME-Version:1.0" . "\r\n";
    $headers .= "Content-type:text/html;charset=UTF-8" . "\r\n";
    $headers .='From: <******>' . "\r\n";

    mail($to,$subject,$message,$headers);
    echo "
Your CAB Approval Form has been submitted. You will receive an email confirmation shortly.<br><br>";
    }
    }
    ?>

我没有收到任何错误，但我收到的电子邮件中包含正确的信息。此问题仅与更新查询有关。有什么想法突然停止了吗？

Answer 1

您宣布$query两次，因此请使用以下UPDATE覆盖SELECT查询。

Answer 2

$query = "UPDATE change_request_tbl SET approval_disposition='$approval_disposition', approval_reason='$approval_reason', approval_impact='$approval_impact', approval_name='$user', approval_date='$approval_date', approval_expiry_date='$approval_expiry_date' WHERE change_id='$change_id'";
$result = mysqli_query($connect,$query);

$query = "SELECT `change_requestor` FROM `change_request_tbl` WHERE `change_id` = $change_id";

$result1 = mysqli_query($connect,$query);

在问题中重写了可查询的SQL查询

Answer 3

我相信你的问题在于：

$query = "UPDATE change_request_tbl SET approval_disposition='$approval_disposition', approval_reason='$approval_reason', approval_impact='$approval_impact', approval_name='$user', approval_date='$approval_date', approval_expiry_date='$approval_expiry_date' WHERE change_id='$change_id'";

$query = "SELECT `change_requestor` FROM `change_request_tbl` WHERE `change_id` = $change_id";

$result1 = mysqli_query($connect,$query);

UPDATE $查询被后续的SELECT $查询覆盖，因此它只是选择数据但不会事先更新。

PHP - 代码突然没有更新mysql行

3 个答案: