Question

我正在尝试使用html和php更新数据库中的条目。但是，我一直收到错误消息，说我的sql语法错误。

这是php文件中的代码：

<?php 
$server = "127.0.0.1";
$dbUsername = "root";
$dbPassword = "";
//create connection
$dbconn = new mysqli($server, $dbUsername, $dbPassword, $dbname);

 $email_follow = $_POST['email_follow'];
 $follow = $_POST['follow'];

 $update = mysqli_query($dbconn, "UPDATE CustomerDetails SET Follow Up = '$follow' WHERE Email = '$email_follow'");

if ($dbconn->query($update) === TRUE) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . $dbconn->error;
}
 ?>

这是html表单：

<form action="cust_details_followup.php" method="post">       

Email:
<input type="email" name="email_follow" id="email_follow">

Enter Follow Up Details:
<input type="text" name="follow" id="follow">


<input type="submit" value="Update">

</form>

感谢您的帮助。谢谢！

Answer 1

您的表格属性Follow Up有一个空格，您需要添加`来包装它

所以改变

UPDATE CustomerDetails SET Follow Up = '$follow' WHERE Email = '$email_follow'

到

UPDATE CustomerDetails SET `Follow Up` = '$follow' WHERE Email = '$email_follow'

Answer 2

需要使用点运算符，并且需要使用刻度线而不是单引号

勾选-Follow Up-正确。

$query = "UPDATE CustomerDetails
    SET `Follow Up` = '".$follow."'
    WHERE Email = '".$email_follow."'";

$update = mysqli_query($dbconn, $query);

Answer 3

尝试以下代码

UPDATE CustomerDetails SET `Follow Up` = '".$follow."' WHERE Email = '".$email_follow."'

什么是正确的语法和PHP中的SQL更新？

3 个答案: