Question

我在这里和其他网站上检查了几十个线程，我无法弄清楚为什么我的代码无效。我正在尝试使用PHP来使用变量来更新MySQL以识别WHERE。如果我将变量换成数字，那么我可以使用的代码，并且该变量适用于我的脚本中的其他位置。只有这一行没有。

有问题的一行是：

$change = "UPDATE reg_info SET fname='$fname', lname='$lname', email='$email', explevel='$experience', addinfo='$additional', event='$regEvent' where id='$id'";

我也尝试了以下内容：

$change = mysqli_query("UPDATE reg_info SET fname='$fname', lname='$lname', email='$email', explevel='$experience', addinfo='$additional', event='$regEvent' where id='$id'");

$change = "UPDATE reg_info SET fname='$fname', lname='$lname', email='$email', explevel='$experience', addinfo='$additional', event='$regEvent' where id=".$id;

$change = 'UPDATE reg_info SET fname="'.$fname.'", lname="'.$lname.'", email="'.$email.'", explevel="'.$experience.'", addinfo="'.$additional.'", event="'.$regEvent.'" where id='.$id;

从我在其他主题上看到的情况来看，其中至少有一个对我有用。

有人能指出我正确的方向吗？

如果它有助于整个PHP代码串：

    <?php
$fnameErr = $lnameErr = $emailErr = $experienceErr = $regEventErr = "";
$fname = $lname = $email = $experience = $regEvent = "";
$id = $_GET["id"];
$errors = "yes";
$servername = "localhost";
$username = "root";
$password = "5tTtFzaz6dIO";
$database = "project2db";
$conn = new mysqli($servername, $username, $password, $database);
$query = mysqli_query($conn, "SELECT * FROM reg_info where id=".$id);
$row = mysqli_fetch_array($query, MYSQLI_NUM);

    if ($_SERVER["REQUEST_METHOD"] == "POST") {
     if (empty($_POST["fname"])) {
       $fnameErr = "First name is required";
       $errors = "yes";
     } else {
       $fname = test_input($_POST["fname"]);
       if (!preg_match("/^[a-zA-Z ]*$/",$fname)) {
         $fnameErr = "Only letters and white space allowed";
         $errors = "yes";
       }
       else {
        $errors = "no";
       }
     }

     if (empty($_POST["lname"])) {
       $lnameErr = "Last name is required";
       $errors = "yes";
     } else {
       $lname = test_input($_POST["lname"]);
       if (!preg_match("/^[a-zA-Z ]*$/",$lname)) {
        $lnameErr = "Only letters and white space allowed";
        $errors = "yes";
       }
       else {
        $errors = "no";
       }
     }

     if (empty($_POST["email"])) {
       $emailErr = "Email is required";
       $errors = "yes";
     } else {
       $email = test_input($_POST["email"]);
       if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
        $emailErr = "Invalid email address";
        $errors = "yes";
       }
       else {
        $errors = "no";
       }
     }

     if (empty($_POST["experience"])) {
       $experienceErr = "Experience level is required";
       $errors = "yes";
     } else {
       $experience = test_input($_POST["experience"]);
       $errors = "no";
     }

     if (empty($_POST["additional"])) {
       $regEvent = "";
     } else {
       $additional = test_input($_POST["additional"]);
     }

     if (empty($_POST["regEvent"])) {
       $regEventErr = "Event is required";
       $errors = "yes";
     } else {
       $regEvent = test_input($_POST["regEvent"]);
       $errors = "no";
     }

     if($errors == "no") {
        $change = 'UPDATE reg_info SET fname="'.$fname.'", lname="'.$lname.'", email="'.$email.'", explevel="'.$experience.'", addinfo="'.$additional.'", event="'.$regEvent.'" where id='.$id;
        $result=$conn->query($change);

        if ($result) {
             echo '<script language="javascript">';
             echo 'alert("New record created successfully.")';
             echo '</script>';
             header('Location: regtable.php');
        } else {
             echo '<script language="javascript">';
             echo 'alert("Error. New record not created.")';
             echo '</script>';
             header('Location: regtable.php');
        }
     }

    }

    function test_input($data) {
    $data = trim($data);
    $data = stripslashes($data);
    $data = htmlspecialchars($data);
    return $data;
    }
    ?>

Answer 1

我想出了这个问题！无论何时提交表单，新的POST数据都没有分配给html id =“id”的任何内容，该内容已传递到PHP代码中以创建$ id变量。

由于表单中没有任何内容，$ id为null，因此查询未更新数据库，即使查询和连接完全有效。

感谢所有发表意见和建议的人，我真的很感激。

Answer 2

由于查询本身是有效的，我只能猜测数据导致问题。尝试以下操作，它将转义将在查询中使用的每个值：

$fname = mysqli_real_escape_string( $conn, $fname );
$lname = mysqli_real_escape_string( $conn, $lname );
$email = mysqli_real_escape_string( $conn, $email );
$experience = mysqli_real_escape_string( $conn, $experience );
$additional = mysqli_real_escape_string( $conn, $additional );
$regEvent = mysqli_real_escape_string( $conn, $regEvent );
$id = mysqli_real_escape_string( $conn, $id );

$change = "UPDATE reg_info SET fname='$fname', lname='$lname', email='$email', explevel='$experience', addinfo='$additional', event='$regEvent' where id='$id'";

在WHERE子句中使用变量时，PHP MySQL Update无法正常工作

2 个答案: