在WHERE子句中使用變量時，PHP MySQL Update無法正常工作

Question

我已經在這里和其他站點檢查了數十個線程，但無法弄清楚為什么我的代碼無法正常工作。 我正在嘗試使用PHP通過變量來標識位置更新MySQL。 如果將變量交換為數字，則我擁有的代碼將起作用，並且該變量在腳本中的其他所有地方都可以工作。 只是這一行沒有。

有問題的行是：

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

我也嘗試了以下方法：

$change = mysqli_query("UPDATE reg_info SET fname='$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;

根據我在其他線程上看到的內容，至少其中一個應該對我有用。

有人能指出我正確的方向嗎？

如果有幫助，則PHP代碼的整個字符串為：

    <?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

我發現了問題！ 無論何時提交表單，新的POST數據都沒有分配給html id =“ id”的任何內容，這些內容已傳遞到PHP代碼中以創建$ id變量。

由於格式中沒有任何內容，因此$ id為null，因此即使查詢和連接完全有效，查詢也不會更新數據庫。

感謝所有發表評論和建議的人，我真的很感激。

Answer 2

由於查詢本身是有效的，因此我只能猜測數據是由某種原因引起的。 請嘗試以下操作，該操作將轉義查詢中將使用的每個值：

$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'";