Question

当我按下更改按钮时，我想从我的输入字段中获取新值并覆盖数据库中的当前值。但出现错误500

<?php session_start();
if (isset($_POST['change-btn'])) {


require 'datalink.php'; // this links the website to the database

$user_name = $_POST['user_Name']; //this is username

$fName = $_POST['Fname'];
$lName = $_POST['Lname']

$fName = ucfirst($fName);
$lName = ucfirst($lName);
$user_name = ucfirst($user_name)


$sql = "SELECT User_name FROM users WHERE User_name=?";
$stmt = mysqli_stmt_init($conn);
  if (!mysqli_stmt_prepare($stmt, $sql)) {
    header("location: ../signup.php?error=sqlerror");
    exit();
  }
  else {
    mysqli_stmt_bind_param($stmt , "s", $user_name);
    mysqli_stmt_execute($stmt);
    mysqli_stmt_store_result($stmt);
    $resultCheck = mysqli_stmt_num_rows($stmt);
        if ($resultCheck > 0) {
      header("location: ../signup.php?error=usertaken&mail=".$email);
      exit();
        }
        else {

        $sql = "UPDATE users (User_name, Fname, Sname) VALUES (?, ?, ?)";
        $stmt = mysqli_stmt_init($conn);
        if (!mysqli_stmt_prepare($stmt, $sql)) {
          header("location: ../signup.php?error=sqlerror");
          exit();
        }

        // this if the user successfully signs up

        else {
          mysqli_stmt_bind_param($stmt , "sss", $user_name, $fName ,$lName);
          mysqli_stmt_execute($stmt);
          header("location: ../accounts.php?update=success");
          exit();
        }
      }
  }

}
mysqli_stmt_close($stmt);
mysqli_close($conn);
}

// this stops the user accessing the signupinc.php file.

else {
  header("location: ../accounts.php");
  exit();
}

我尽一切努力使代码工作均失败并出现错误500，我要做的就是让网站更改名字，姓氏和用户名

试图让我的网站更新我的数据库

0 个答案: