Question

我不确定问题出在哪里。所以我将在这里给出一个简短的描述：我试图用PHP从头开始构建一个Web应用程序。有了这个，我希望页面几乎每个表单都是ajax。我创建了2个类（用户和客户端）。代码应该处理一个表单，通过另一个PHP文件提交给数据库。

class User{
  public static function edit ($id, $user, $pass, $is_admin, $client){
      global $db;
      $id = mysqli_real_escape_string($db, $id);
      $user = mysqli_real_escape_string($db, $user);
      $pass = mysqli_real_escape_string($db, $pass);
      $is_admin = mysqli_real_escape_string($db, $is_admin);
      $client = mysqli_real_escape_string($db, $client);
      if ($_SESSION['user']['is_admin'] == true){
        if ($client == '')
          $client = 'NULL';
        $sql = "UPDATE users SET users.username = '{$user}', users.password = '{$pass}', users.is_admin = {$is_admin}, ";
        $sql .= "users.client = '{$client}' WHERE users.id = {$id};";
        $result = $db->query($sql);
        if ($db->affected_rows > 0) {
          $_SESSION['success'] = array( 1 => "All changes with the User has been saved." );
        } else {
          $_SESSION['errors'] = array( 1 => "No changes are made. Please contact the Administration for further assistance if needed." );
        }
      }
    }
}
class Client{
  public static function edit ($id, $name, $logo, $street, $city, $state, $zip, $phone, $email, $active){
    global $db;
    $sql = "UPDATE clients SET name=?, logo=?, street=?, city=?, state=?, zip_code=?, phone=?, email=?, active=? WHERE id=?;";
    if ($stmt = $db->prepare($sql) and $_SESSION['user']['is_admin'] == true) {
      $stmt->bind_param('s', $name);
      $stmt->bind_param('s', $logo);
      $stmt->bind_param('s', $street);
      $stmt->bind_param('s', $city);
      $stmt->bind_param('d', $state);
      $stmt->bind_param('s', $zip);
      $stmt->bind_param('s', $phone);
      $stmt->bind_param('s', $email);
      $stmt->bind_param('d', $active);
      $stmt->bind_param('d', $id);
      $stmt->excute();
      if ($db->affected_rows > 0) {
        $_SESSION['success'] = array( 1 => "All changes with the Client has been saved." );
      } else {
        $_SESSION['errors'] = array( 1 => $db->error );
      }
    } else $_SESSION['errors'] = array( 1 => $db->error );
  }
}

处理ajax表单的javascript代码在这里：

function editUser() {
  $.ajax({
    type: 'POST',
    url: 'editUser.php',
    data: $('#editUserForm').serialize(),
    cache: false,
    dataType:'json'
  }).always(function(){
    renderUsers();
    $('body').removeClass('modal-open');
    $('.modal-backdrop').remove();
  });
  return false;
}

function editClient() {
  $.ajax({
    type: 'POST',
    url: 'editClient.php',
    data: $('#editClientForm').serialize(),
    cache: false,
    dataType:'json',
    error: function(data){
      console.log(data);
    }
  }).always(function(){
    renderUsers();
    $('body').removeClass('modal-open');
    $('.modal-backdrop').remove();
  });;
  return false;
}

用于执行用户aka editUser.php文件的ajax提交的PHP代码：

<?php
  require_once('class_lib.php');
  if($_SESSION['user']['is_admin'] == true){
    if(isset($_POST['id'])){
      User::edit($_POST['id'], $_POST['username'], $_POST['password'], $_POST['is_admin'], $_POST['client']);
    }
  }
?>

这是editClient.php

<?php
  require_once('class_lib.php');
  if($_SESSION['user']['is_admin'] == true){
    if(isset($_POST['id'])){
      Client::edit($_POST['id'], $_POST['name'], $_POST['logo'], $_POST['street'], $_POST['city'], $_POST['state'], $_POST['zip_code'], $_POST['phone'], $_POST['email'], $_POST['active']);
    }
    if($_POST['process'] == 'newClient'){
      Client::add();
    }
  }
?>

代码几乎完全相同（我复制并粘贴），我首先创建了用户功能并且它工作了所以我继续使用客户端，这就出现了错误。用户功能仍然完美，但客户端不是。我创建了2个虚拟客户端来测试编辑。第一个客户端打印没有错误（$db->error），但第二个虚拟客户端输出此You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' clients.phone = '',clients.email = '', clients.active = 1 WHERE clients.id = 2' at line 1 我现在在这里呆了一天，但仍然不知道为什么它不起作用。更多信息：我测试了类功能，它工作正常。测试发送表单正常方式，它工作。是因为ajax没有发送正确的数据吗？为什么第一个客户没有显示错误？提前谢谢。

编辑：使用prepare语句而不是escape_string更改查询。

没有显示错误，但声明未执行。

Ajax不工作还是PHP查询？

0 个答案: