Question

我想使用session（使用PHP + ajax）创建登录表单，我使用json从控制器发送用户名，但它不起作用。我不知道什么是错的，请帮助
这是控制器中的功能：

public function actionLogin()
    {
        $username = isset($_POST['username'])?$_POST['username']:null;
        $password = isset($_POST['password'])?sha1($_POST['password']):null;

        $json = new JsonHelper();
        $result = array();

            if($username && $password !=''){
                $checkLogin = Administrator::model()->findByAttributes(
                            array('username'=>$username, 'password'=>$password));
                $checkUser = Administrator::model()->findByAttributes(
                            array('username'=>$username));          
                $checkPass = Administrator::model()->findByAttributes(
                            array('password'=>$password));

                $login = count($checkLogin);
                $user = count($checkUser);
                $pass= count($checkPass);

                if($login==1)
                {
                    $result['status'] = 'success';
                    $result['username'] = $username;
                    $json->addData('ajax', $result);
                }
                elseif($user == 1 && $pass == 0)
                {
                    $result['status'] = 'wrongPass';
                    $json->addData('ajax', $result);
                }
                elseif($user == 0 && $pass == 1)
                {
                    $result['status'] = 'wrongUser';
                    $json->addData('ajax', $result);
                }
            }

        echo json_encode($json->getJson());
    }

这是form_login.js文件：

function login(){
    var form = $('#login-form');
    var formId = form.attr('id');
    var action = form.attr('data-action');
    var method = form.attr('data-method');
    var formData = serializer(form);  //don't mind this function
    $.ajax(
    {
        url: action,
        cache: false,
        processData: false,
        contentType: false,
        type: method,
        data: formData,
        success: function(json)
        {
            // AJAX SUCCESS
            var json = JSON.parse(result);
            if(json['result']['ajax']['status']=='success')
            {
                //$_SESSION['username'] =json['username'];
                window.location = baseUrl + "/appsterize/dashboard/index";
            }
            else if(json['result']['ajax']['status']=='wrongPass')
            {
                // Password wrong
                alert("The password you entered is incorrect.");
            }
            else if(json['result']['ajax']['status']=='wrongUser')
            {
                // Username isn't exist
                alert("Username isn't exist");
            }
        },
        error: function(xhr, status, error)
        {
            // AJAX ERROR
            var string = "<strong>Error!</strong> " + xhr['responseText'];
            $(alertError).attr('data-text', string);
            $(alertError).click();
        },
    });
}

某些错误是'未捕获的ReferenceError：未定义alertError'

Answer 1

有一个id =＆＃39; alertError＆＃39 ;?的元素这可能是解决方案：

$("#alertError").attr('data-text', string);
...

Answer 2

基本上，@ serakfalcon上面说的是：

...
    error: function(xhr, status, error)
            {
                // AJAX ERROR
                var errorMsg = "<strong>Error!</strong> " + xhr['responseText'];
                alert(errorMsg);
            },
...

php ajax登录表单

2 个答案: