0

我的 angularjs 表单中有一个错误。当来自 angularjs 表单的输入有效时,我想更改元素的类,如果无效则将其更改回来。

我得到了这个代码:

<li data-ng-class="{ success: userFormRegistration.email.$valid, danger: userFormRegistration.email.$invalid }" class="danger">valid email</li>

现在这是电子邮件的输入:

<input type="email" name="email" data-ng-model="userFormRegistration.email" class="form-control ng-pristine ng-invalid ng-invalid-required ng-invalid-email" placeholder="Email" required="">

一开始添加的课程是danger,但是当我输入有效的电子邮件时,它正在删除课程但没有添加success课程..

这是angularjs代码:

// app.js
            // create angular app
            var FormApp = angular.module('FormApp', []); ;


            // create angular controller
            /*FormApp.controller('registrationFormController', function ($scope, $http) {*/
            function registrationFormController($scope, $http) {
                // create a blank object to hold our form information
                // $scope will allow this to pass between controller and view
                $scope.userFormRegistration = {};

                // function to submit the form after all validation has occurred            
                $scope.submitForm = function (form) {
                    // check to make sure the form is completely valid
                    if (form.$valid) {
                        $scope.userFormRegistration = { actionname: "registration", email: $scope.userFormRegistration.email, password: $scope.userFormRegistration.password }
                        // start ajax call
                        $http({
                            method: 'POST',
                            url: 'Registration.php',
                            data: $.param($scope.userFormRegistration),  // pass in data as strings
                            headers: { 'Content-Type': 'application/x-www-form-urlencoded'}  // set the headers so angular passing info as form data (not request payload)
                        })
                            .success(function (data) {
                                console.log(data);
                                if (!data.success) {
                                    // if not successful, bind errors to error variables
                                    $scope.errorRegistrationEmail = data.errors.email2;
                                    $scope.errorRegistrationPassword = data.errors.password;
                                    $scope.errorRegistrationActionname = data.errors.actionname;
                                    $scope.registrationMessage = data.message;
                                }
                                else {
                                    // if successful, bind success message to message
                                    $scope.message = data.message;
                                }
                            });
                        //end ajax call
                    }
                    else {
                        if (!$scope.userFormRegistration.email.$valid) {
                            var error = { message: "please enter valid email" };
                            $scope.errorRegistrationMessage = error.message;
                        }
                        else if (!$scope.userFormRegistration.password.$valid) {
                            var error = { message: "please enter valid password " };
                            $scope.errorRegistrationMessage = error.message;
                        }
                    }
                };
            };

请你的建议。

4

1 回答 1

1

$valid 和 $invalid 属性来自表单而不是来自输入的值。您需要一个带有名称的表单元素,然后在输入中也需要一个名称,然后使用 formName.inputName.$valid 或 $invalid。

于 2015-11-24T06:08:48.627 回答