javascript - 在 AngularJS ng-submit 中提交表单后如何绑定(bind) View ？

Question

我有一个简单的问题。在 AngularJS 中提交表单后如何绑定(bind) View ？我的 Controller 有这个功能来测试提交是否有效:

var spinnrApp = angular.module('spinnrApp', ['ngAnimate', 'ui.router', 'slick']);

spinnrApp.config(function ($stateProvider, $urlRouterProvider) {
  //
  // Now set up the states
  $stateProvider
    .state('registration', {
      url: "/registration",
      templateUrl: "app/components/registration/registration.php",
      controller: "FormController"
    })
    .state('registration.spinnrapp', { // nested state for the registration form
      url: "/spinnrapp", // url will be nested /registration/artist
      templateUrl: "app/components/registration/partials/registration-profile.php"
    })
    .state('registration.artist', { // nested state for the registration form
      url: "/artist", // url will be nested /registration/artist
      templateUrl: "app/components/registration/partials/registration-artist.php"
    })
    .state('registration.share', { // each nested state will have their own view
      url: "/share", // url will be nested /registration/share
      templateUrl: "app/components/registration/partials/registration-share.php"
    });
    //
    // For any unmatched url, redirect to /
    $urlRouterProvider.otherwise("/registration/spinnrapp");
});
spinnrApp.controller('FormController', ['$scope', '$http', function (scope, http){

  // get list of cities and store it to select
  http.get('cities.json').success(function(data){
    scope.cities = data;
  })

  // we will store all our form data in this object
  scope.formData = {};

  // function to process the form
  scope.processForm = function() {
    $state.go('registration.share');
  };
}]);

在ng-submit通过后如何把alert("awesome!");改成ui-view？我有一个名为 registration-share.php 的 ui-view。

更新:我在提交按钮中调用 processForm，如下所示:

<div class="form-group row">
  <div class="col-xs-3 col-xs-offset-1">
    <button type="submit" class="submit btn">Next Section</button>
  </div>
</div>

我的 formApp 包装器是这样的:

<form id="signup-form" ng-submit="processForm()">
  <!-- our nested state views will be injected here -->
  <div id="form-views" ui-view></div>
</form>

Answer 1

当您使用 ui-router 时，将 $state 注入(inject) Controller 并使用以下内容:

$state.go('newUiRouterStateYouWishToDisplay');

---

这在这一点上不适用，因为你澄清说你正在使用 ui-router，但是如果您使用的是 ngRoute，请将 $location 注入(inject) Controller 并使用以下命令:

$location.path('/newNgRouteYouWishToDisplay');