如何在 angularjs ajax 成功后更改 ui-view

Question

现在我正在尝试触发 ajax 进行搜索。在 $http( may be $resourse just ajax) 成功之后，我需要更新 index.page 中的 ui-view 以显示结果。这是我的代码：

angular.module('ecom').config(['$stateProvider','$urlRouterProvider',function($stateProvider,$urlRouterProvider){
    //for any unmatched url,send to /dashboard
    $urlRouterProvider.otherwise("/dashboard");
    $stateProvider
     .state('home',
             {
            url:'/home',
            templateUrl:'submitter/home/home.html'
             }) 
      .state('expenses',
              {
            url:'/expenses',
            templateUrl:'submitter/expenses/expense-list.html',
            controller:'SubmitterExpensesCtrl'
              })
        .state('receipts',
              {
            url:'/receipts',
            templateUrl:'submitter/receipts/receipts.html'
              })
        .state('settings',
             {
            url:'/settings',
            templateUrl:'settings/settings.html',
            controller:'SettingCtrl'
             })
        .state('logout',
              {
            url:'/logout',
            templateUrl:'logout/logout.html',
            controller:'LogoutCtrl'
              })
        .state('searchResult',
                {
            url:'/searchresult',
            templateUrl:'ExpenseResult/searchresult.html',
            controller:'resultCtrl'     
                })
}]);
This is index.page :
<nav><search></search></nav> // When click search, it will fire a ajax call, I will display the result in searchresult.html
<div class="body-container">
        <div class="col-md-1"></div>
        <div class="container app-wrapper col-md-12" ui-view></div> //the main content will display here.
        <div class="col-md-1"></div>
    </div>
<footer></footer>

如何将 url 更改为 /result 并在索引页中显示 searchresult.html？我可以在 ajax seccuee 方法中尝试 $location.path('/searchresult') 吗？;

Answer 1

将 $state 注入您的控制器并在 $http/$resource 调用返回承诺后调用 $state.go(statename)。

所以在你的情况下，像...

$http.post("/ajax")
.then(function(data, status, headers, config) {
    $state.go("searchResult");
})

如何在 angularjs ajax 成功后更改 ui-view

How to change the ui-view after ajax success in angularjs

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