目前我有一个Angular.js页面,允许搜索和显示结果。用户单击搜索结果,然后单击返回按钮。我想再次显示搜索结果,但我不知道如何触发搜索执行。细节如下:

My Angular.js page is a search page, with a search field and a search button. The user can manually type in a query and press a button and and ajax query is fired and the results are displayed. I update the URL with the search term. That all works fine. User clicks on a result of the search and is taken to a different page - that works fine too. User clicks back button, and goes back to my angular search page, and the correct URL is displayed, including the search term. All works fine. I have bound the search field value to the search term in the URL, so it contains the expected search term. All works fine.

我如何让搜索功能再次执行,而不需要用户按下“搜索按钮”?如果是jquery,那么我会在documentready函数中执行一个函数。我找不到Angular.js的等价物。


当前回答

试试这个?

$scope.$on('$viewContentLoaded', function() {
    //call it here
});

其他回答

如果你有一个特定于该页面的控制器,还有另一种选择:

(function(){
    //code to run
}());

当使用$routeProvider时,你可以解析.state并引导你的服务。也就是说,你要加载控制器和视图,只有在解析你的服务之后:

ui-routes

 .state('nn', {
        url: "/nn",
        templateUrl: "views/home/n.html",
        controller: 'nnCtrl',
        resolve: {
          initialised: function (ourBootstrapService, $q) {

            var deferred = $q.defer();

            ourBootstrapService.init().then(function(initialised) {
              deferred.resolve(initialised);
            });
            return deferred.promise;
          }
        }
      })

服务

function ourBootstrapService() {

 function init(){ 
    // this is what we need
 }
}

我遇到了同样的问题,只有这个解决方案对我有效(它在加载完整的DOM后运行一个函数)。我使用这个滚动锚后页面已加载:

angular.element(window.document.body).ready(function () {

                        // Your function that runs after all DOM is loaded

                    });

如果你想观察viewContentLoaded DOM对象的变化,然后做一些事情,你可以这样做。使用美元的范围。$on也可以,但不同的是,当你的路由是单页模式时。

 $scope.$watch('$viewContentLoaded', function(){
    // do something
 });

你可以使用angular的$window对象:

$window.onload = function(e) {
  //your magic here
}