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 = $q.defer();
ourBootstrapService.init().then(function(initialised) {
deferred.resolve(initialised);
});
return deferred.promise;
}
}
})
服务
function ourBootstrapService() {
function init(){
// this is what we need
}
}
当使用$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
}
}
