Question

我会用一个简单的例子来解释我的问题。

＆＃13;

Parse.initialize("nlxy5xYYZQ1fLfFkzcyLHOifkie1dOU0ZSxoxw1w", "IRBJO7nyd1vQquhMvnyMd298ZVJ0qWg1AjxBY5nr");
var People = Parse.Object.extend("People");

var app = angular.module('app', []);

app.controller("MyCtrl", ["$scope", "PeopleService", function($scope, PeopleService){
  $scope.people = PeopleService.getPeople();
}]);

app.service("PeopleService", function(){
  var people = null;
  
  return {
    getPeople: function(){
      people = [];
      var queryObject = new Parse.Query(People);
      queryObject.find({
        success: function (results) {
          for (var i = 0; i < results.length; i++) {
            var result = results[i];
            people.push(result.get("name"));
          }
          return people;
        },
        error: function (error) {
          console.error("Error: " + error.code + " " + error.message);
        }
      });
    }
  }
});

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js">
</script>
<script src="https://www.parsecdn.com/js/parse-1.3.2.min.js"></script>
<div ng-app="app">
  <ul ng-controller="MyCtrl">
    <li ng-repeat="person in people">{{person}}</li>
  </ul>
</div>

＆＃13;

现在，当然，当Parse返回数据时，控制器中的范围不会更新。如何使这项服务正常运作？

Answer 1

正如@james所述，使用$ q服务中的promises是您正在寻找的。以下是您的例子。

getPeople: function(){
          var deferred = $q.defer();
          people = [];
          var queryObject = new Parse.Query(People);
          queryObject.find({
            success: function (results) {
              for (var i = 0; i < results.length; i++) {
                var result = results[i];
                people.push(result.get("name"));
              }
              deferred.resolve(people);
            },
            error: function (error) {
              deferred.reject(error);
            }
          });
          return deferred.promise;
        }

这会返回一个你可以像这样使用的承诺

$scope.people = PeopleService.getPeople();

或

$scope.people = PeopleService.getPeople()
.then(function(people) {
    //modify data as necessary
    return people
  }, function(reason) {
    alert('Failed: ' + reason);
  }
);

如何在Angular.js服务中使用Parse JS SDK？

1 个答案: