遗漏的类型错误：无法调用未定义的方法“推” - Angularjs

Question

可有人请看看这个代码，并帮助找出原因，我不断收到此错误。我是angularjs的新手（特别是使用服务/工厂功能）。当函数和局部变量在控制器（而不是$ service）中被定义为$ scope变量时，它工作得很好，但是由于这个函数返回的数据是多个其他控制器所需要的，我想要减少代码冗余。

app.service('SupportService', function() 
{ 
    var locations = []; //local variable to which other elements should be added 



//function to get data and update the local variable (this.locations) 

this.setLocations = function() 
    { 
     var temp_location = {}; 

     $.ajax 
     ({ 
      type: 'get', 
      url: 'myurl', 
      success: function(response) 
      { 

       for(var i = 0; i < response.length; i++) 
       { 
        temp_location = 
        { 
         id: response[i].id, 
         name: response[i].name 

        }; 
        locations.push(temp_location); //this is where the error is 
       } 
      }, 
      error: function(response) 
      { 

      } 

     }); 
    } 

用于访问来自服务变量的位置的附加功能是

this.getLocations = function() 
    { 
     return locations; 
    } 

，我使用绑定数据如下

app.controller('RSVPController', function($scope,SupportService,AuthService) 
{ 

    $scope.init = function() 
    { 
     SupportService.setLocations(); 
    } 

    $scope.locations = SupportService.getLocations(); 
}); 

控制器和在视图中我有一个调用该控制器的init函数，并将值附加到select如下

<select ng-model="location" ng-options="item.name for item in locations track by item.id" required></select> 

谢谢。

Answer 1

app.service('SupportService', function() 
{ 
    this.locations = []; //local variable to which other elements should be added 
    var self = this; 

    enter code here 

//function to get data and update the local variable (this.locations) 

this.setLocations = function() 
    { 
     var temp_location = {}; 

     $.ajax 
     ({ 
      type: 'get', 
      url: 'myurl', 
      success: function(response) 
      { 

       for(var i = 0; i < response.length; i++) 
       { 
        temp_location = 
        { 
         id: response[i].id, 
         name: response[i].name 

        }; 
        self.locations.push(temp_location); //this is where the error is 
       } 
      }, 
      error: function(response) 
      { 

      } 

     }); 
    } 

Answer 2

当你使用“这个”，你必须要小心，因为在不同的范围可能指向不同的对象，你可以简单地删除这个放在一个封闭的位置。服务是单身，它会记住。