遺漏的類型錯誤：無法調用未定義的方法「推」 - Angularjs

Question

可有人請看看這個代碼，並幫助找出原因，我不斷收到此錯誤。我是angularjs的新手（特別是使用服務/工廠功能）。當函數和局部變量在控制器（而不是$ service）中被定義爲$ scope變量時，它工作得很好，但是由於這個函數返回的數據是多個其他控制器所需要的，我想要減少代碼冗餘。

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) 
      { 

      } 

     }); 
    } 

用於訪問來自服務變量的位置的附加功能是

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

，我使用綁定數據如下

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

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

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

控制器和在視圖中我有一個調用該控制器的init函數，並將值附加到select如下

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

謝謝。

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

當你使用「這個」，你必須要小心，因爲在不同的範圍可能指向不同的對象，你可以簡單地刪除這個放在一個封閉的位置。服務是單身，它會記住。