Question

我有以下代码，我需要创建一个已更新并保存的单个实例，而不会对先前的范围进行任何更改。

console.log("Trying to save playlist");
                    console.log($rootScope.selectedSounds);
                    $scope.playlistInstance = $rootScope.selectedSounds;
                    var preparedSoundsForSave = [];
                    angular.forEach($scope.playlistInstance, function (selectedSound, key){
                      console.log("Sound to playlist is following:");
                      console.log(selectedSound);
                      var selectedSoundInstance = selectedSound;
                      selectedSoundInstance.state = 0;
                      delete selectedSoundInstance.mediaInstance ;
                      preparedSoundsForSave.push(selectedSoundInstance);
                    });
                    // Create the new object which has been saved
                    $scope.playlist.name = $scope.playlistName;
                    $scope.playlist.tracks = preparedSoundsForSave;
                    console.log($scope.playlist);
                    existingPlaylists.push($scope.playlist);
                    localStorageService.set("playlists", existingPlaylists);
                    $ionicLoading.show({
                        duration: 1000,
                        template: $translate.instant('SAVED')
                    });
                    $scope.playlistName = "";

所以我创建了单个实例，更改了一些值并保存了更改的数组。问题是，更改会自动传递给：

$rootScope.selectedSounds

没有任何理由。

我想问一下，我怎么能正确地在AngularJS中做到这一点？

非常感谢您的任何建议。

Answer 1

在javascript对象中通过引用传递。因此，当您在另一个变量中复制对象时，引用保持不变，因此对任何一个的任何更改都将应用于该对象。

要创建具有不同引用的副本，您可以使用

// Shallow copy
var newObject = jQuery.extend({}, oldObject);

// Deep copy
var newObject = jQuery.extend(true, {}, oldObject);

或者使用loadash或下划线，如

_.extend({},oldObj)

AngularJS：如何避免将引用从一个范围复制到另一个范围？

1 个答案: