-1

问题是这样的:

$scope.model1 = [1,2,3,4,5,6,7];
$scope.model2 =  $scope.model1;

$scope.model2.splice(2, 1);

 <pre>{{model1}}</pre>
 <pre>{{model2}}</pre>

返回:

[1,2,4,5,6,7]
[1,2,4,5,6,7]

需要:

 [1,2,3,4,5,6,7]
 [1,2,4,5,6,7]

为什么会这样?

更新: 解决方案:到处使用angular.copy($scope.val) 我的代码很糟糕:

$scope.$watch('checkedImages', function (newVal) {
        if (newVal !== undefined && newVal[0] !== undefined) {
            if ($scope.model.curSupplier === undefined) {
                $scope.model.curSupplier = newVal[0].supplier_id;
                $scope.model.curCheckedImages = newVal;
            }
            $scope.supplier = newVal[0].supplier_id;
        }
    });

对比

$scope.$watch('checkedImages', function (newVal) {
    if (newVal !== undefined && newVal[0] !== undefined) {
        if ($scope.model.curSupplier === undefined) {
            $scope.model.curSupplier = angular.copy(newVal[0].supplier_id);
            $scope.model.curCheckedImages =  angular.copy(newVal);
        }
        $scope.supplier =  angular.copy(newVal[0].supplier_id);
    }
});
4

2 回答 2

2

通过将一个列表分配给另一个列表,您只是在复制参考。

这意味着两个模型实际上都引用了同一个列表。因此,改变其中任何一个都会影响另一个。

而是试试这个:

$scope.model2 =  angular.copy($scope.model1);

更新:

$scope.$watch('checkedImages', function (newVal) {
    if (newVal !== undefined && newVal[0] !== undefined) {
        var newObj = angular.copy(newVal);
        if ($scope.model.curSupplier === undefined) {
            $scope.model.curSupplier = newObj[0].supplier_id;
            $scope.model.curCheckedImages =  newObj;
        }
        $scope.supplier =  newObj[0].supplier_id;
    }
});
于 2013-11-21T06:14:23.480 回答
1

这取决于您想要数组中数据的深拷贝还是浅拷贝。对于深拷贝,您可以使用angular.copy,对于浅层深拷贝,您可以使用array.slice()

于 2013-11-21T06:38:12.330 回答