Question

使用AngularJS，假设数组decodeURIComponent具有以下结构的项目：

$scope.data.children

是否有一种简单的方法可以引用所有{ "sku" : "<sku>", "selected" : "<boolean>" }个孩子，理想情况下会在selected = true中表示？

所以，例如，如果

$scope.data.components

然后

$scope.data.children = [
    {"sku" : "A","selected" : "true"},
    {"sku" : "B","selected" : "false"},
    {"sku" : "C","selected" : "true"},
    {"sku" : "D","selected" : "false"}
]

如果$scope.data.components = [ {"sku" : "A","selected" : "true"}, {"sku" : "C","selected" : "true"}, ]更新为$scope.data.children

然后[{"sku" : "D","selected" : "true"}]，

$scope.data.components =

Answer 1

$scope.$watch("data.children", setSelected);

var setSelected = function () {
    var selected = []

    for (var i = 0; i < $scope.data.children; i++) {
        if ($scope.data.children[i].selected == "true") {
            selected.push($scope.data.children[i];
        }
    }

    $scope.data.components = selected;
}

Answer 2

尝试在项目中使用underscore.js。它让你的生活更容易500％！

并且您的item.selected值存储为字符串，而不是布尔值。

http://underscorejs.org/#filter

提供相同数据的不同绑定

2 个答案: