javascript - 处理 Rally 数据存储中的空异常

Question

我陷入了如何使用结果中 null 中的嵌套对象来处理 null 异常的问题。我将这些值返回到网格。

示例是迭代。我需要获取迭代对象的 name 属性。

    _onDataLoaded: function(store, data){
                var stories = [];

                _.each(data, function(story) {
                            var s  = {
                                FormattedID: story.get('FormattedID'),
                                Name: story.get('Name'),
                                Project: story.get('Project').Name,
                                ScheduleState: story.get('ScheduleState'),
                                TaskRemainingTotal: story.get('TaskRemainingTotal'),
                                Blocked: story.get('Blocked'),
                                Iteration: story.get('Iteration').Name,
                                Release: story.get('Release').Name,
                                Predecessor: []
                            };

                            var predecessors = story.getCollection('Predecessors', {

                     fetch:['Rank','FormattedID','Name','Predecessors','Successors','Project','ScheduleState','FormattedID','Blocked','Iteration','Release','TaskRemainingTotal']});
                            predecessors.load({
                                callback: function(records, operation, success){
                                    _.each(records, function(predecessors){
                                        s.Predecessor.push({Predecessorformatid: predecessors.get('FormattedID'),
                                                        PredecessorsBlocked: predecessors.get('Blocked'),
                                                        Predecessorteam: predecessors.get('Project').Name,
                                                        PredecessorsState: predecessors.get('ScheduleState'),
                                                        PredecessorsTaskActualTotal: predecessors.get('TaskRemainingTotal'),
                                                        PredecessorsIteration: predecessors.get('Iteration').Name,
                                                        PredecessorsRelease: predecessors.get('Release').Name,
                                                        Predecessorname: predecessors.get('Name')

                                                    });
                                    }, this);

                                    this._createGrid(stories);
                                },
                                scope: this
                            });
                            stories.push(s);
                }, this);

Answer 1

如果有可能为空，我通常会这样做:

var iterationName = story.get('Iteration') && story.get('Iteration').Name;

这是有效的，因为如果 && 运算符的左侧为假(即 null)，那么该运算符将短路并且不会评估右侧。如果左侧为真，那么它将评估并返回右侧。