相关疑难解决方法(0)

snapshot如此CodePen所示，我能够获取我感兴趣的对象

以下是代码片段：

$scope.post = {};
        var postsRef = new Firebase('https://docs-examples.firebaseio.com/web/saving-data/fireblog/posts');

        $scope.searchPost = function () {
            console.log('searched for author : ' + $scope.post.authorName);

            postsRef.orderByChild('author')
                .equalTo($scope.post.authorName)
                .once('value', function (snapshot) {
                    var val = snapshot.val();
                    console.log("Searched Post is : ");
                    console.log(val);
                    console.log("(From Val) Title is : " + val.title);

                    var exportVal = snapshot.exportVal();
                    console.log("Export Value is : ");
                    console.log(exportVal);
                    console.log("(From Export Val) Title is : " + exportVal.title);
                });
        }

这是我使用的Firebase 数据集。

当我搜索 author: 时gracehop，我得到了正确的快照，但是，我无法访问title …