Firebase：函数的返回值（快照）

Question

我想获取返回值的人，并使用它。

如果我在 .on 内执行 var somthing = function(snapshot){ 会崩溃，说第二个值应该是函数。

我不能放弃orderBy，只能这样做

var results = function(snapshot)

这是我的代码。

我如何获得人们的价值？

peopleDataBase.orderByChild("mCalculateFaceSizeWidth")
                        .startAt(mCalculateFaceSizeWidth())
                        .limitToFirst(1)
                        .on("child_added", function(snapshot) {
                         var person = {
                                "name":snapshot.val().name,
                                "age":snapshot.val().age,
                                "id":snapshot.val().id,
                                "children":snapshot.val().children,
                                "address":snapshot.val().address,
                                "image":snapshot.val().image
                             }

                             if(key != snapshot.key){
                                people += person;
                                key = snapshot.key;

                                 }
                            return people;
                        });

Answer 1

在您的示例中，您将该值作为参数返回给导致错误的 .on 函数。试试这个（更多信息：https://firebase.googleblog.com/2016/01/keeping-our-promises-and-callbacks_76.html）：

var result = peopleDataBase.orderByChild("mCalculateFaceSizeWidth")
                    .startAt(mCalculateFaceSizeWidth())
                    .limitToFirst(1)
                    .on("value").then(function(snapshot) {

                         //This forEach brings at most one node
                         snapshot.forEach(snapshotChild){

                           var person = {
                               "name":snapshotChild.val().name,
                               "age":snapshotChild.val().age,
                               "id":snapshotChild.val().id,
                               "children":snapshotChild.val().children,
                               "address":snapshotChild.val().address,
                               "image":snapshotChild.val().image
                            }

                            if(key != snapshotChild.key){
                                people += person;
                                key = snapshotChild.key;
                            }
                         }
                         return people;
                       });