No, that will not work; he will not have a request.
ref$ FirebaseListObservable ref, . ( ref $ref):
return new FirebaseListObservable(ref, subscriber => {
...
let queried: firebase.database.Query = ref;
if (query.orderByChild) {
queried = queried.orderByChild(query.orderByChild);
} else if (query.orderByKey) {
queried = queried.orderByKey();
} else if (query.orderByPriority) {
queried = queried.orderByPriority();
} else if (query.orderByValue) {
queried = queried.orderByValue();
}
, child_added , :
this.curUserPosts = this.af.database.list('/posts', {
query: {
orderByChild: 'user/id',
equalTo: id
}
});
this.curUserPosts.$ref
.orderByChild('user/id')
.equalTo(id)
.on("child_added", (child) => {
console.log(child);
});
, , on $ref, this.curUserPosts , ; , on.