在 AngularFire 中查询集合时不能使用 "not equal" 运算符

Question

我正在尝试从 Firebase 中提取一个与用户无关的集合。到目前为止，我已经成功查询了用户的集合，但我似乎无法使用 != 运算符获取与用户无关的集合。

这是我的代码：

  // Does not work!
  eventCollection: AngularFirestoreCollection<Event> = this.angularFirestore.collection("events", ref =>
    ref.where("uid", "!=", this.firebaseAuth.auth.currentUser.uid)
  );
  eventCollection$: Observable<Event[]> = this.eventCollection.valueChanges();

  // Works!
  myCollection: AngularFirestoreCollection<Event> = this.angularFirestore.collection("events", ref =>
    ref.where("uid", "==", this.firebaseAuth.auth.currentUser.uid)
  );
  myCollection$: Observable<Event[]> = this.myCollection.valueChanges();

抛出的错误是：

ERROR in src/app/event/events.service.ts(35,22): error TS2345: Argument of type '"!"' is not assignable to parameter of type 'WhereFilterOp'.

我试图寻找对应于 != 的逻辑运算符，但似乎找不到。

如有任何帮助，我们将不胜感激！

Answer 1

It looks like 没有 != 运算符。我不熟悉 Firebase，但也许您可以对 < 和 > 进行单独查询并合并结果？或者因为你也想要相同的结果，只做一个不带 where 的查询并自己分开相等和不相等的结果？

Answer 2

根据 Firestore doc：

Queries with a != clause. In this case, you should split the query into a greater-than query and a less-than query. For example, although the query clause where("age", "!=", "30") is not supported, you can get the same result set by combining two queries, one with the clause where("age", "<", "30") and one with the clause where("age", ">", 30).

因此您必须同时比较“<”和“>”条件

eventCollection: AngularFirestoreCollection<Event> = 
  this.angularFirestore.collection("events", ref =>
    ref.where("uid", ">", this.firebaseAuth.auth.currentUser.uid)
       .where("uid", "<", this.firebaseAuth.auth.currentUser.uid)
  );

在 AngularFire 中查询集合时不能使用 "not equal" 运算符

Cannot use the "not equal" operator when querying for a collection in AngularFire

firebase

typescript

angularfire2

angular

google-cloud-firestore