我只是想执行“OR”运算并将两个查询的结果放入一个流中。
这是我的带有单个流的代码
StreamBuilder(
stream: Firestore.instance
.collection('list')
.where('id', isEqualTo: 'false')
.orderBy('timestamp')
.snapshots(),
builder: (context, snapshot) {
if (!snapshot.hasData)
return Column(
mainAxisAlignment: MainAxisAlignment.center,
children: <Widget>[
Center(
child: CircularProgressIndicator(),
)
],
);
if (snapshot.data.documents.length == 0)
return const Center(
child: Text(
"Not Available",
style: TextStyle(fontSize: 30.0, color: Colors.grey),
),
);
return ListView.builder(
padding: EdgeInsets.all(5.0),
key: Key(randomString(20)),
itemCount: snapshot.data.documents.length,
itemBuilder: (BuildContext context, int index) {
return ListCard(snapshot.data.documents[index]);
},
);
}),
现在我想将两个流提供给同一个流构建器,而不是单个流。
我尝试了 StreamGroup,但自从 Widgets 重建后它就不起作用了
StreamGroup.merge([streamOne, streamTwo]).asBroadcastStream();
我也尝试了以下方法
Stream<List<DocumentSnapshot>> searchResult() {
List<Stream<List<DocumentSnapshot>>> streamList = [];
Firestore.instance
.collection('room-list')
.where('id', isEqualTo: 'false')
.snapshots()
.forEach((snap) {
streamList.add(Observable.just(snap.documents));
});
Firestore.instance
.collection('room-list')
.where('id', isEqualTo: 'pending')
.snapshots()
.forEach((snap) {
streamList.add(Observable.just(snap.documents));
});
var x = Observable.merge(streamList)
.scan<List<DocumentSnapshot>>((acc, curr, i) {
return acc ?? <DocumentSnapshot>[]
..addAll(curr);
});
return x;
}
这里我收到错误,应该至少有一个要合并的流。这是因为 Observable.merge(streamList) 在将项目添加到streamList之前被调用。
我只是想将两个查询的结果放入一个流中。
这应该有效。
//Change your streams here
Stream<List<QuerySnapshot>> getData() {
Stream stream1 = Firestore.instance.collection('list').where('id', isEqualTo: 'false').orderBy('timestamp').snapshots();
Stream stream2 = Firestore.instance.collection('list').where('id', isEqualTo: 'true').orderBy('timestamp').snapshots();
return StreamZip([stream1, stream2]);
}
@override
Widget build(BuildContext context) {
return new Scaffold(
body: StreamBuilder(
stream: getData(),
builder: (BuildContext context, AsyncSnapshot<List<QuerySnapshot>> snapshot1) {
List<QuerySnapshot> querySnapshotData = snapshot1.data.toList();
//copy document snapshots from second stream to first so querySnapshotData[0].documents will have all documents from both query snapshots
querySnapshotData[0].documents.addAll(querySnapshotData[1].documents);
if (querySnapshotData[0].documents.isEmpty)
return Column(
mainAxisAlignment: MainAxisAlignment.center,
children: <Widget>[
Center(
child: CircularProgressIndicator(),
)
],
);
if (querySnapshotData[0].documents.length == 0)
return const Center(
child: Text(
"Not Available",
style: TextStyle(fontSize: 30.0, color: Colors.grey),
),
);
return new ListView(
children: querySnapshotData[0].documents.map((DocumentSnapshot document){
// put your logic here. You will have access to document from both streams as "document" here
return new ListCard(document);
}).toList()
);
}
),
);
}
希望这有帮助!!!
我还尝试合并来自 firestore 的两个流(因为查询不支持 OR)并像这样进行:
import 'package:rxdart/rxdart.dart';
Rx.combineLatest2(
StreamQuerySnapshot1, //a snapshot from firestore
StreamQuerySnapshot2, //another snapshot from firestore
(var stream1, var stream2) {
return [...stream1.docs, ...stream2.docs]; //Concatenated list
}
)
无论哪个流发生变化,这都会发出更改,与我发现的其他解决方案相反,只有当两个流都发生更改时,该支持才会发出。
我使用 RxDart 包来组合两个流,如下所示
final Stream<DocumentSnapshot> user = Firestore.instance
.collection("users")
.document(firebaseUser.uid)
.snapshots();
final Stream<QuerySnapshot> cards =
Firestore.instance.collection("cards").snapshots();
CombineLatestStream.list([user, cards]).listen((data) {
add(LoadedHomeEvent(
data.elementAt(0),
data.elementAt(1),
));
});
我不确定你为什么使用 forEach 和 Observable.just()。
您可以直接合并两个 firestore 流,例如:
Observable.merge([stream1,stream2]).pipe(combineStream);
stream1/2 只是您的 Firestore 快照。
好吧,我迟到了,但还是要把它放在那里。
您可以在查询中添加 whereIn 子句,如下所示:
Firestore.instance.collection("collection_name").where("field",whereIn:["false","true"]).snapshots();
我发现的最好方法是使用
RxDart中的
MergeStream
Stream<QuerySnapshot> searchResult() {
final falseRoomStream = FirebaseFirestore.instance
.collection('room-list')
.where('id', isEqualTo: 'false')
.snapshots();
final pendingRoomStream = FirebaseFirestore.instance
.collection('room-list')
.where('id', isEqualTo: 'pending')
.snapshots();
return MergeStream([falseRoomStream, pendingRoomStream]);
}
这是我在没有任何包的情况下解决这个问题的方法。在我的特殊情况下,同一个文档有可能会被重复,为了避免重复,我使用一组作为中介。
static Stream<List<CleaningEvent>> streamForCleaner(
String cleanerID,
DateTime rawDate,
) {
final date = DateTime.utc(rawDate.year, rawDate.month, rawDate.day);
StreamController<List<CleaningEvent>> controller = StreamController();
Iterable<CleaningEvent> eventsS1 = [], eventsS2 = [];
Set<CleaningEvent> joinedSet = {};
update() {
joinedSet = {};
joinedSet.addAll(eventsS1);
joinedSet.addAll(eventsS2);
controller.add(joinedSet.toList());
}
Stream<QuerySnapshot> stream1 = _collection
.where("cleanersIDs", arrayContains: cleanerID)
.where("reqDate", isEqualTo: date)
.snapshots();
Stream<QuerySnapshot> stream2 = _collection
.where("cleanersIDs", arrayContains: cleanerID)
.where("schDate", isEqualTo: date)
.snapshots();
stream1.forEach((query) {
eventsS1 = query.docs.map(CleaningEvent.fromDoc);
update();
});
stream2.forEach((query) {
eventsS2 = query.docs.map(CleaningEvent.fromDoc);
update();
});
return controller.stream;
}