我只想执行“或”运算并将两个查询的两个结果都放入一个流中。
这是我的单流代码
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]);
},
);
}),
我现在想将两个流而不是单个流馈给同一流生成器。
我尝试了StreamGroup,但是由于小部件重建,它无法正常工作
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;
}
在这里我得到一个错误,至少应该合并一个流。这是因为Observable.merge(streamList)在项目添加到streamList之前被调用。
我只是想将两个查询的两个结果都放入一个流中。
答案 0 :(得分:2)
我迟到了,但要把它放在那里。
您可以像这样在查询中添加whereIn子句:
Firestore.instance.collection("collection_name").where("field",whereIn:["false","true"]).snapshots();
答案 1 :(得分:1)
我不确定您为什么要使用forEach和Observable.just()。
您可以直接合并两个Firestore流,例如:
Observable.merge([stream1,stream2])。pipe(combineStream);
Werre stream1 / 2只是您的Firestore快照。
答案 2 :(得分:1)
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()
);
}
),
);
}
希望这会有所帮助!
答案 3 :(得分:0)
我使用RxDart包将两个流合并,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),
));
});