我只是想执行“或”运算，并将两个查询的两个结果放入一个流中。

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]);
        },
      );
    }),

现在我想将两个流提供给同一个流生成器，而不是单个流。

我试过 StreamGroup，但自 Widgets 重建后它就没用了

StreamGroup.merge([streamOne, streamTwo]).asBroadcastStream();

我也试过下面的方法

 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;
}

这里我收到错误消息，应该至少有一个流要合并。这是因为在将项目添加到 streamList 之前调用了 Observable.merge(streamList)。

我只是想将两个查询的两个结果合并到一个流中。

最佳答案

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!!!

关于firebase - Flutter 将两个 firestore 流合并为一个流，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/53287717/