Question

我想在statefulwidget触发构建之前获取用户ID。如果这样做，则构建将在获得我的ID之前呈现。如果我将其置于setstate中，则我的构建将首先使用空字符串，然后再次使用我的id重新渲染它，但这将导致不必要的行为。

那我该如何解决呢？

String _userid = '';
Future<Null> setUserid() async {
    SharedPreferences pref = await SharedPreferences.getInstance();

    _userid = pref.getString('FB_USER');

  }
  initState() {
    super.initState();
    setUserid();
  }

构建

// Widget build
new Flexible(
          child: new StreamBuilder<QuerySnapshot>(
            stream: Firestore.instance
                .collection('users')
                .document(_userid)
                .collection('rooms')
                .snapshots(),
            builder:
                (BuildContext context, AsyncSnapshot<QuerySnapshot> snapshot) {
              if (!snapshot.hasData) return new Text('Loading...');
              return new ListView(
                children: snapshot.data.documents
                    .map(
                      (DocumentSnapshot document) => new Text('lol'),
                      // )
                      //new OverviewPresentation(presentation: document),
                    )
                    .toList(),
              );
            },
          ),
        ),

Answer 1

您不能，但是可以防止它为null。将StreamBuilder移至initState

也更容易

String _userid = '';
dynamic _data;
Future<Null> setUserid() async {
    SharedPreferences pref = await SharedPreferences.getInstance();

    _userid = pref.getString('FB_USER');

    _data = await Firestore.instance
            .collection('users')
            .document(_userid)
            .collection('rooms')
            .snapshots().first;
    setState(() {});
  }
  initState() {
    super.initState();
    setUserid();
  }

  return new Flexible(
      child: 
          if(_data == null) return new Text('Loading...');
          return new ListView(
            children: _data.documents
                .map(
                  (DocumentSnapshot document) => new Text('lol'),
                  // )
                  //new OverviewPresentation(presentation: document),
                )
                .toList(),
          );
        },
      ),
    ),

Answer 2

您可以使用FutureBuilder

Future<String> setUserid() async {
  SharedPreferences pref = await SharedPreferences.getInstance();

  _userid = pref.getString('FB_USER');
  return _userid;
}

@override
Widget build(BuildContext context) {
  return FutureBuilder(
  future: setUserid(),
  builder: (BuildContext context, AsyncSnapshot<String> snapshot) {
    if (snapshot.hasData) {
      return ... // your widget
    } else return CircularProgressIndicator();
  });

类似这样的东西

在构建小部件之前获取SharedPreferences

2 个答案: