Question

     @override
      Widget build(BuildContext context) {
        widget.groupid;
        widget.event_id;
        var futureBuilder = new FutureBuilder(
          future: _getAllTickets(),

          builder: (BuildContext context, AsyncSnapshot snapshot) {
            print(snapshot.connectionState);
            switch (snapshot.connectionState) {
              case ConnectionState.none:
              case ConnectionState.waiting:
                return new Text('...');
              default:
                if (snapshot.hasError)
                  return new Text('Error: ${snapshot.error}');
                else
                  return createListTickets(context, snapshot);
            }
          },
        );



    return new Scaffold(

      body: futureBuilder,

    );
  }

  Widget createListTickets(BuildContext context, AsyncSnapshot snapshot) {
    List values = snapshot.data;

     child: new Card(

                child:
                new Column(mainAxisSize: MainAxisSize.min, children: <Widget>[
                  new Text(
                      values[index]["ticket_type_id"].toString(), style:
                  const TextStyle(
                      fontFamily: 'Poppins',
                      fontWeight: FontWeight.w600,
                      fontSize: 25.0)),
}


  _getAllTickets() async {
    final response = await http.get(
        "https...}"
        , headers: {
      HttpHeaders.AUTHORIZATION: access_token
    });

    returnTickets = json.decode(response.body);
    return returnTickets;
  }


  _getTicketType() async {
    for (i = 0; i < (returnTickets?.length ?? 0); i++) {
      /*print("https....);*/
     final responseType = await http.get(
          "https...}"
          , headers: {
        HttpHeaders.AUTHORIZATION: access_token
      });

      Map<String, dynamic> hey = json.decode(responseType.body);


    }

大家好，我有一个问题。当我发送多个API请求并动态构建具有我得到的响应的卡时，我想知道我是否可以在将来包含更多的那个方法：_getAllTickets（），+（另一种方法），因为我想使用值[index] [＆＃34; name＆＃34;]替换值[index] [＆＃34; ticket_type_id＆＃34;]，该名称是我通过方法_getTicketType（）获得的新索引响应。提前谢谢！

Answer 1

您可以使用Future.wait(Future[])将期货列表合并为一个。

class Merged {
  final String foo;
  final String bar;

  Merged({this.foo, this.bar});
}

Future<String> foo;
Future<int> bar;
new FutureBuilder(
  future: Future.wait([bar, foo]).then(
    (response) => new Merged(foo: response[0], bar: response[1]),
  ),
  builder: (context, AsyncSnapshot<Merged> snapshot) {

  },
);

我可以在未来的构建器上使用多种方法吗？

1 个答案: