使用API​​的showSearch

Question

我正在尝试实现搜索功能，并希望从API中获取结果。

在方法buildResults（）下，您会发现我的注释//数据为空，但是问题是我正在从API调用中获取数据。我在这里想念东西吗？

在buildsResults（）下，我正在调用Future _getResults并返回接收到的数据。我记录了您可以看到的数据。

    class SearchBar extends StatefulWidget {
      @override
      _SearchBarState createState() => new _SearchBarState();
    }

    class _SearchBarState extends State<SearchBar> {
      @override
      Widget build(BuildContext context) {
        return Scaffold(
          appBar: AppBar(
            iconTheme: new IconThemeData(color: Theme.of(context).hintColor),
            elevation: 1,
            backgroundColor: Theme.of(context).primaryColor,
            actions: <Widget>[
              IconButton(
                  autofocus: true,
                  icon: Icon(Icons.search),
                  onPressed: () async {
                    final results = await showSearch<SearchModel>(context: context, delegate: DataSearch(context));
                  })
            ],
            centerTitle: true,
            title: Text('Search content'),
          ),
        );
      }
    }

    class DataSearch extends SearchDelegate<SearchModel> {

      final BuildContext parentContext;
      final Logger logger = new Logger();

      DataSearch(this.parentContext);

      @override
      List<Widget> buildActions(BuildContext context) {
        return [
          IconButton(
            icon: Icon(Icons.clear),
            onPressed: () {
              query = "";
            },
          )
        ];
      }

      @override
      Widget buildLeading(BuildContext context) {
        return IconButton(
          icon: AnimatedIcon(
            icon: AnimatedIcons.menu_arrow,
            progress: transitionAnimation,
          ),
          onPressed: () {
            Navigator.pop(context);
            Navigator.pop(parentContext);
          },
        );
      }

      @override
      Widget buildResults(BuildContext context) {
        return FutureBuilder<List<SearchModel>>(
          future: _getResults(),
          builder: (context, AsyncSnapshot<List<SearchModel>> snapshot) {
            if (snapshot.connectionState == ConnectionState.done) {
              logger.d(snapshot.hasData);
              return ListView.builder(
                itemBuilder: (context, index) {
                  return ListTile(
                    title: Text(snapshot.data[index].title),
                    onTap: () {
                      close(context, snapshot.data[index]);
                    },
                  );
                },
                itemCount: snapshot.data.length, // data is null
              );
            } else {
              return Center(
                child: CircularProgressIndicator(),
              );
            }
          },
        );
      }

      @override
      Widget buildSuggestions(BuildContext context) {
        return Container();
      }

      Future<List<SearchModel>> _getResults() async {
        SharedPreferences prefs = await SharedPreferences.getInstance();
        String language = prefs.getString('language');
        var data;
        await http.get(Constants.BASE_URL + "/search/" + language + "/" + query,).then((response) {
              data = convert.jsonDecode(response.body) as List;
        });
        logger.d(data);
        return data.map((model) => SearchModel.fromJson(model)).toList();
      }
    }

Answer 1

我认为是这样的：

onTap: () async {
  final results = await showSearch(context: context, delegate: SearchBar(),query:query);
}

结果获取返回值 查询是传递的参数