flutter - 在 Flutter 中存储缓存

Question

我有一个用于获取 gmys 并将结果加载到颤振应用程序的 api。我希望颤振应用程序加载数据并存储在缓存中。当用户访问该页面时，flutter 应用程序应该检查是否有互联网连接，如果有，应该从在线 api 加载数据，如果没有互联网连接，则应该从缓存 json 文件中加载。

目前，该应用程序仅在有互联网时才从在线 api 获取。贝娄是我的代码：

import 'package:flutter/material.dart';
import 'package:http/http.dart' as http;
import 'dart:convert';

    class GymList extends StatelessWidget {
      final String apiUrl ="apitogetgyms";
    
      Future<List<dynamic>> fetchGyms() async {
        var result = await http.get(apiUrl);
        return json.decode(result.body)['gyms'];
      }
    
      String _name(dynamic gyms) {
        return gyms['name'];
      }  
    
    
    String _location(dynamic gyms) {
        return gyms['location'];
      }
    
      String _phone(dynamic gyms) {
        return gyms['phone'];
      }
    
      String _email(dynamic gyms) {
        return gyms['email'];
      }
    
      String _img(dynamic gyms){
        return gyms['img'];
      }
    
      @override
      Widget build(BuildContext context) {
        return Scaffold(
          appBar: AppBar(
            title: Text('Gyms List'),
          ),
          body: Container(
            child: FutureBuilder<List<dynamic>>(
              future: fetchGyms(),
              builder: (BuildContext context, AsyncSnapshot snapshot) {
                if (snapshot.hasData) {
                  //print(_name(snapshot.data[]));
                  return ListView.builder(
                      padding: EdgeInsets.all(8),
                      itemCount: snapshot.data.length,
                      itemBuilder: (BuildContext context, int index) {
                        return Card(
                          child: Column(
                            children: <Widget>[
                              ListTile(
                                leading: CircleAvatar(
                                    radius: 30,
                                      backgroundImage:
                                        NetworkImage(snapshot.data[index]['img'])),
                                title: Text(_name(snapshot.data[index])),
                                subtitle: Text(_location(snapshot.data[index])),
                                trailing: Text(_email(snapshot.data[index])),
                              )
                            ],
                          ),
                        );
                      });
                } else if (snapshot.hasError) {
                  return Text("${snapshot.error}");
                } else {
                  return Center(child: CircularProgressIndicator());
                }
              },
            ),
          ),
        );
      }
    }

标签： flutter

Answer 1

您可以通过使用 2 个包shared_preferences和connectivity来实现此结果。

使用连接，您可以检查您的连接状态并知道您是否可以访问互联网。

import 'package:connectivity/connectivity.dart';

final connectivityResult = await (Connectivity().checkConnectivity());
if (connectivityResult == ConnectivityResult.none) {
  // I am not connected.
} else {
  // I am connected to internet.
}

然后使用 shared_preferences 您可以轻松保存 JSON 并在未连接到 Internet 时加载它。

import 'dart:convert';
import 'package:shared_preferences/shared_preferences.dart';

final jsonKey = 'json_key'; // Key used to save and get your data from SharedPreferences.
final prefs = await SharedPreferences.getInstance();

// Save your JSON as a String by encoding it.
await prefs.setString(jsonKey, jsonEncode(myJsonData));

// Get your JSON from SharedPreferences and decode it.
final json = jsonDecode(prefs.getString(jsonKey));

例子

import 'package:connectivity/connectivity.dart';
import 'dart:convert';
import 'package:shared_preferences/shared_preferences.dart';

Future<List<dynamic>> fetchGyms() async {
   final jsonKey = 'json_key';
   final prefs = await SharedPreferences.getInstance();
   final connectivityResult = await (Connectivity().checkConnectivity());

   if (connectivityResult == ConnectivityResult.none) {
      final json = jsonDecode(prefs.getString(jsonKey));
      return json['gyms'];
   } else {
      final result = await http.get(apiUrl);
      await prefs.setString(jsonKey, jsonEncode(result.body.toString()));
      return jsonDecode(result.body)['gyms'];
   }
}