在 dart (List<List<Tuple2>>) 中序列化/反序列化嵌套 Tuple2 数组的最佳方法是什么答案

【问题标题】：What is the best way to Serialize/Deserialize nested Tuple2 arrays in dart (List<List<Tuple2>>)在 dart (List<List<Tuple2>>) 中序列化/反序列化嵌套 Tuple2 数组的最佳方法是什么
【发布时间】：2021-01-13 14:52:25
【问题描述】：

我有一个嵌套的 Tuple2 数组，我正在寻找一种快速而优雅的方法来序列化/反序列化 dart 中的嵌套 Tuple2 数组，这是结构：List<List<Tuple2<String,String>>>

[

  [

    [atd, OK],

    [atkw0, ?],

    [atsp0, OK], 

    [ate0, OK], 

    [ati, ELM327v15d], 

    [ath1, OK], 

    [ats0, OK], 

    [atl0, OK],

    [atat1, OK], 

    [0100, 7E9124100FFFFC012], 

    [0120, 7E9124120E01FE01E],

    [0140, 7E91241406FECC078], 

    [0160, 7E912416000000000], 

    [0180, 7E9124180FFFFFFFF], 

    [01A0, 7E91241A0FFFFFFFF], 

    [01C0, NODATA], 

    [0902, 7E98449020031004300340048004A00570043003500300043004C003100320031003200350033],

    [atdp, ISO15765���4(CAN11/250)]],

  [

    [ate0, OK], 

    [ath0, OK], 

    [atal, ?], 

    [ati, ELM327v15d],

    [atl0, OK], 

    [ats0, OK], 

    [atat1, OK], 

    [atcra7E9, ?], 

    [0100, 4100FFFFC012], 

    [0120, 4120E01FE01E], 

    [0140, 41406FECC078], 

    [0160, 416000000000], 

    [0180, 4180FFFFFFFF],

    [01A0, 41A0FFFFFFFF], 

    [01C0, NODATA], 

    [0900, 490040400000], 

    [0902, 49020031004300340048004A00570043003500300043004C003100320031003200350033]

  ], 

  [

    [ate0, OK], 

    [ath0, OK], 

    [atal, ?], 

    [ati, ELM327v15d], 

    [atl0, OK], 

    [ats0, OK], 

    [atat1, OK], 

    [atcra7E9, ?], 

    [0100, 4100FFFFC012], 

    [0120, 4120E01FE01E], 

    [0140, 41406FECC078], 

    [0160, 416000000000], 

    [0180, 4180FFFFFFFF], 

    [01A0, 41A0FFFFFFFF], 

    [01C0, NODATA], 

    [0900, 490040400000], 

    [0902, 49020031004300340048004A00570043003500300043004C003100320031003200350033]

  ], 

  [

    [atdp, ISO15765���4(CAN11/250)]

  ], 

  [

    [010D, 410D6C], 

    [010C, 410C3D8D]

  ], 

  [

    [010D, 410D6C], 

    [010C, 410C3D8D]

  ], 

  [

    [010D, 410D6C], 

    [010C, 410C3D8D]

  ], 

  [

    [010D, 410D6C],

    [010C, 410C3D8D]

  ], 

  [

    [010D, 410D6C], 

    [010C, 410C3D8D]

  ], 

  [

    [010D, 410D6C], 

    [010C, 410C3D8D]

  ], 

  [

    [010D, 410D6C],

    [010C, 410C3D8D], 

    [01A6, 41A6072B]

  ],

【问题讨论】：

这些是什么数据类型？
Tuple2 内部是字符串
如果所有的值都是Dart可以识别的类型，你可以使用jsonEncode()或者jsonDecode()，你可以阅读更多关于它的内容here。
我觉得最优雅的方式是使用json_serializable包，需要build_runner生成代码。

标签： flutter dart

【解决方案1】：

更新答案。

import 'dart:convert' as convert;

import 'package:tuple/tuple.dart';

class Model {
  final String string1;
  final String string2;

  Model(this.string1, this.string2);

  factory Model.fromTuple(Tuple2 tuple2) {
    final String string1 = tuple2.item1;
    final String string2 = tuple2.item2;
    return Model(string1, string2);
  }

  Tuple2 toTuple() => Tuple2(string1, string2);
}

void main(List<String> arguments) {
  final tuples = <List<Tuple2>>[
    [
      Tuple2('a', 'b'),
      Tuple2('c', 'd'),
      Tuple2('e', 'f'),
    ],
    [
      Tuple2('g', 'h'),
      Tuple2('i', 'j'),
      Tuple2('k', 'l'),
    ],
    [
      Tuple2('m', 'n'),
      Tuple2('o', 'p'),
      Tuple2('q', 'r'),
    ]
  ];

  /// converts tuples to Generic List
  List list = tuples
      .map((e) => e
          .map((e) => {
                'item1': e.item1,
                'item2': e.item2,
              })
          .toList())
      .toList();

  print(tuples is List<List<Tuple2>>);
  print(list is List<List<Tuple2>>);
  print(tuples.length);

  /// converts generic list to json string
  String json = convert.json.encode(list);
  print(json);

  /// converts json string to generic list
  Iterable iterable = convert.json.decode(json);
  List newList = iterable.toList();
  /// converts generic list to List<List<Tuple2>>
  final newTuple = newList
      .map((e) => (e as Iterable)
          .map((e) => Tuple2(
                e['item1'],
                e['item2'],
              ))
          .toList())
      .toList();

  print(newList is List<List<Tuple2>>);
  print(newTuple is List<List<Tuple2>>);
  print(newTuple.length);
}

【讨论】：

对不起，我的描述可能不是很清楚。我想将 >[] 序列化为字符串（可能是 JSON 或上述格式）并将其从字符串反序列化回 >[]
是的，这是非常不同的场景
如果是对的别忘了标记为接受
谢谢，看起来很完美。我认为可以删除 Model 类。
我会维护它，以供有相同问题的新开发人员参考。

【解决方案2】：

只是为了补充上面的答案，我创建了这两个静态方法并将它们添加到 Tuple2 类中。

 test(
    'Tuple2 serialization test with Tuple infrastructure',
    () async {
      final tuples = <List<Tuple2>>[
        [
          Tuple2('a', 'b'),
          Tuple2('c', 'd'),
          Tuple2('e', 'f'),
        ],
        [
          Tuple2('g', 'h'),
          Tuple2('i', 'j'),
          Tuple2('k', 'l'),
        ],
        [
          Tuple2('m', 'n'),
          Tuple2('o', 'p'),
          Tuple2('q', 'r'),
        ]
      ];

      String json = Tuple2.toJsonFromTupleList(tuples);
      List<List<Tuple2>> result = Tuple2.fromJsonToTupleList(json);
      expect(result.toString(), tuples.toString());
    },
  );
}

在 Tuple2 上使用这两个静态方法：

static List<List<Tuple2>> fromJsonToTupleList(String json) {
    Iterable iterable = convert.json.decode(json);
    List newList = iterable.toList();

    /// converts generic list to List<List<Tuple2>>
    final newTuples = newList
        .map((e) => (e as Iterable)
            .map(
              (e) => Tuple2<String,String>(
                e['1'],
                e['2'],
              ),
            )
            .toList())
        .toList();

    return newTuples as List<List<Tuple2>>;
  }

  static String toJsonFromTupleList(List<List<Tuple2>> tuples) {
    List list = tuples
        .map((e) => e
            .map(
              (e) => {
                '1': e.item1,
                '2': e.item2,
              },
            )
            .toList())
        .toList();

    String json = convert.json.encode(list);
    return json;
  }

【讨论】：