Question

我想将列表List<Word> myList转换为String并将其放入sharedPreference，稍后我也想将该字符串（从sharedPreference）转换回List<Word>。

这是我的模型课程Word

class Word {
  int id;
  String word;
  String meaning;
  String fillInTheGapSentence;

  Word.empty();

  Word(int id, String word, String meaning, String fillInTheGapSentence){
    this.id = id;
    this.word = word;
    this.meaning = meaning;
    this.fillInTheGapSentence = fillInTheGapSentence;
  }
}

我可以将List<Word> myList转换为String

var myListString = myList.toString();

但是无法从List<Word> myListFromString中获得myListString。

谢谢。

Answer 1

您将需要某种序列化，并且其中有很多。最受欢迎的一种是JSON序列化。

Flutter的文档很好，如何做到这一点： https://flutter.dev/docs/development/data-and-backend/json

您要：

将您的对象转换为Map
将您的地图编码为JSON（是字符串）
保存
将其检索为字符串
将您的JSON解码为Map
将地图转换为对象

Answer 2

首先，myList.toString()并非JSON格式，除非您覆盖toString()方法。您要做的是手动将对象转换为Dictionary，然后将其编码为JSON字符串。相反，您需要将字符串转换为字典，然后将其转换为对象。像这样：

import 'dart:convert';

class Word {
  int id;
  String word;
  String meaning;
  String fillInTheGapSentence;

  Word.empty();

  Word(int id, String word, String meaning, String fillInTheGapSentence) {
    this.id = id;
    this.word = word;
    this.meaning = meaning;
    this.fillInTheGapSentence = fillInTheGapSentence;
  }

  Map<String, dynamic> toMap() {
    return {
      'id': this.id,
      'word': this.word,
      'meaning': this.meaning,
      'fillInTheGapSentence': this.fillInTheGapSentence,
    };
  }

  factory Word.fromMap(Map<String, dynamic> map) {
    return new Word(
      map['id'] as int,
      map['word'] as String,
      map['meaning'] as String,
      map['fillInTheGapSentence'] as String,
    );
  }
}

String convertToJson(List<Word> words) {
  List<Map<String, dynamic>> jsonData =
      words.map((word) => word.toMap()).toList();
  return jsonEncode(jsonData);
}

List<Word> fromJSon(String json) {
  List<Map<String, dynamic>> jsonData = jsonDecode(json);
  return jsonData.map((map) => Word.fromMap(map)).toList();
}

将列表转换为Json字符串，然后将此字符串转换回Dart中的List

2 个答案: