Dart/Flutter 如何比较并合并到两个模型列表中的一个？答案

【问题标题】：Dart/Flutter How to compare and merge into one from two List of model?Dart/Flutter 如何比较并合并到两个模型列表中的一个？
【发布时间】：2021-02-26 09:44:39
【问题描述】：

如何在 Dart/Flutter 中比较两个模型列表并将答案值合并为一个？

class Questions {
  dynamic id;
  final String label;
  final String description;
  String answerText;

  Questions({
    this.id,
    this.label,
    this.description,
    this.answerText,
  });

 List<Questions> q1 = [Question(1, 'This is question 1', 'first_question',''), Question(2, 'This is question 2','second_question',''),Question(3, 'This is question 3', 'third',''), Question(4, 'This is question 4', 'fourth',''),Question(5, 'This is question 5', 'fifth',''), Question(6, 'This is question 6', 'sixth','')]; 

 List<Questions> q2 = [Question(1,'answer for question 1'), Question(3,'answer for question 3'), Question(5,'answer for question 5')];

【问题讨论】：

标签： flutter dart

【解决方案1】：

我想您想将答案合并到问题中，如下所示：

1: This is question 1
Desc: first_question
Answer:answer for question 1

2: This is question 2
Desc: second_question
Answer:Missing answer

3: This is question 3
Desc: third
Answer:answer for question 3

4: This is question 4
Desc: fourth
Answer:Missing answer

5: This is question 5
Desc: fifth
Answer:answer for question 5

6: This is question 6
Desc: sixth
Answer:Missing answer

这是在Question 的List 上使用map 的解决方案：

void main() {
  List<Question> combined = questions
      .map((question) => question
          ..answerText =
            answers.firstWhere((answer) => answer.id == question.id, orElse: () => null)?.answerText ?? 'Missing answer')
      .toList();
  combined.forEach((question) => print(question));
}

class Question {
  dynamic id;
  final String label;
  final String description;
  String answerText;

  Question({
    this.id,
    this.label,
    this.description,
    this.answerText,
  });
  
  toString() => '$id: $label\nDesc: $description\nAnswer:$answerText\n';
}

List<Question> questions = [
  Question(id: 1, label: 'This is question 1', description: 'first_question'),
  Question(id: 2, label: 'This is question 2', description: 'second_question'),
  Question(id: 3, label: 'This is question 3', description: 'third'),
  Question(id: 4, label: 'This is question 4', description: 'fourth'),
  Question(id: 5, label: 'This is question 5', description: 'fifth'),
  Question(id: 6, label: 'This is question 6', description: 'sixth'),
];

List<Question> answers = [
  Question(id: 1, answerText: 'answer for question 1'),
  Question(id: 3, answerText: 'answer for question 3'),
  Question(id: 5, answerText: 'answer for question 5'),
];

【讨论】：

我尝试了您的解决方案如下：List<Question> combined = q1.map((question) => question.answerText = q2.firstWhere((answer) => answer.id == question.id, orElse: () => null)?.answerText ?? '').toList(); 它说如下。不能将“List”类型的值分配给“List”类型的变量。尝试更改变量的类型，或将右侧类型转换为“List”。
注意，我用的是 Dart 的cascade notation。 question.answeredText = 'answer' 将返回字符串 'answer'，而 question..answeredText = 'answer' 将返回实体 question。

【解决方案2】：

使用darq包，有set操作有except、intersect、union方法。

例子

var listA = [1, 2, 3, 4];
var listB = [3, 4, 5, 6];

var exclusion = listA.except(listB);       // [1, 2]
var intersection = listA.intersect(listB); // [3, 4]
var union = listA.union(listB);            // [1, 2, 3, 4, 5, 6]

【讨论】：

谢谢，但我不想为此使用额外的包。