Question

我遇到了问题，我试图根据对象列表的排序方式对字符串列表进行排序。

这是我的代码：

void main() => runApp(MaterialApp(home: App()));

List<objects> items = List<objects>();
List<String> filteritems = List();

class App extends StatefulWidget {
  @override
  _AppState createState() => _AppState();
}

class _AppState extends State<App> {


  void berechnung() {
    for (var i = 0; i < items.length; i++) {
      items[i].result = items[i].number1 + items[i].number2;
    }
    final Map<String, String> mappings = {
      for (int i = 0; i < items.length; i++)
        items[i].title : filteritems[i]
    };
    Comparator<objects> sortById = (a, b) => a.result.compareTo(b.result);
    items.sort(sortById);

    filteritems = [
      for (objects item in items) mappings[item]
    ];

    print(filteritems);
  }
  @override
  void initState() {
    super.initState();
    items = [
      objects(title: 'Gummibärchen', number1: 1, number2: 3, result: 0),
      objects(title: 'Schokolade', number1: 5, number2: 2, result: 0),
      objects(title: 'Eis', number1: 4, number2: 6, result: 0),
      objects(title: 'Nutella', number1: 2, number2: 0, result: 0),
    ];
    filteritems = ['Gummibärchen','Schokolade','Eis','Nutella' ];
    berechnung();
  }
  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text('Mapping'),
      ),
      body: ListView.builder(
        itemCount: items.length,
        itemBuilder: (context, index) {
          return Card(
            child: Container(
              height: MediaQuery.of(context).size.height / 20,
              width: MediaQuery.of(context).size.width,
              child: Column(
                children: <Widget>[
                  Text(items[index].title),
                  Text(items[index].result.toString()),
                ],
              ),
            ),
          );
        },
      ),
    );
  }
}


class objects {
  String title;
  int number1;
  int number2;
  int result;

  objects({this.title,this.number1,this.number2,this.result});
}

我对编码非常陌生，所以我不理解自己编写的所有代码都很好，我只是复制了代码并希望做到最好，但这对我没有帮助。我的行print(filteritems);的输出为[null，null，null，null]，我希望得到： [Nutella，Gummibärchen，Schokolade，Eis]，因为这是对其他列表进行排序的方式。

我希望有人能提供帮助，我将非常感谢。

Answer 1

您可以在下面复制粘贴运行完整代码
您需要使用mappings[item.title]而不是mappings[item]

filteritems = [for (objects item in items) mappings[item.title]];

输出

I/flutter (20843): [Nutella, Gummibärchen, Schokolade, Eis]

完整代码

import 'package:flutter/material.dart';

void main() => runApp(MaterialApp(home: App()));

List<objects> items = List<objects>();
List<String> filteritems = List();

class App extends StatefulWidget {
  @override
  _AppState createState() => _AppState();
}

class _AppState extends State<App> {
  void berechnung() {
    for (var i = 0; i < items.length; i++) {
      items[i].result = items[i].number1 + items[i].number2;
    }
    print(items.toString());
    final Map<String, String> mappings = {
      for (int i = 0; i < items.length; i++) items[i].title: filteritems[i]
    };

    print(mappings.toString());
    Comparator<objects> sortById = (a, b) => a.result.compareTo(b.result);
    items.sort(sortById);

    print(items.toString());
    filteritems = [for (objects item in items) mappings[item.title]];

    print(filteritems);
  }

  @override
  void initState() {
    super.initState();
    items = [
      objects(title: 'Gummibärchen', number1: 1, number2: 3, result: 0),
      objects(title: 'Schokolade', number1: 5, number2: 2, result: 0),
      objects(title: 'Eis', number1: 4, number2: 6, result: 0),
      objects(title: 'Nutella', number1: 2, number2: 0, result: 0),
    ];
    filteritems = ['Gummibärchen', 'Schokolade', 'Eis', 'Nutella'];
    berechnung();
  }

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text('Mapping'),
      ),
      body: ListView.builder(
        itemCount: items.length,
        itemBuilder: (context, index) {
          return Card(
            child: Container(
              height: MediaQuery.of(context).size.height / 20,
              width: MediaQuery.of(context).size.width,
              child: Column(
                children: <Widget>[
                  Text(items[index].title),
                  Text(items[index].result.toString()),
                ],
              ),
            ),
          );
        },
      ),
    );
  }
}

class objects {
  String title;
  int number1;
  int number2;
  int result;

  objects({this.title, this.number1, this.number2, this.result});
}

根据对象列表的排序方式对字符串列表进行排序

1 个答案: