颤动:根据某些条件筛选列表

Question

我有电影列表.这包含所有动画和非动画电影.要识别它的动画是否有一个名为isAnimated的标志.

我只想展示动画电影.我编写的代码只过滤掉了动画电影,但却出现了一些错误.

import 'package:flutter/material.dart';

void main() => runApp(new MyApp());

class MyApp extends StatelessWidget {
  // This widget is the root of your application.
  @override
  Widget build(BuildContext context) {
    return new MaterialApp(
      title: 'Flutter Demo',
      theme: new ThemeData(

        primarySwatch: Colors.blue,
      ),
      home: new HomePage(),
    );
  }
}

class Movie {
  Movie({this.movieName, this.isAnimated, this.rating});
  final String movieName;
  final bool isAnimated;
  final double rating;
}

List<Movie> AllMovies = [
  new Movie(movieName: "Toy Story",isAnimated: true,rating: 4.0),
  new Movie(movieName: "How to Train Your Dragon",isAnimated: true,rating: 4.0),
  new Movie(movieName: "Hate Story",isAnimated: false,rating: 1.0),
  new Movie(movieName: "Minions",isAnimated: true,rating: 4.0),
];



class HomePage extends StatefulWidget{
  @override
  _homePageState createState() => new _homePageState();
}


class _homePageState extends State<HomePage> {

  List<Movie> _AnimatedMovies = null;

  @override
  void initState() {
    super.initState();
    _AnimatedMovies = AllMovies.where((i) => i.isAnimated);
  }

  @override
  Widget build(BuildContext context) {
    return new Scaffold(
      body: new Container(
        child: new Text(
            "All Animated Movies here"
        ),
      ),
    );
  }
}

Answer 1

toList() 缺少结果的物化器

_AnimatedMovies = AllMovies.where((i) => i.isAnimated).toList();

Answer 2

解决方案就在这里

只需尝试使用此函数 getCategoryList()，
这里的条件将是列表中的 catogory_id == '1'

List<dynamic> getCategoryList(List<dynamic> inputlist) {
    List outputList = inputlist.where((o) => o['category_id'] == '1').toList();
    return outputList;
  }

Answer 3

您可以将其用于特定条件

List<String> strings = ['one', 'two', 'three', 'four', 'five'];
List<String> filteredStrings  = strings.where((item) {
   return item.length == 3;
});

Answer 4

您可以使用toList()方法来获得您想要的输出，如下所示

toList()将此流的所有元素收集到一个List.

要解决上述问题：

添加一个toList()（此代码创建一个List<dynamic>）

_AnimatedMovies = AllMovies.where((i) => i.isAnimated).toList();

代替

_AnimatedMovies = AllMovies.where((i) => i.isAnimated);

的输出where()是另一个 Iterable，您可以使用它来迭代它或应用其他 Iterable 方法。在下一个示例中，where()直接在for-in循环内部使用的输出。

var evenNumbers = AllMovies.where((i) => i.isAnimated).toList();
for (var i in evenNumbers) {
  print('$i');
}

2. 您还可以使用 takeWhile

这些方法takeWhile()还可以帮助您从 Iterable 中过滤元素。

 _AnimatedMovies = AllMovies.takeWhile((i) => i.isAnimated).toList();