【发布时间】:2019-07-10 04:28:45
【问题描述】:
我需要帮助将列表视图传递到另一个屏幕。
我是 Flutter 和 Dart 的新手,但对一般编程很熟悉。我想出了如何从 api 调用创建列表。但是,我当前的代码,当我按下按钮时,它会显示在我当前所在的屏幕上。我想将它传递到另一个屏幕以显示在列表中(主页上将有文本字段将更改 url 以从中获取数据)。我想我在逻辑上试图做的是当用户按下按钮时,它会调用获取数据,构建我的列表并将其显示在不同的屏幕上。
import 'package:flutter/material.dart';
import 'package:http/http.dart';
import 'dart:convert';
void main(){
runApp(
MaterialApp(
home: MyApp(),
),
);
}
class MyApp extends StatefulWidget {
@override
State<StatefulWidget> createState() {
return HomeScreen();
}
}
//List is here, had to make global so I could access in 2 widgets
List<TrailModel> trails=[];
class HomeScreen extends State<MyApp> {
Future<List<TrailModel>> fetchData() async {
var response = await get('https://www.hikingproject.com/data/get-trails?lat=40.0274&lon=-105.2519&maxDistance=10&key=200419778-6a46042e219d019001dd83b13d58aa59');
final trailModels = List<TrailModel>();
final trailModel = TrailModel.fromJson(json.decode(response.body));
trails.add(trailModel);
/*setState(() {
trails.add(trailModel);
});*/
return trailModels;
}
@override
Widget build(BuildContext context) {
return MaterialApp(
home: Scaffold(
appBar: AppBar(
title: Text("HikeLocator"),
),
body:
new RaisedButton(
child: Text("click me"),
onPressed: () async {
final trails = await fetchData();
Navigator.push(
context,
new MaterialPageRoute(builder: (context) => new ListScreen(trails)),
);
}
),
)
);
}
}
//Display ListView in here
class ListScreen extends StatelessWidget {
final List<TrailModel> trails;
ListScreen(this.trails);
//ListScreen(this.trails);
@override
Widget build (BuildContext ctxt) {
return new Scaffold(
appBar: new AppBar(
title: new Text("This is where the list lies"),
),
body: TrailList(trails),
);
}
}
//create ListView here which I want displayed on 2nd page
class TrailList extends StatelessWidget {
final List<TrailModel> trails;
TrailList(this.trails);
Widget build(context) {
return ListView.builder(
itemCount: trails.length,
itemBuilder: (context, int index) {
String myText = trails[index].trails.toString();
var splitString = myText.split("\, type");
var splitString2 = splitString[0];
var splitString3 = splitString2.split("name: ");
String name = splitString3[1];
splitString = myText.split("\, name");
splitString2 = splitString[0];
splitString3 = splitString2.split("id: ");
String id = splitString3[1];
splitString = myText.split("\, conditionStatus");
splitString2 = splitString[0];
splitString3 = splitString2.split("latitude: ");
String latitude = splitString3[1];
splitString = myText.split("\, latitude");
splitString2 = splitString[0];
splitString3 = splitString2.split("longitude: ");
String longitude = splitString3[1];
splitString = myText.split("\, url");
splitString2 = splitString[0];
splitString3 = splitString2.split(" location: ");
String location = splitString3[1];
return Text("name: $name,\n location: $location, \nlatitude: $latitude, \nlongitude: $longitude\n");
},
);
}
}
class TrailModel{
Object trails;
TrailModel(this.trails);
TrailModel.fromJson(Map<String, dynamic> parsedJson) {
trails = parsedJson['trails'];
}
}
自从我将一些小部件更改为有状态后,我单击以转到下一页的按钮不再起作用。我传递数据的尝试是传入 trails 变量,这是一个从主屏幕到下一页的列表,在 List Screen 中创建构造函数,然后在那里创建 ListView。但是,无法转到下一页。我不知道它是否有效。 .我最初并不打算发布我的所有代码,但我认为通过全貌查看类之间的交互会有所帮助。感谢您的帮助
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