Question

通过将Xamarin.Forms与MVVM一起使用，我具有带有StoryName，StoryCategories和Popup（按钮）的故事的ListView。单击“弹出”按钮时，将出现一个弹出窗口，用户可以在其中为故事选择不同的类别。并且，当用户选择其他类别时，StoryCategories标签应使用所有选定类别（逗号分隔）进行更新。但是我不知道如何从单击弹出按钮的位置找到并更新StoryCategories标签？

过去8年来，我一直在开发.net mvc，我始终能够在StackOverflow上找到我的问题的答案（谢谢！），但是由于某些原因，我似乎无法找到该问题的答案。

希望您能提供帮助！：）

弹出窗口是Rg.Plugins.Popup，我正在使用James Montemagno的MvvmHelpers。

型号

    public class Story : ObservableObject
    {
        string name;
        public string Name
        {
            get { return name; }
            set { SetProperty(ref name, value); }
        }

        string categories;
        public string Categories
        {
            get { return categories; }
            set { SetProperty(ref categories, value); }
        }
    }

ViewModel

public class StoriesViewModel : BaseViewModel
    {
        public ObservableCollection<Story> Stories { get; set; }

        public StoriesViewModel()
        {
            Stories = new ObservableCollection<Story>();

            Stories.Add(new Story { Name = "Hitchhiker's Guide to the Galaxy", Categories = "a, b" });
            Stories.Add(new Story { Name = "The Restaurant at the End of the Universe", Categories = "b, c, y" });
            Stories.Add(new Story { Name = "Life, the Universe and Everything", Categories = "e" });
            Stories.Add(new Story { Name = "So Long, and Thanks for All the Fish", Categories = "a, c, e" });
        }

        public ICommand OpenPopupCommand
        {
            get
            {
                return new Command<Story>((x) =>
                {
                    Debug.WriteLine("Name: " + x.Name);
                });
            }
        }
    }

查看：

<ListView ItemsSource="{Binding Stories}">
    <ListView.ItemTemplate>
        <DataTemplate>
            <ViewCell>
                <StackLayout>
                    <Label Text="{Binding Name}" />
                    <Label Text="{Binding Categories} "/>
                    <Button Text="Click" Command="{ Binding Path=BindingContext.OpenPopupCommand, Source={ x:Reference Stories } }" CommandParameter="{ Binding . }" />
                </StackLayout>
            </ViewCell>
        </DataTemplate>
    </ListView.ItemTemplate>
</ListView>

Answer 1

如果您使用clicked事件，则可以执行以下操作，我只是猜测您使用的类的名称，但您应该得到它：

var story = ((Button)sender).BindingContext as Story;

Answer 2

如果您使用的是MVVM模式，请尝试以下操作：

{
  "name": "acbook_angular",
  "private": true,
  "dependencies": {
    "@angular/common": "^7.2.8",
    "@angular/compiler": "^7.2.8",
    "@angular/core": "^7.2.8",
    "@angular/platform-browser": "^7.2.8",
    "@angular/platform-browser-dynamic": "^7.2.8",
    "@angular/router": "^7.2.10",
    "@rails/webpacker": "^4.0.2",
    "core-js": "^2.6.5",
    "html-loader": "^0.5.5",
    "rxjs": "^6.4.0",
    "ts-loader": "^5.3.3",
    "typescript": "^3.3.3333",
    "zone.js": "^0.8.29"
  },
  "devDependencies": {
    "css-loader": "^2.1.1",
    "resolve-url-loader": "^3.0.1",
    "sass-loader": "^7.1.0",
    "to-string-loader": "^1.1.5",
    "webpack-dev-server": "^3.2.1",
    "webpack-merge": "^4.2.1"
  }
}

在视图模型中

<ListView x:Name="Stories">
    <ListView.ItemTemplate>
        <DataTemplate>
            <ViewCell>
                <StackLayout>
                    <Label Text="{Binding StoryName}" />
                    <Label Text="{Binding StoryCategories} "/>
                    <Button Command="{ Binding Path=BindingContext.OpenPopupCommand, Source={ x:Reference Stories } }" CommandParameter="{ Binding . }" />
                </StackLayout>
            </ViewCell>
        </DataTemplate>
    </ListView.ItemTemplate>
</ListView>

Answer 3

我过去解决此问题的方法是将整个故事页面ViewModel传递到弹出页面，然后在更改类别时，可以从弹出页面视图模型更新故事页面中的标签。我不知道这是否是最好的解决方案，但是它确实有效。

Xamarin从外部访问ListView

3 个答案: