Question

从模型更新动态创建的复选框状态的最佳做法是什么？复选框的实际值保存在主模型的子模型中，并根据其逻辑进行更改。复选框的属性绑定到其各自的FooViewModel。但是如何更改FooViewModel的属性呢？

1种方式：主模型触发特殊事件 - ＆gt;主VM处理它并找到目标FooViewModel使用事件args更新 - ＆gt;主VM使用事件args中指定的值设置目标FooViewModel属性 - ＆gt;复选框通过绑定到FooViewModel

更新

2路：主模型包含实现INPC的FooModel的可观察集合，每个都用FooViewModel包装（使用主VM中的CollectionChanged事件）。主模型设置了一些FooModel的属性 - ＆gt; FooViewModel处理PropertyChanged并进一步传输它自己的PropertyChanged事件 - ＆gt;复选框通过绑定更新到FooViewModel。 FooViewModel中的转移代码：

this._model.PropertyChanged += (s, a) => this.RaisePropertyChangedEvent(a.PropertyName);

我的第二种方式的实施是下一步：

// MainModel class that holds collection of extra models (CfgActionModel):
class MainModel: BindableBase
{
    ObservableCollection<CfgActionModel> _actionsColl
        = new ObservableCollection<CfgActionModel>();

    public ObservableCollection<CfgActionModel> ActionCollection
    {
        get => this._actionsColl;
    }

    public void AddAction(ConfigEntry cfgEntry, bool isMeta)
    {
        CfgActionModel actionModel = new CfgActionModel()
        {
            CfgEntry = cfgEntry,
            Content = cfgEntry.ToString(),
            IsEnabled = true,
            IsChecked = false
        };

        this._actionsColl.Add(actionModel);
    }
}

// Extra model that is wrapped with CfgActionViewModel:
class CfgActionModel: BindableBase
{
    ConfigEntry _cfgEntry; // Custom enumeration value unique for each checkbox
    string _content;
    bool _isEnabled = false;
    bool _isChecked = false;

    public ConfigEntry CfgEntry
    {
        get => this._cfgEntry;
        set
        {
            if (this._cfgEntry == value) return;

            this._cfgEntry = value;
            this.RaisePropertyChangedEvent(nameof(CfgEntry));
        }
    }

    public string Content
    {
        get => this._content;
        set
        {
            if (this._content == value) return;

            this._content = value;
            this.RaisePropertyChangedEvent(nameof(Content));
        }
    }

    public bool IsEnabled
    {
        get => this._isEnabled;
        set
        {
            if (this._isEnabled == value) return;

            this._isEnabled = value;
            this.RaisePropertyChangedEvent(nameof(IsEnabled));
        }
    }

    public bool IsChecked
    {
        get => this._isChecked;
        set
        {
            if (this._isChecked == value) return;

            this._isChecked = value;
            this.RaisePropertyChangedEvent(nameof(IsChecked));
        }
    }
}

// CfgActionViewModel that is checkbox in UI is bound to:
class CfgActionViewModel: BindableBase
{
    CfgActionModel _model;

    public CfgActionViewModel(CfgActionModel model)
    {
        this._model = model;

        this._model.PropertyChanged += (s, a) => this.RaisePropertyChangedEvent(a.PropertyName);
    }

    public string Content
    {
        get => this._model.Content;
        set => this._model.Content = value;
    }

    public bool IsEnabled
    {
        get => this._model.IsEnabled;
        set => this._model.IsEnabled = value;
    }

    public bool IsChecked
    {
        get => this._model.IsChecked;
        set => this._model.IsChecked = value;
    }
}

// MainViewModel where we fill the model with data:
class MainViewModel
{
    MainModel model;

    readonly ObservableCollection<CfgActionViewModel> _actionVMColl = new ObservableCollection<CfgActionViewModel>();
    public ObservableCollection<CfgActionViewModel> ActionVMCollection => this._actionVMColl;

    public MainViewModel()
    {
        this.model = new MainModel();

        this.model.ActionCollection.CollectionChanged += (s, a) =>
        {
            // when new model is created we create new ViewModel wrapping it
            if (a.Action == NotifyCollectionChangedAction.Add)
            {
                CfgActionModel newModel = (CfgActionModel) a.NewItems[0];
                CfgActionViewModel actionViewModel = new CfgActionViewModel(newModel);

                _actionVMColl.Add(actionViewModel);
            }
        };

        model.AddAction(ConfigEntry.AutoBuy, false);
        model.AddAction(ConfigEntry.Bomb, false);
    }
}

View中的DataTemplate看起来像这样：

<DataTemplate DataType="{x:Type mvvm:CfgActionViewModel}">
    <CheckBox
        IsChecked="{Binding Path=IsChecked, Mode=TwoWay}" 
        IsEnabled="{Binding Path=IsEnabled, Mode=TwoWay}" 
        Content="{Binding Path=Content, Mode=OneWay}"/>
</DataTemplate>

MVVM是否可以避免在某处（第二路）避免与MainViewModel交互，或者每个subViewModel的属性必须由MainViewModel（第一路）设置？

Answer 1

这两种方法都是可以接受的。但就个人而言，我会采用＃1来保持我的模型尽可能薄。

您可以参考示例代码，了解如何进行＃1。

    [HttpPut("UpdateControlLinePointSet")]
    public async Task<IActionResult> UpdateControlLinePointSet([FromBody] JArray pointSetJson)
    {
        if (!ModelState.IsValid) return BadRequest(ModelState);
        foreach (JToken p in pointSetJson)
        {
            ControlLinePoint clp = _context.ControlLinePoint.First(x => x.ControlLinePointId == (int)p["ControlLinePointId"]);
            JsonConvert.PopulateObject(p.ToString(), clp);
            _context.Entry(clp).State = EntityState.Modified;
        }

        await _context.SaveChangesAsync();

        return NoContent();
    }

正如您所看到的，我甚至不需要在示例代码中包含任何模型，这意味着此处的所有逻辑都清楚地是表示层的一部分。现在，您可以专注于模型中的业务/域逻辑。

如何处理主模型（MVVM）中子模型的更改

1 个答案: