MVVM Light Relay命令参数

发布于 2024-10-22 03:29:15 字数 823 浏览 2 评论 0原文

我在使用 GalaSoft MVVM Light 框架将参数传递给中继命令时遇到问题。我知道 mvvm light 的relaycommand 实现不使用 lambda 参数，所以我做了一些研究，找到了一种人们通过执行以下操作来解决它的方法：

public RelayCommand ProjMenuItem_Edit
{
    get
    {
        if (_projmenuItem_Edit == null)
        {
            //This should work....
            _projmenuItem_Edit = new RelayCommand(ProjEditNode);
        }
        return _projmenuItem_Edit;
    }
}

private void ProjEditNode(object newText)
{
    var str = newText as string;
    OrganLocationViewModel sel = 
        ProjectOrganLocationView.GetExtendedTreeView().GetTopNode();

    //Console.WriteLine(sel.OrganDisplayName);
    sel.OrganDisplayName = str;
}

但是，我不断收到错误 _projmenuItem_Edit = new RelayCommand(ProjEditNode); 上面写着参数 1：无法从“方法组”转换为“System.Action”

我缺少什么？

原文

I'm having an issue with passing a parameter to a relaycommand using the GalaSoft MVVM Light framework. I know that mvvm light's implementation of relaycommand doesn't use lambda parameters, so I did some research and found a way that people worked around it by doing something like this:

public RelayCommand ProjMenuItem_Edit
{
    get
    {
        if (_projmenuItem_Edit == null)
        {
            //This should work....
            _projmenuItem_Edit = new RelayCommand(ProjEditNode);
        }
        return _projmenuItem_Edit;
    }
}

private void ProjEditNode(object newText)
{
    var str = newText as string;
    OrganLocationViewModel sel = 
        ProjectOrganLocationView.GetExtendedTreeView().GetTopNode();

    //Console.WriteLine(sel.OrganDisplayName);
    sel.OrganDisplayName = str;
}

However, I keep getting an error on the line _projmenuItem_Edit = new RelayCommand(ProjEditNode); that says Argument 1: cannot convert from 'method group' to 'System.Action'

What am I missing?

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暮光沉寂 2024-10-29 03:29:15

我相信这会起作用：

_projmenuItem_Edit = new RelayCommand<object>((txt)=>ProjEditNode(txt));

--编辑--

您还需要使用类型定义 RelayCommand：

例如

public RelayCommand<string> myCommand { get; private set; }
myCommand = new RelayCommand<string>((s) => Test(s));

private void Test(string s)
{
    throw new NotImplementedException();
}

I believe this will work:

_projmenuItem_Edit = new RelayCommand<object>((txt)=>ProjEditNode(txt));

-- EDIT --

You'll need to define your RelayCommand with the type as well:

e.g.

public RelayCommand<string> myCommand { get; private set; }
myCommand = new RelayCommand<string>((s) => Test(s));

private void Test(string s)
{
    throw new NotImplementedException();
}

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夏见 2024-10-29 03:29:15

我认为 RelayCommand() 没有一个非空的构造函数。你试图向它传递错误的方法。

如果您希望 RelayCommand 支持命令参数，则应使用 RelayCommand，其中 T 可以是任何类型的参数。在您的情况下，它将是 RelayCommand ，它将接受带有 void(string) 签名的方法。（因此也将是强类型的并且不会使用丑陋的对象）

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抠脚大汉 2024-10-29 03:29:15

另一种声明中继命令的方法将有助于减少代码

public RelayCommand ChartCommand
{
    set
    {
        RelayCommand<string> chartCommand = 
            new RelayCommand<string>(e => ExecuteChartCommand(e));               
    }
}

public void ExecuteChartCommand(string vendor)
{

}

Another way to declare relay commands, will help to reduce your code

public RelayCommand ChartCommand
{
    set
    {
        RelayCommand<string> chartCommand = 
            new RelayCommand<string>(e => ExecuteChartCommand(e));               
    }
}

public void ExecuteChartCommand(string vendor)
{

}

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~没有更多了~