是否可以重载表单的 ShowDialog 方法并返回不同的结果? [英] Is it possible to overload the ShowDialog method for forms and return a different result?

问题描述

这种方法实际上效果很好，我问了它，然后找到了解决方案.我在重载的 ShowDialog() 方法中添加了正确的调用(它不是完全重载，甚至不是覆盖，但它的工作原理相同.我的新问题是底部的问题.

我有一个表单，您可以在其中单击三个按钮之一.我为返回的结果定义了一个枚举.我想打电话:

I have a form in which you click one of three buttons. I have defined an enum for the returned results. I want to make the call:

MyFormResults res = MyForm.ShowDialog();

我可以用这段代码添加一个新的 ShowDialog 方法:

I can add a new ShowDialog method with this code:

public new MyFormResults ShowDialog()
{
    //Show modal dialog
    base.ShowDialog(); //This works and somehow I missed this

    return  myResult; //Form level variable (read on)
}

我为单击按钮时的结果设置了一个表单级变量:

I set a form-level variable for the result when the buttons are clicked:

MyFormResults myResult;

private void btn1_click(object sender, EventArgs e)
{
    myResult = MyFormsResults.Result1;
    this.DialogResult = DialogResult.OK; //Do I need this for the original ShowDialog() call?
    this.Close(); //Should I close the dialog here or in my new ShowDialog() function?
}

//Same as above for the other results

我唯一缺少的是显示对话框(模式)然后返回结果的代码.没有base.ShowDialog()函数，我该怎么做呢?

The only thing I'm missing is the code to show the dialog (modal) and then return my result. There is no base.ShowDialog() function, so how do I do this?

有一个base.ShowDialog()"，它可以工作.这是我的新问题:

另外，这是做这一切的最佳方式吗?为什么?

Also, is this the best way to do all this and Why?

谢谢.

推荐答案

更改 ShowDialog() 的功能可能不是一个好主意，人们希望它返回一个 DialogResult 并显示表单，我建议如下我的建议.因此，仍然允许 ShowDialog() 以正常方式使用.

It's proberly not a good idea to change the functionality of ShowDialog(), people expect it to return a DialogResult and show the form, I suggest something like my suggestion below. Thus, still allowing ShowDialog() to be used the normal manner.

您可以在 MyForm 上创建一个静态方法，例如 DoShowGetResult()

You could create a static method on your MyForm, something like DoShowGetResult()

看起来像这样

public static MyResultsForm DoShowGetResult()
{
   var f = new MyForm();
   if(f.ShowDialog() == DialogResult.OK)
   {
      return f.Result;   // public property on your form of the result selected
   }
   return null;
}

那你就可以用这个了

MyFormsResult result = MyForm.DoShowGetResult();

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