如何在Windows窗体中为弹出窗口设置位置? [英] How to set a location for a popup window in windows form?

问题描述

在一种形式中，我有一个datagridview，其中包含我的xml文件的内容.我在单元格上单击鼠标右键打开一个弹出设置窗口.弹出设置窗口是另一种形式.

我已经编写了用于打开弹出式设置窗口表单的代码.但它显示在屏幕的最上方.如何在datagridview中的currentcell下方设置弹出设置窗口窗体的位置.

我尝试过的事情:

In a form I have a datagridview which has a contents of my xml file. I am opening a popup settings window to on cell right click. Here popup setting window is a another form.

I have written a code for opening a popup setting window form. but it is showing on the topmost corner of the screen. how to set the location of the popup setting window form below the currentcell in a datagridview.

What I have tried:

private void dataGridView1_CellMouseUp(object sender, DataGridViewCellMouseEventArgs e)
        {
 if (e.Button == MouseButtons.Right)
  {
private frmDGVColList objform = new frmDGVColList();
objform.StartPosition = FormStartPosition.Manual;
objform.Location = dataGridView1.PointToScreen(dataGridView1.GetCellDisplayRectangle(e.ColumnIndex, e.RowIndex, false).Location);
objform.Show(this.dataGridView1);
}
}

推荐答案

这似乎是您正在寻找的内容: ^ ]

This seems to be what you are looking for: c# - Open a popup from just below of current selected cell of datagrid view - Stack Overflow[^]

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