如何設置另一個控件按比例位置

Question

我的問題是：我有2個面板（panel1，panel2）其中panel1.Size = new Size(200, 200);和Panel2.Size = new Size(600, 600);其中兩個面板都在一個CustomControl中可以拖動並更改Possition（szbControl1，szbControl2）。

我的問題是，如何設置szbControl2.Location正確（按比例）的基礎上szbControl1.Location其中szbControl1父panel1和szbControl2父panel2，就像如果我在底部移動szbControl1也szbControl2應該在底部。 到目前爲止，我想這：

private void sizeAbleCTR2_LocationChanged(object sender, EventArgs e) 
     { 
      int smallX = (sizeAbleCTR2.Location.X * panel1.Size.Width)/100; 
      int smallY = (sizeAbleCTR2.Location.Y * panel1.Size.Height)/100; 

      int largeX = (smallX * panel2.Width)/100; 
      int largeY = (smallY * panel2.Height)/100; 

      sizeAbleCTR1.Location = new Point(largeX,largeY); 
     } 

喜歡用百分比，但它不工作。

Answer 1

您提供的代碼沒有考慮到szbControls的大小。 （位置/大小的差異）的比例應該相等。

private void sizeAbleCTR2_LocationChanged(object sender, EventArgs e) 
{ 
    float srcHeightDiff = panel2.Height - sizeAbleCTR2.Height; 
    float dstHeightDiff = panel1.Height - sizeAbleCTR1.Height; 

    int locY = (int)(dstHeightDiff * (sizeAbleCTR2.Location.Y/srcHeightDiff)); 

    float srcWidthDiff = panel2.Width - sizeAbleCTR2.Width; 
    float dstWidthDiff = panel1.Width - sizeAbleCTR1.Width; 

    int locX = (float)(dstWidthDiff * (sizeAbleCTR2.Location.X/srcWidthDiff)); 

    sizeAbleCTR1.Location = new Point(locX, locY); 
}