我想在PictureBox
内绘制一条线的旋转动画。我使用pictureBox1.CreateGraphics()
画线,但我听说这种方法不适合PictureBox
。另外,我在PictureBox
窗口上遇到了大量的闪烁,有什么建议吗?这是一段代码:
private void OnTimedEvent(object source, PaintEventArgs e)
{
try
{
if (comport.BytesToRead > 0)
{
X = comport.ReadByte();
Y = comport.ReadByte();
Z = comport.ReadByte();
}
Graphics g = pictureBox1.CreateGraphics();
Pen red = new Pen(Color.Red, 2);
g.TranslateTransform(100, 80 - X);
g.RotateTransform(120);
g.DrawLine(red, new Point(100, 0 + X), new Point(100, 80 - X));
}
catch (Exception ex)
{
MessageBox.Show(ex.Message);
}
}
private void timer1_Tick(object sender, EventArgs e)
{
try
{
timer1.Interval = 1;
pictureBox1.Invalidate();
}
catch (Exception ex)
{
MessageBox.Show(ex.Message);
}
}
答案 0 :(得分:1)
尝试在pictureBox1.Paint事件处理程序中移动绘图代码,然后调用pictureBox1.Invalidate whenewer,您需要重绘pictureBox1。这是如何绘制的真实方式。目前你正在闪烁,因为你每秒都会重绘你的pictureBox1,但那时你没有原始绘制。
byte X;
byte Y;
byte Z;
private void OnTimedEvent(object source, PaintEventArgs e)
{
try
{
if (comport.BytesToRead > 0)
{
X = comport.ReadByte();
Y = comport.ReadByte();
Z = comport.ReadByte();
}
pictureBox1.Invalidate();
}
catch (Exception ex)
{
MessageBox.Show(ex.Message);
}
}
private void timer1_Tick(object sender, EventArgs e)
{
try
{
timer1.Interval = 1;
pictureBox1.Invalidate();
}
catch (Exception ex)
{
MessageBox.Show(ex.Message);
}
}
private void pictureBox1_Paint(object sender, PaintEventArgs e)
{
Graphics g = e.Graphics;
Pen red = new Pen(Color.Red, 2);
g.TranslateTransform(100, 80 - X);
g.RotateTransform(120);
g.DrawLine(red, new Point(100, 0 + X), new Point(100, 80 - X));
}