背景:
摇杆-Razor锐机大师
视觉工作室 2019
C# 中的 Windows 窗体
我试图弄清楚如何按下物理操纵杆按钮,使用 C# 在 Visual Studio 中按下 Windows 窗体 UI 上的按钮。
物理摇杆代码:
private void joystickTimer_Tick_1(object sender, EventArgs e)
{
try
{
joystick.UpdateStatus();
joystickButtons = joystick.buttons;
if (joystick.Xaxis == 0)
output.Text+="Left\n";
if (joystick.Xaxis == 65534)
output.Text+="Right\n";
if (joystick.Yaxis == 0)
output.Text+="Up\n";
if (joystick.Yaxis == 65535)
output.Text+="Down\n";
for (int i = 0; i < joystickButtons.Length; i++)
{
// Assigning buttons
if (joystickButtons[0] == true)
output.Text += "Cross\n";
else if(joystickButtons[1] == true)
output.Text += "Circle\n";
else if(joystickButtons[2] == true)
output.Text += "Square\n";
else if (joystickButtons[3] == true)
output.Text += "Triangle\n";
Btn_Emergency_Stop_Click(sender,e); //Attempt at button press
//Btn_Emergency_Stop_Click.Click += joystickButtons[3];
break;
}
}
catch
{
joystickTimer.Enabled = false;
connectToJoystick(joystick);
}
}
要按下的虚拟按钮:
private void Btn_Emergency_Stop_Click(object sender, EventArgs e)
{
bool showed = false;
if (!showed)
{
MessageBox.Show("Emergency stop pressed!");
showed = true;
}
}
如果我在提供时遗漏了任何信息,请通知我,谢谢!