让马达前进一段时间？覆盆子皮

Question

我最近购买了4 4tronix的Pi2go lite。作为提供的库的一部分，您可以向前或向后移动，但这是永远发生的，我希望，在键盘上按下就会发生，然后当按钮不再按下时，就不会发生任何动作。在做了一些研究后，这并不是一个众所周知的过程，或者根本可以做到，但是我的解决办法是按下键，操作只需半秒钟，就可以模拟按住键的过程。如何才能做到这一点？谢谢你的进阶。下面是由GitHub中的raspberry Pi Guy提供的代码，但是一旦按下'W‘键，就很难控制它，因为当您松开键时，它不会停止。

import pi2go, time

# Reading a button press from your keyboard, don't worry about this too much!
import sys
import tty
import termios

UP = 0
DOWN = 1
RIGHT = 2
LEFT = 3

def readchar():
fd = sys.stdin.fileno()
old_settings = termios.tcgetattr(fd)
try:
    tty.setraw(sys.stdin.fileno())
    ch = sys.stdin.read(1)
finally:
    termios.tcsetattr(fd, termios.TCSADRAIN, old_settings)
if ch == '0x03':
    raise KeyboardInterrupt
return ch

def readkey(getchar_fn=None):
getchar = getchar_fn or readchar
c1 = getchar()
if ord(c1) != 0x1b:
    return c1
c2 = getchar()
if ord(c2) != 0x5b:
    return c1
c3 = getchar()
return ord(c3) - 65  # 0=Up, 1=Down, 2=Right, 3=Left arrows


speed = 30

pi2go.init()

try:
while True:
    keyp = readkey()
    if keyp == 'w' or keyp == UP:
        pi2go.forward(speed)
        print 'Forward', speed
    elif keyp == 's' or keyp == DOWN:
        pi2go.reverse(speed)
        print 'Backward', speed
    elif keyp == 'd' or keyp == RIGHT:
        pi2go.spinRight(speed)
        print 'Spin Right', speed
    elif keyp == 'a' or keyp == LEFT:
        pi2go.spinLeft(speed)
        print 'Spin Left', speed

    elif keyp == '.' or keyp == '>':
        speed = min(100, speed+10)
        print 'Speed+', speed
    elif keyp == ',' or keyp == '<':
        speed = max (0, speed-10)
        print 'Speed-', speed

    elif keyp == ' ':
        pi2go.stop()
        print 'Stop'
    elif ord(keyp) == 3:
        break


except KeyboardInterrupt:
   pi2go.cleanup()

Answer 1

这是我的密码。

elif key == 'w':
        speed = min(100, speed+10)
        pi2go.forward(speed)
        print 'Forward', speed
        time.sleep(.100)
    pi2go.stop()
elif key == 's':
    speed = min(100, speed+10)
        pi2go.reverse(speed)
        print 'Reverse', speed
        time.sleep(.100)
    pi2go.stop()
elif key == 'd':
    speed = min(100, speed+10)
        pi2go.spinRight(speed)
        print 'Spin Right', speed
        time.sleep(.100)
    pi2go.stop()
elif key == 'a':
    speed = min(100, speed+10)
        pi2go.spinLeft(speed)
        print 'Spin Left', speed
    time.sleep(.100)
    pi2go.stop()
    elif key == 'b':
        speed = min(100, speed+10)
        print 'Speed+', speed
    elif key == 'v':
        speed = max (0, speed-10)
        print 'Speed-', speed
    elif key == ' ':
        pi2go.stop()
        print 'Stop'
    elif ord(key) == 3:
    break

Answer 2

最简单的方法是使用检测按键的Gtk gui制作一个简单的python程序。

import gtk

class DK:
    def __init__(self):
        window = gtk.Window()
        window.add_events(gtk.gdk.KEY_PRESS_MASK)
        window.connect("key-press-event", self.forward)
        window.show()
    def forward(self, widget, event):
        if event.keyval == 119: #W key
            print "do what you want with this"
            forward(100)
        else:
            stop()

DK()
gtk.main()

Answer 3

我有从同一个站点购买的初始化，您可以在pi2go.forward(速度)下的行中添加一个pi2go.forward(0.5秒)，或者在键按下或在pi2go.py中的实际前向函数中添加一个