• <small id='evVnt'></small><noframes id='evVnt'>

    <legend id='evVnt'><style id='evVnt'><dir id='evVnt'><q id='evVnt'></q></dir></style></legend>
    <i id='evVnt'><tr id='evVnt'><dt id='evVnt'><q id='evVnt'><span id='evVnt'><b id='evVnt'><form id='evVnt'><ins id='evVnt'></ins><ul id='evVnt'></ul><sub id='evVnt'></sub></form><legend id='evVnt'></legend><bdo id='evVnt'><pre id='evVnt'><center id='evVnt'></center></pre></bdo></b><th id='evVnt'></th></span></q></dt></tr></i><div id='evVnt'><tfoot id='evVnt'></tfoot><dl id='evVnt'><fieldset id='evVnt'></fieldset></dl></div>
  • <tfoot id='evVnt'></tfoot>

        • <bdo id='evVnt'></bdo><ul id='evVnt'></ul>

      1. Python键盘库方向键问题

        时间:2023-09-12
            <i id='ccfjj'><tr id='ccfjj'><dt id='ccfjj'><q id='ccfjj'><span id='ccfjj'><b id='ccfjj'><form id='ccfjj'><ins id='ccfjj'></ins><ul id='ccfjj'></ul><sub id='ccfjj'></sub></form><legend id='ccfjj'></legend><bdo id='ccfjj'><pre id='ccfjj'><center id='ccfjj'></center></pre></bdo></b><th id='ccfjj'></th></span></q></dt></tr></i><div id='ccfjj'><tfoot id='ccfjj'></tfoot><dl id='ccfjj'><fieldset id='ccfjj'></fieldset></dl></div>
          1. <legend id='ccfjj'><style id='ccfjj'><dir id='ccfjj'><q id='ccfjj'></q></dir></style></legend>

            <tfoot id='ccfjj'></tfoot>

            <small id='ccfjj'></small><noframes id='ccfjj'>

              <tbody id='ccfjj'></tbody>
              • <bdo id='ccfjj'></bdo><ul id='ccfjj'></ul>

                  本文介绍了Python键盘库方向键问题的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

                  问题描述

                  我正在编写一个脚本,它会截取屏幕截图并解码以图像名称命名的特定按键,如下所示.我的问题是,当我按下左键盘箭头时,也按下了数字 4.我在谷歌或键盘库的文档中找不到任何东西.我正在使用 Windows 和 Python 3.6.5

                  I was writing a script, which takes a screenshot and decodes specific key presses in the name of the image as seen below. My problem is that when I press the left keyboard arrow, also the number 4 is pressed. I can't find anything on google or in the documentation of the keyboard library. I am using Windows and Python 3.6.5

                  (75,)
                  left arrow pressed
                  (5, 75)
                  4 pressed
                  

                  向下箭头也会发生同样的事情,但数字 3 是这样的.

                  The same thing happens with the down arrow, but with the number 3.

                  (80,)
                  down arrow pressed
                  (3, 80)
                  2 pressed
                  

                  代码:

                  from PIL import ImageGrab
                  import keyboard  # using module keyboard
                  import time
                  
                  keys = [
                      "down arrow",
                      "up arrow",
                      "left arrow",
                      "right arrow",
                      "w",
                      "s",
                      "a",
                      "d",
                      "1",
                      "2",
                      "3",
                      "4",
                      "q",
                      
                  

                  e",f"]

                  if __name__ == "__main__":
                      while True:
                          code = []
                          try:
                              for key in keys:
                                  if keyboard.is_pressed(key):
                                      print(keyboard.key_to_scan_codes(key))
                                      print(f"{key} pressed")
                                      code.append(1)
                                  else:
                                      code.append(0)
                                      
                              if keyboard.is_pressed('esc'):
                                  print(key + " pressed")
                                  break
                                  
                              c = "".join(map(str, code))
                              snapshot = ImageGrab.grab()
                              save_path = str(int(time.time()*1000)) + "-" + c + ".jpg"
                              snapshot.save("tmp\" + save_path)
                  
                          except:
                              break
                  

                  推荐答案

                  keyboard 模块对于此类实例有简单的解决方案,它们使用 event-triggered 激活而不是polling 在您的尝试中使用.

                  The keyboard module has simple solutions for instances like these, they use event-triggered activation rather than polling as is used in your attempt.

                  示例代码:

                  import keyboard
                  
                  def handleLeftKey(e):
                      if keyboard.is_pressed("4"):
                          print("left arrow was pressed w/ key 4")
                          # work your magic
                  
                  keyboard.on_press_key("left", handleLeftKey)
                  # self-explanitory: when the left key is pressed down then do something
                  
                  keyboard.on_release_key("left", handleLeftKey02)
                  # also self-explanitory: when the left key is released then do something
                  
                  # don't use both ...on_release & ...on_press or it will be
                  # triggered twice per key-use (1 up, 1 down)
                  

                  替换下面的代码并根据您的需要进行更改.

                  Replace the code below and change it to suit your needs.

                  if __name__ == "__main__":
                      while True:
                          code = []
                          try:
                              for key in keys:
                                  if keyboard.is_pressed(key):
                                      print(keyboard.key_to_scan_codes(key))
                                      print(f"{key} pressed")
                                      code.append(1)
                                  else:
                                      code.append(0)
                  

                  另一种更动态的方法如下所示:

                  Another, more dynamic approach would look like:

                  import keyboard
                  
                  keys = [
                      "down",
                      "up",
                      "left",
                      "right",
                      "w",
                      "s",
                      "a",
                      "d",
                      "1",
                      "2",
                      "3",
                      "4",
                      "q",
                      "e",
                      "f"
                  ]
                  
                  def kbdCallback(e):
                      found = False
                      for key in keys:
                          if key == keyboard.normalize_name(e.name):
                              print(f"{key} was pressed")
                              found = True
                              # work your magic
                  
                      if found == True:
                          if e.name == "left":
                              if keyboard.is_pressed("4"):
                                  print("4 & left arrow were pressed together!")
                                  # work your magic
                  
                  keyboard.on_press(kbdCallback)
                  # same as keyboard.on_press_key, but it does this for EVERY key
                  

                  我注意到的另一个问题是您使用 "left arrow" 而实际上它被识别为 "left" (至少在我的系统上,它可能会有所不同在你的,但我假设你希望它在所有系统上工作,所以使用 "left" 会更安全)

                  Another issue I noticed was that you were using "left arrow" when really it was recognized as "left" (at least on my system, it may be different on yours, but I assume you want it to work on all systems so it'd be safer using "left" instead)

                  您可以使用的最后一种方法是非常静态类型的并且没有动态功能,但可以在 "4+left""left+4"

                  The last method you could use is very statically typed and has no dynamic capabilities, but would work in the case of "4+left" or "left+4"

                  import keyboard
                  
                  def left4trigger:
                      print("the keys were pressed")
                  
                  keyboard.add_hotkey("4+left", left4trigger)
                  # works as 4+left or left+4 (all of the examples do)
                  

                  你看起来很聪明,可以从那里弄清楚其余的事情.

                  You seem smart enough to figure out the rest from there.

                  这篇关于Python键盘库方向键问题的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持html5模板网!

                  上一篇:如何在没有按键的情况下更改大写锁定状态 下一篇:TypeError:列表索引必须是整数,而不是 str Python

                  相关文章

                  最新文章

                  <legend id='VLibj'><style id='VLibj'><dir id='VLibj'><q id='VLibj'></q></dir></style></legend>
                • <i id='VLibj'><tr id='VLibj'><dt id='VLibj'><q id='VLibj'><span id='VLibj'><b id='VLibj'><form id='VLibj'><ins id='VLibj'></ins><ul id='VLibj'></ul><sub id='VLibj'></sub></form><legend id='VLibj'></legend><bdo id='VLibj'><pre id='VLibj'><center id='VLibj'></center></pre></bdo></b><th id='VLibj'></th></span></q></dt></tr></i><div id='VLibj'><tfoot id='VLibj'></tfoot><dl id='VLibj'><fieldset id='VLibj'></fieldset></dl></div>
                  <tfoot id='VLibj'></tfoot>

                        <bdo id='VLibj'></bdo><ul id='VLibj'></ul>
                    1. <small id='VLibj'></small><noframes id='VLibj'>