防止python睡眠模式(python上的Wakelock)
我如何在不使用其他操作系统(Ubuntu、Windows…)的情况下防止python上的睡眠模式,但在大多数情况下,我需要Linux解决方案 我正在制作一款能在大量时间内运行的应用程序。它使用了80%的CPU,所以用户只需启动这个应用程序,离开键盘。所以我想我需要像系统api或库这样的东西来锁定睡眠模式。我确信,它是存在的。例如,如果您在操作系统上打开任何视频播放器,您的(PC、笔记本电脑)将不会进入睡眠模式,在浏览器中也是如此防止python睡眠模式(python上的Wakelock),python,python-3.x,linux,windows,sleep-mode,Python,Python 3.x,Linux,Windows,Sleep Mode,我如何在不使用其他操作系统(Ubuntu、Windows…)的情况下防止python上的睡眠模式,但在大多数情况下,我需要Linux解决方案 我正在制作一款能在大量时间内运行的应用程序。它使用了80%的CPU,所以用户只需启动这个应用程序,离开键盘。所以我想我需要像系统api或库这样的东西来锁定睡眠模式。我确信,它是存在的。例如,如果您在操作系统上打开任何视频播放器,您的(PC、笔记本电脑)将不会进入睡眠模式,在浏览器中也是如此 另外,在Android()或Windows中也有同样的情况,我遇到
另外,在Android()或Windows中也有同样的情况,我遇到过类似的情况,即一个进程花了足够长的时间自行执行,Windows将休眠。为了克服这个问题,我写了一个脚本 下面的简单代码可以防止此问题。使用时,它会要求windows在脚本运行时不要睡眠。(在某些情况下,例如电池耗尽时,Windows将忽略您的请求。)
类窗口抑制器:
''在windows中防止操作系统睡眠/休眠;代码来源:
https://github.com/h3llrais3r/Deluge-PreventSuspendPlus/blob/master/preventsuspendplus/core.py
API文档:
https://msdn.microsoft.com/en-us/library/windows/desktop/aa373208(v=vs.85).aspx''
ES_连续=0x8000000
ES_系统_必需=0x00000001
定义初始化(自):
通过
def抑制(自):
导入ctypes
打印(“防止Windows进入睡眠状态”)
ctypes.windell.kernel32.SetThreadExecutionState(
3.e|u连续|\
WindowsInhibitor.ES_系统(需要)
def解除抑制(自我):
导入ctypes
打印(“允许Windows进入睡眠状态”)
ctypes.windell.kernel32.SetThreadExecutionState(
(不连续)
要运行脚本,只需执行以下操作:
导入操作系统
osSleep=无
#在Windows中,防止操作系统在运行时休眠
如果os.name='nt':
osSleep=WindowsInhibitor()
osSleep.inhibit()
#做慢动作
如果是睡眠:
osSleep.unhibit()
Keep.Awake中的逻辑将在ubuntu或任何运行Gnome(包括Unity在内的新旧版本)的Linux发行版上解决您的问题,该发行版可在Wayland和X上运行。它易于使用
我在这里发布了一个类似问题的解决方案:
因此,在逻辑上,请执行以下操作:
nohup ./keepawake.py -c 13 -r > /dev/null 2>&1 &
要作为后台服务运行并在确定用户空闲之前将15分钟(900秒)设置为用户活动空闲时间,请执行以下操作:
nohup ./keepawake.py -u 900 -r > /dev/null 2>&1 &
要作为后台服务运行并将最小网络流量设置为5KB(5120字节):
要作为后台服务运行并将计划设置为1小时后休眠/挂起(仅当用户活动、cpu和网络流量都被确定为空闲时,才设置此值):
要一次性运行上述所有设置(网络、CPU、用户空闲、睡眠计划),并将日志文件路径设置为“/home/$User/sleep/log/Keep.Awake/”,并提供详细输出:
nohup ./keepawake.py -s 5120 -c 13 -u 900 -w 3600 -l /home/$USER/sleep/log/Keep.Awake/ -v Detail -r > /dev/null 2>&1 &
基于我在互联网上找到的多种方法,我提出了以下模块。特别感谢@mishsx提供的
Windows
解决方案
使用它非常简单。您可以使用standby\u lock
或通过standbyllock
作为上下文管理器选择decorator方法:
## decorator
@standby_lock
def foo(*args, **kwargs):
# do something lazy here...
pass
## context manager
with StandbyLock():
# ...or do something lazy here instead
pass
执行foo
时,系统将保持清醒
注意:仍然有一些警告,因为Linux
可能需要sudo
特权,而且osx
(Darwin
)尚未测试
from functools import wraps
import platform
class MetaStandbyLock(type):
"""
"""
SYSTEM = platform.system()
def __new__(cls, name: str, bases: tuple, attrs: dict) -> type:
if not ('inhibit' in attrs and 'release' in attrs):
raise TypeError("Missing implementations for classmethods 'inhibit(cls)' and 'release(cls)'.")
else:
if name == 'StandbyLock':
cls._superclass = super().__new__(cls, name, bases, attrs)
return cls._superclass
if cls.SYSTEM.upper() in name.upper():
if not hasattr(cls, '_superclass'):
raise ValueError("Class 'StandbyLock' must be implemented.")
cls._superclass._subclass = super().__new__(cls, name, bases, attrs)
return cls._superclass._subclass
else:
return super().__new__(cls, name, bases, attrs)
class StandbyLock(metaclass=MetaStandbyLock):
"""
"""
_subclass = None
@classmethod
def inhibit(cls):
if cls._subclass is None:
raise OSError(f"There is no 'StandbyLock' implementation for OS '{platform.system()}'.")
else:
return cls._subclass.inhibit()
@classmethod
def release(cls):
if cls._subclass is None:
raise OSError(f"There is no 'StandbyLock' implementation for OS '{platform.system()}'.")
else:
return cls._subclass.release()
def __enter__(self, *args, **kwargs):
self.inhibit()
return self
def __exit__(self, *args, **kwargs):
self.release()
class WindowsStandbyLock(StandbyLock):
"""
"""
ES_CONTINUOUS = 0x80000000
ES_SYSTEM_REQUIRED = 0x00000001
INHIBIT = ES_CONTINUOUS | ES_SYSTEM_REQUIRED
RELEASE = ES_CONTINUOUS
@classmethod
def inhibit(cls):
import ctypes
ctypes.windll.kernel32.SetThreadExecutionState(cls.INHIBIT)
@classmethod
def release(cls):
import ctypes
ctypes.windll.kernel32.SetThreadExecutionState(cls.RELEASE)
class LinuxStandbyLock(metaclass=MetaStandbyLock):
"""
"""
COMMAND = 'systemctl'
ARGS = ['sleep.target', 'suspend.target', 'hibernate.target', 'hybrid-sleep.target']
@classmethod
def inhibit(cls):
import subprocess
subprocess.run([cls.COMMAND, 'mask', *cls.ARGS])
@classmethod
def release(cls):
import subprocess
subprocess.run([cls.COMMAND, 'unmask', *cls.ARGS])
class DarwinStandbyLock(metaclass=MetaStandbyLock):
"""
"""
COMMAND = 'caffeinate'
BREAK = b'\003'
_process = None
@classmethod
def inhibit(cls):
from subprocess import Popen, PIPE
cls._process = Popen([cls.COMMAND], stdin=PIPE, stdout=PIPE)
@classmethod
def release(cls):
cls._process.stdin.write(cls.BREAK)
cls._process.stdin.flush()
cls._process.stdin.close()
cls._process.wait()
def standby_lock(callback):
""" standby_lock(callable) -> callable
This decorator guarantees that the system will not enter standby mode while 'callable' is running.
"""
@wraps(callback)
def new_callback(*args, **kwargs):
with StandbyLock():
return callback(*args, **kwargs)
return new_callback
创建示例TK应用程序以保持windows处于唤醒状态
import tkinter as tk
import ctypes
import sys
def display_on():
print("Always On")
ctypes.windll.kernel32.SetThreadExecutionState(0x80000002)
def display_reset():
ctypes.windll.kernel32.SetThreadExecutionState(0x80000000)
sys.exit(0)
root = tk.Tk()
root.geometry("200x60")
root.title("Display App")
frame = tk.Frame(root)
frame.pack()
button = tk.Button(frame,
text="Quit",
fg="red",
command=display_reset)
button.pack(side=tk.LEFT)
slogan = tk.Button(frame,
text="Always ON",
command=display_on)
slogan.pack(side=tk.LEFT)
root.mainloop()
在谷歌搜索解决方案时,没有可用的包,所以我决定将其打包以将其放入PyPI:。我得到了跨平台支持的PRs,目前wakepy支持Windows、Linux和macOS CLI 使用可选的
-s
标志也将保持屏幕打开
Python API
从wakepy导入集\u keepawake,取消设置\u keepawake
设置唤醒(保持屏幕唤醒=错误)
#做需要很长时间的事情
unset_keepawake()
这是什么类型的应用程序?是否使用tkinter
?您需要提供帮助您的人的详细信息。@mishsx不,它不使用tkinter,只是在终端上运行的简单应用程序。我不希望有跨平台的解决方案可用。您可能需要自己实现一个特定于平台的解决方案,该解决方案使用WakeLock
或SetThreadExecutionState
或其他东西,具体取决于运行它的操作系统。我想您提到了windows
。这就是为什么我的答案是专门为windows
量身定做的。对于Linux,你可以参考-很棒的解决方案!你介意将其作为PyPI包发布吗?这非常好。您测试过linux/osx版本了吗?decorator和context manager方法中的一个警告是,如果您碰巧两次使用它,即使是在不同的进程中,第一次退出也会移除“锁”。谢谢您的建议!我的方法确实非常简单,考虑到多进程的情况,我没有想到这一点。我想我会按照@BenHagen的要求,考虑到多个应用程序,很快构建一个pypi包
from functools import wraps
import platform
class MetaStandbyLock(type):
"""
"""
SYSTEM = platform.system()
def __new__(cls, name: str, bases: tuple, attrs: dict) -> type:
if not ('inhibit' in attrs and 'release' in attrs):
raise TypeError("Missing implementations for classmethods 'inhibit(cls)' and 'release(cls)'.")
else:
if name == 'StandbyLock':
cls._superclass = super().__new__(cls, name, bases, attrs)
return cls._superclass
if cls.SYSTEM.upper() in name.upper():
if not hasattr(cls, '_superclass'):
raise ValueError("Class 'StandbyLock' must be implemented.")
cls._superclass._subclass = super().__new__(cls, name, bases, attrs)
return cls._superclass._subclass
else:
return super().__new__(cls, name, bases, attrs)
class StandbyLock(metaclass=MetaStandbyLock):
"""
"""
_subclass = None
@classmethod
def inhibit(cls):
if cls._subclass is None:
raise OSError(f"There is no 'StandbyLock' implementation for OS '{platform.system()}'.")
else:
return cls._subclass.inhibit()
@classmethod
def release(cls):
if cls._subclass is None:
raise OSError(f"There is no 'StandbyLock' implementation for OS '{platform.system()}'.")
else:
return cls._subclass.release()
def __enter__(self, *args, **kwargs):
self.inhibit()
return self
def __exit__(self, *args, **kwargs):
self.release()
class WindowsStandbyLock(StandbyLock):
"""
"""
ES_CONTINUOUS = 0x80000000
ES_SYSTEM_REQUIRED = 0x00000001
INHIBIT = ES_CONTINUOUS | ES_SYSTEM_REQUIRED
RELEASE = ES_CONTINUOUS
@classmethod
def inhibit(cls):
import ctypes
ctypes.windll.kernel32.SetThreadExecutionState(cls.INHIBIT)
@classmethod
def release(cls):
import ctypes
ctypes.windll.kernel32.SetThreadExecutionState(cls.RELEASE)
class LinuxStandbyLock(metaclass=MetaStandbyLock):
"""
"""
COMMAND = 'systemctl'
ARGS = ['sleep.target', 'suspend.target', 'hibernate.target', 'hybrid-sleep.target']
@classmethod
def inhibit(cls):
import subprocess
subprocess.run([cls.COMMAND, 'mask', *cls.ARGS])
@classmethod
def release(cls):
import subprocess
subprocess.run([cls.COMMAND, 'unmask', *cls.ARGS])
class DarwinStandbyLock(metaclass=MetaStandbyLock):
"""
"""
COMMAND = 'caffeinate'
BREAK = b'\003'
_process = None
@classmethod
def inhibit(cls):
from subprocess import Popen, PIPE
cls._process = Popen([cls.COMMAND], stdin=PIPE, stdout=PIPE)
@classmethod
def release(cls):
cls._process.stdin.write(cls.BREAK)
cls._process.stdin.flush()
cls._process.stdin.close()
cls._process.wait()
def standby_lock(callback):
""" standby_lock(callable) -> callable
This decorator guarantees that the system will not enter standby mode while 'callable' is running.
"""
@wraps(callback)
def new_callback(*args, **kwargs):
with StandbyLock():
return callback(*args, **kwargs)
return new_callback
import tkinter as tk
import ctypes
import sys
def display_on():
print("Always On")
ctypes.windll.kernel32.SetThreadExecutionState(0x80000002)
def display_reset():
ctypes.windll.kernel32.SetThreadExecutionState(0x80000000)
sys.exit(0)
root = tk.Tk()
root.geometry("200x60")
root.title("Display App")
frame = tk.Frame(root)
frame.pack()
button = tk.Button(frame,
text="Quit",
fg="red",
command=display_reset)
button.pack(side=tk.LEFT)
slogan = tk.Button(frame,
text="Always ON",
command=display_on)
slogan.pack(side=tk.LEFT)
root.mainloop()
python -m wakepy [-s]