[英]How to get active window title using Python in Mac?
我正在尝试编写 Python 脚本,它在 Mac OS 中使用 python 打印活动 window 的标题。
这是我的代码:
from AppKit import NSWorkspace
active_app_name = NSWorkspace.sharedWorkspace().frontmostApplication().localizedName()
print active_app_name
此代码仅打印应用程序的名称,如 Google chrome 或 firefox,但不打印标题。 如何获得 window 的标题?
这是我使用 Quartz API 在 Mac OS X 上使用 Python 查找活动应用程序名称和窗口标题时使用的内容。
首先,我们需要根据需要添加导入:
if sys.platform == "darwin":
import applescript
from AppKit import NSWorkspace
from Quartz import (
CGWindowListCopyWindowInfo,
kCGWindowListOptionOnScreenOnly,
kCGNullWindowID
)
然后我们可以通过下面的代码获取活动的应用程序名称和窗口标题:
def getActiveInfo(event_window_num):
try:
if sys.platform == "darwin":
app = NSWorkspace.sharedWorkspace().frontmostApplication()
active_app_name = app.localizedName()
options = kCGWindowListOptionOnScreenOnly
windowList = CGWindowListCopyWindowInfo(options, kCGNullWindowID)
windowTitle = 'Unknown'
for window in windowList:
windowNumber = window['kCGWindowNumber']
ownerName = window['kCGWindowOwnerName']
# geometry = window['kCGWindowBounds']
windowTitle = window.get('kCGWindowName', u'Unknown')
if windowTitle and (
event_window_num == windowNumber
or ownerName == active_app_name
):
# log.debug(
# 'ownerName=%s, windowName=%s, x=%s, y=%s, '
# 'width=%s, height=%s'
# % (window['kCGWindowOwnerName'],
# window.get('kCGWindowName', u'Unknown'),
# geometry['X'],
# geometry['Y'],
# geometry['Width'],
# geometry['Height']))
break
return _review_active_info(active_app_name, windowTitle)
if sys.platform == "win32":
(active_app_name, windowTitle) = _getActiveInfo_Win32()
return _review_active_info(active_app_name, windowTitle)
except:
log.error('Unexpected error: %s' % sys.exc_info()[0])
log.error('error line number: %s' % sys.exc_traceback.tb_lineno)
return 'Unknown', 'Unknown'
无法从NSWorkspace.sharedWorkspace().activeApplication()
访问应用程序标题。
但是您可以通过其 PID 找到当前窗口标题:
例如:
from AppKit import NSWorkspace
pid = NSWorkspace.sharedWorkspace().activeApplication()['NSApplicationProcessIdentifier']
然后使用下面的代码(它存储在kCGWindowOwnerPID
)找到正确的窗口,如下代码所示:
这是一个基于@JakeW 脚本的完整 shell 示例:
#!/usr/bin/python
# Prints list of windows in the current workspace.
import sys
if sys.platform == "darwin":
from AppKit import NSWorkspace
from Quartz import (
CGWindowListCopyWindowInfo,
kCGWindowListOptionOnScreenOnly,
kCGNullWindowID
)
if sys.platform == "darwin":
curr_app = NSWorkspace.sharedWorkspace().frontmostApplication()
curr_pid = NSWorkspace.sharedWorkspace().activeApplication()['NSApplicationProcessIdentifier']
curr_app_name = curr_app.localizedName()
options = kCGWindowListOptionOnScreenOnly
windowList = CGWindowListCopyWindowInfo(options, kCGNullWindowID)
for window in windowList:
pid = window['kCGWindowOwnerPID']
windowNumber = window['kCGWindowNumber']
ownerName = window['kCGWindowOwnerName']
geometry = window['kCGWindowBounds']
windowTitle = window.get('kCGWindowName', u'Unknown')
if curr_pid == pid:
print("%s - %s (PID: %d, WID: %d): %s" % (ownerName, windowTitle.encode('ascii','ignore'), pid, windowNumber, geometry))
elif sys.platform == "win32":
(active_app_name, windowTitle) = _getActiveInfo_Win32()
它将列出当前活动窗口的详细信息,包括其标题。
从 macOS 10.6 及更高版本开始,最好使用: frontmostApplication ,如果您想列出所有应用程序,您可以调用runningApplications方法。
您可以在https://developer.apple.com/documentation/appkit/nsworkspace#overview查看更多详细信息
例如:
from AppKit import NSWorkspace
NSWorkspace.sharedWorkspace().runningApplications() // for getting all applications
NSWorkspace.sharedWorkspace().frontmostApplication() // for active window
我尝试了@kenorb 的代码,但不幸的是它只能获取应用程序名称,但没有标题内容。
我终于找到了使用 AppleScript 执行此操作的方法:您可以在这里找到答案: MacOSX:获得最重要的 window 标题
首先创建一个appleScript GetNameAndTitleOfActiveWindow.scpt
global frontApp, frontAppName, windowTitle
set windowTitle to ""
tell application "System Events"
set frontApp to first application process whose frontmost is true
set frontAppName to name of frontApp
tell process frontAppName
tell (1st window whose value of attribute "AXMain" is true)
set windowTitle to value of attribute "AXTitle"
end tell
end tell
end tell
return {frontAppName, windowTitle}
你可以先在你的mac终端测试一下:
osascript GetNameAndTitleOfActiveWindow.scpt
然后在python中这样写:
title = subprocess.check_output(['osascript', 'GetNameAndTitleOfActiveWindow.scpt'])
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