I have simple script that launches Windows File Explorer
import subprocess
subprocess.call(["start", "explorer.exe"],shell=True)
I'd like to check if Windows File Explorer is already opened so that I don't open another instance. How can I do it? Solutions without external libraries are preferred.
Found the solution
import win32gui
explorerWindows = []
def handler( hwnd, list ):
# only explorer windows have class 'CabinetWClass'
if 'CabinetWClass' in win32gui.GetClassName(hwnd):
list.append(hwnd)
win32gui.EnumWindows(handler, explorerWindows)
explorerOpened = len(explorerWindows) > 0
EDIT: easier method
import win32gui
explorerOpened = win32gui.FindWindow('CabinetWClass', None) != 0
Related
I am using this code to obtain current window
from typing import Optional
from ctypes import wintypes, windll, create_unicode_buffer
def getForegroundWindowTitle() -> Optional[str]:
hWnd = windll.user32.GetForegroundWindow()
length = windll.user32.GetWindowTextLengthW(hWnd)
buf = create_unicode_buffer(length + 1)
windll.user32.GetWindowTextW(hWnd, buf, length + 1)
return buf.value if buf.value else None
print(getForegroundWindowTitle())
output:
Videos
git
Downloads
Python check if current window is file explorer - Stack Overflow - Google Chrome
While google chrome tabs can be easily identified using this, problem is there's no way to know whether Videos, git, Downloads are windows folder (opened using file explorer).
So, is there a way to get the output in this format Videos - File Explorer ?
/ check whether the current window is a windows folder/file explorer window ?
From the same Question I modified the Code of Nuno Andre
https://stackoverflow.com/a/56572696/2532695
import ctypes
from ctypes import wintypes
import psutil
user32 = ctypes.windll.user32
h_wnd = user32.GetForegroundWindow()
pid = wintypes.DWORD()
user32.GetWindowThreadProcessId(h_wnd, ctypes.byref(pid))
print(psutil.Process(pid.value).name())
This one should do the trick, but you need psutil (pip install psutil).
You should see something like "Explorer.exe" if the active Window is an Explorer-Window.
This question already has answers here:
How to get Desktop location?
(9 answers)
Closed 6 months ago.
I am new to the os library and I was wondering how I could find the path for any user who uses windows and access their desktop directory using python. Thanks in advance!
You can do it by using os.environ mapping and add the Desktop path
import os
print(os.environ['USERPROFILE'] + '\Desktop')
Previous solutions won't work if the Desktop folder was manually changed by the user (to a OneDrive folder or whatever...).
This will work:
from win32com.shell import shell, shellcon
desktop = shell.SHGetFolderPath (0, shellcon.CSIDL_DESKTOP, 0, 0)
You can also try to query the registry.
import subprocess
import sys
import os
if sys.platform == "win32":
command = r'reg query "HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\Explorer\User Shell Folders" /v "Desktop"'
result = subprocess.run(command, stdout=subprocess.PIPE, text = True)
desktop = result.stdout.splitlines()[2].split()[2]
else:
desktop = os.path.expanduser("~/Desktop")
print(desktop)
#D:\Desktop
I already found this question that suggests to use os.path.expanduser(path) to get the user's home directory.
I would like to achieve the same with the "Downloads" folder. I know that this is possible in C#, yet I'm new to Python and don't know if this is possible here too, preferable platform-independent (Windows, Ubuntu).
I know that I just could do download_folder = os.path.expanduser("~")+"/Downloads/", yet (at least in Windows) it is possible to change the Default download folder.
from pathlib import Path
downloads_path = str(Path.home() / "Downloads")
This fairly simple solution (expanded from this reddit post) worked for me
import os
def get_download_path():
"""Returns the default downloads path for linux or windows"""
if os.name == 'nt':
import winreg
sub_key = r'SOFTWARE\Microsoft\Windows\CurrentVersion\Explorer\Shell Folders'
downloads_guid = '{374DE290-123F-4565-9164-39C4925E467B}'
with winreg.OpenKey(winreg.HKEY_CURRENT_USER, sub_key) as key:
location = winreg.QueryValueEx(key, downloads_guid)[0]
return location
else:
return os.path.join(os.path.expanduser('~'), 'downloads')
The GUID can be obtained from Microsoft's KNOWNFOLDERID docs
This can be expanded to work more generically other directories
For python3+ mac or linux
from pathlib import Path
path_to_download_folder = str(os.path.join(Path.home(), "Downloads"))
Correctly locating Windows folders is somewhat of a chore in Python. According to answers covering Microsoft development technologies, such as this one, they should be obtained using the Vista Known Folder API. This API is not wrapped by the Python standard library (though there is an issue from 2008 requesting it), but one can use the ctypes module to access it anyway.
Adapting the above answer to use the folder id for downloads shown here and combining it with your existing Unix code should result in code that looks like this:
import os
if os.name == 'nt':
import ctypes
from ctypes import windll, wintypes
from uuid import UUID
# ctypes GUID copied from MSDN sample code
class GUID(ctypes.Structure):
_fields_ = [
("Data1", wintypes.DWORD),
("Data2", wintypes.WORD),
("Data3", wintypes.WORD),
("Data4", wintypes.BYTE * 8)
]
def __init__(self, uuidstr):
uuid = UUID(uuidstr)
ctypes.Structure.__init__(self)
self.Data1, self.Data2, self.Data3, \
self.Data4[0], self.Data4[1], rest = uuid.fields
for i in range(2, 8):
self.Data4[i] = rest>>(8-i-1)*8 & 0xff
SHGetKnownFolderPath = windll.shell32.SHGetKnownFolderPath
SHGetKnownFolderPath.argtypes = [
ctypes.POINTER(GUID), wintypes.DWORD,
wintypes.HANDLE, ctypes.POINTER(ctypes.c_wchar_p)
]
def _get_known_folder_path(uuidstr):
pathptr = ctypes.c_wchar_p()
guid = GUID(uuidstr)
if SHGetKnownFolderPath(ctypes.byref(guid), 0, 0, ctypes.byref(pathptr)):
raise ctypes.WinError()
return pathptr.value
FOLDERID_Download = '{374DE290-123F-4565-9164-39C4925E467B}'
def get_download_folder():
return _get_known_folder_path(FOLDERID_Download)
else:
def get_download_folder():
home = os.path.expanduser("~")
return os.path.join(home, "Downloads")
A more complete module for retrieving known folders from Python is available on github.
Some linux distributions localize the name of the Downloads folder. E.g. after changing my locale to zh_TW, the Downloads folder became /home/user/下載. The correct way on linux distributions (using xdg-utils from freedesktop.org) is to call xdg-user-dir:
import subprocess
# Copy windows part from other answers here
try:
folder = subprocess.run(["xdg-user-dir", "DOWNLOAD"],
capture_output=True, text=True).stdout.strip("\n")
except FileNotFoundError: # if the command is missing
import os.path
folder = os.path.expanduser("~/Downloads") # fallback
Note that the use of capture_output requires Python ≥3.7.
If you already use GLib or don't mind adding more dependencies, see also
these approaches using packages.
For python3 on windows try:
import os
folder = os.path.join(os.path.join(os.environ['USERPROFILE']), 'folder_name')
print(folder)
I'm writing a PyGTK GUI application in Ubuntu to browse some images, and I'd like to open an image in the default image viewer application when it is double-clicked (like when it is opened in Nautilus).
How can I do it?
I don't know specifically using PyGTK but: xdg-open opens the default app for a file so running something like this should work:
import os
os.system('xdg-open ./img.jpg')
EDIT: I'd suggest using the subprocess module as in the comments. I'm not sure exactly how to use it yet so I just used os.system in the example to show xdg-open.
In GNU/Linux use xdg-open, in Mac use open, in Windows use start. Also, use subprocess, if not you risk to block your application when you call the external app.
This is my implementation, hope it helps: http://goo.gl/xebnV
import sys
import subprocess
import webbrowser
def default_open(something_to_open):
"""
Open given file with default user program.
"""
# Check if URL
if something_to_open.startswith('http') or something_to_open.endswith('.html'):
webbrowser.open(something_to_open)
return 0
ret_code = 0
if sys.platform.startswith('linux'):
ret_code = subprocess.call(['xdg-open', something_to_open])
elif sys.platform.startswith('darwin'):
ret_code = subprocess.call(['open', something_to_open])
elif sys.platform.startswith('win'):
ret_code = subprocess.call(['start', something_to_open], shell=True)
return ret_code
GTK (>= 2.14) has gtk_show_uri:
gtk.show_uri(screen, uri, timestamp)
Example usage:
gtk.show_uri(None, "file:///etc/passwd", gtk.gdk.CURRENT_TIME)
Related
How to open a file with the standard application?
I want to know what is the current url of active tab in running firefox instance from python module. Does FireFox have any API for this and does python know to work with it?
The most convenient way maybe insatll a firefox extension to open up a tcp service, then you can exchange info with firefox.
mozrepl can set up a telnet service, you can call js-like command to get info.
With telnetscript (http: //code.activestate.com/recipes/152043/), you can write:
import telnetscript
script = """rve
w content.location.href;
ru repl>
w repl.quit()
cl
"""
conn = telnetscript.telnetscript( '127.0.0.1', {}, 4242 )
ret = conn.RunScript( script.split( '\n' )).split( '\n' )
print ret[-2][6:]
If on windows you can use win32com
import win32clipboard
import win32com.client
shell = win32com.client.Dispatch("WScript.Shell")
shell.AppActivate('Some Application Title')
Then use shell.SendKeys to do a ctrl+l and a ctrl+c
Then read the string in the clipboard.
It's horkey though it will work, alternatly you can use something like AutoIt an compile the code to an exe that you can work with.
Hope this helps.