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I need to be able to open a document using its default application in Windows and Mac OS. Basically, I want to do the same thing that happens when you double-click on the document icon in Explorer or Finder. What is the best way to do this in Python?
Use the subprocess module available on Python 2.4+, not os.system(), so you don't have to deal with shell escaping.
import subprocess, os, platform
if platform.system() == 'Darwin': # macOS
subprocess.call(('open', filepath))
elif platform.system() == 'Windows': # Windows
os.startfile(filepath)
else: # linux variants
subprocess.call(('xdg-open', filepath))
The double parentheses are because subprocess.call() wants a sequence as its first argument, so we're using a tuple here. On Linux systems with Gnome there is also a gnome-open command that does the same thing, but xdg-open is the Free Desktop Foundation standard and works across Linux desktop environments.
open and start are command-interpreter things for Mac OS/X and Windows respectively, to do this.
To call them from Python, you can either use subprocess module or os.system().
Here are considerations on which package to use:
You can call them via os.system, which works, but...
Escaping: os.system only works with filenames that don't have any spaces or other shell metacharacters in the pathname (e.g. A:\abc\def\a.txt), or else these need to be escaped. There is shlex.quote for Unix-like systems, but nothing really standard for Windows. Maybe see also python, windows : parsing command lines with shlex
MacOS/X: os.system("open " + shlex.quote(filename))
Windows: os.system("start " + filename) where properly speaking filename should be escaped, too.
You can also call them via subprocess module, but...
For Python 2.7 and newer, simply use
subprocess.check_call(['open', filename])
In Python 3.5+ you can equivalently use the slightly more complex but also somewhat more versatile
subprocess.run(['open', filename], check=True)
If you need to be compatible all the way back to Python 2.4, you can use subprocess.call() and implement your own error checking:
try:
retcode = subprocess.call("open " + filename, shell=True)
if retcode < 0:
print >>sys.stderr, "Child was terminated by signal", -retcode
else:
print >>sys.stderr, "Child returned", retcode
except OSError, e:
print >>sys.stderr, "Execution failed:", e
Now, what are the advantages of using subprocess?
Security: In theory, this is more secure, but in fact we're needing to execute a command line one way or the other; in either environment, we need the environment and services to interpret, get paths, and so forth. In neither case are we executing arbitrary text, so it doesn't have an inherent "but you can type 'filename ; rm -rf /'" problem, and if the file name can be corrupted, using subprocess.call gives us little additional protection.
Error handling: It doesn't actually give us any more error detection, we're still depending on the retcode in either case; but the behavior to explicitly raise an exception in the case of an error will certainly help you notice if there is a failure (though in some scenarios, a traceback might not at all be more helpful than simply ignoring the error).
Spawns a (non-blocking) subprocess: We don't need to wait for the child process, since we're by problem statement starting a separate process.
To the objection "But subprocess is preferred." However, os.system() is not deprecated, and it's in some sense the simplest tool for this particular job. Conclusion: using os.system() is therefore also a correct answer.
A marked disadvantage is that the Windows start command requires you to pass in shell=True which negates most of the benefits of using subprocess.
I prefer:
os.startfile(path, 'open')
Note that this module supports filenames that have spaces in their folders and files e.g.
A:\abc\folder with spaces\file with-spaces.txt
(python docs) 'open' does not have to be added (it is the default). The docs specifically mention that this is like double-clicking on a file's icon in Windows Explorer.
This solution is windows only.
Just for completeness (it wasn't in the question), xdg-open will do the same on Linux.
import os
import subprocess
def click_on_file(filename):
'''Open document with default application in Python.'''
try:
os.startfile(filename)
except AttributeError:
subprocess.call(['open', filename])
If you have to use an heuristic method, you may consider webbrowser.
It's standard library and despite of its name it would also try to open files:
Note that on some platforms, trying to open a filename using this
function, may work and start the operating system’s associated
program. However, this is neither supported nor portable.
(Reference)
I tried this code and it worked fine in Windows 7 and Ubuntu Natty:
import webbrowser
webbrowser.open("path_to_file")
This code also works fine in Windows XP Professional, using Internet Explorer 8.
If you want to go the subprocess.call() way, it should look like this on Windows:
import subprocess
subprocess.call(('cmd', '/C', 'start', '', FILE_NAME))
You can't just use:
subprocess.call(('start', FILE_NAME))
because start is not an executable but a command of the cmd.exe program. This works:
subprocess.call(('cmd', '/C', 'start', FILE_NAME))
but only if there are no spaces in the FILE_NAME.
While subprocess.call method enquotes the parameters properly, the start command has a rather strange syntax, where:
start notes.txt
does something else than:
start "notes.txt"
The first quoted string should set the title of the window. To make it work with spaces, we have to do:
start "" "my notes.txt"
which is what the code on top does.
Start does not support long path names and white spaces. You have to convert it to 8.3 compatible paths.
import subprocess
import win32api
filename = "C:\\Documents and Settings\\user\\Desktop\file.avi"
filename_short = win32api.GetShortPathName(filename)
subprocess.Popen('start ' + filename_short, shell=True )
The file has to exist in order to work with the API call.
os.startfile(path, 'open') under Windows is good because when spaces exist in the directory, os.system('start', path_name) can't open the app correctly and when the i18n exist in the directory, os.system needs to change the unicode to the codec of the console in Windows.
I am pretty late to the lot, but here is a solution using the windows api. This always opens the associated application.
import ctypes
shell32 = ctypes.windll.shell32
file = 'somedocument.doc'
shell32.ShellExecuteA(0,"open",file,0,0,5)
A lot of magic constants. The first zero is the hwnd of the current program. Can be zero. The other two zeros are optional parameters (parameters and directory). 5 == SW_SHOW, it specifies how to execute the app.
Read the
ShellExecute API docs for more info.
Here is the answer from Nick, adjusted slightly for WSL:
import os
import sys
import logging
import subprocess
def get_platform():
if sys.platform == 'linux':
try:
proc_version = open('/proc/version').read()
if 'Microsoft' in proc_version:
return 'wsl'
except:
pass
return sys.platform
def open_with_default_app(filename):
platform = get_platform()
if platform == 'darwin':
subprocess.call(('open', filename))
elif platform in ['win64', 'win32']:
os.startfile(filename.replace('/','\\'))
elif platform == 'wsl':
subprocess.call('cmd.exe /C start'.split() + [filename])
else: # linux variants
subprocess.call(('xdg-open', filename))
If you want to specify the app to open the file with on Mac OS X, use this:
os.system("open -a [app name] [file name]")
On windows 8.1, below have worked while other given ways with subprocess.call fails with path has spaces in it.
subprocess.call('cmd /c start "" "any file path with spaces"')
By utilizing this and other's answers before, here's an inline code which works on multiple platforms.
import sys, os, subprocess
subprocess.call(('cmd /c start "" "'+ filepath +'"') if os.name is 'nt' else ('open' if sys.platform.startswith('darwin') else 'xdg-open', filepath))
On mac os you can call open:
import os
os.open("open myfile.txt")
This would open the file with TextEdit, or whatever app is set as default for this filetype.
I think you might want to open file in editor.
For Windows
subprocess.Popen(["notepad", filename])
For Linux
subprocess.Popen(["text-editor", filename])
I built a small library combining the best answers here for cross-platform support:
$ pip install universal-startfile
then launch a file or URL:
from startfile import startfile
startfile("~/Downloads/example.png")
startfile("http://example.com")
I was getting an error when calling my open file() function. I was following along with a guide but the guide was written in windows while I'm on Linux. So the os.statrfile method wasn't working for me. I was able to alleviate this problem by doing the following:
Import libraries
import sys, os, subprocess
import tkinter
import tkinter.filedioalog as fd
import tkinter.messagebox as mb
After the lib imports I then called the subprocess method for opening a file in unix based OS which is "xdg-open" and the file that will be opened.
def open_file():
file = fd.askopenfilename(title='Choose a file of any type', filetypes=[('All files', "*.*")])
subprocess.call(['xdg-open', file])
This question already has answers here:
Open document with default OS application in Python, both in Windows and Mac OS
(17 answers)
Closed 8 years ago.
I'm wondering how to open files in programs such as Notepad and Picture Viewer depending on the extension the file has. I'm using Python 3.3 on Windows.
I've done some research and people have mentioned a module named Image, but when I try and import this module I get an ImportError.
Here's what I have so far:
def openFile():
fileName = listbox_1.get(ACTIVE)
if fileName.endswith(".jpg"):
fileName.open()
I will also have HTML and JSON files that I will need to open in Notepad.
On Windows you could use os.startfile() to open a file using default application:
import os
os.startfile(filename)
There is no shutil.open() that would do it cross-platform. The close approximation is webbrowser.open():
import webbrowser
webbrowser.open(filename)
that might use automatically open command on OS X, os.startfile() on Windows, xdg-open or similar on Linux.
If you want to run a specific application then you could use subprocess module e.g., Popen() allows to start a program without waiting for it to complete:
import subprocess
p = subprocess.Popen(["notepad.exe", fileName])
# ... do other things while notepad is running
returncode = p.wait() # wait for notepad to exit
There are many ways to use the subprocess module to run programs e.g., subprocess.check_call(command) blocks until the command finishes and raises an exception if the command finishes with a nonzero exit code.
Use this to open any file with the default program:
import os
def openFile():
fileName = listbox_1.get(ACTIVE)
os.system("start " + fileName)
If you really want to use a certain program, such as notepad, you can do it like this:
import os
def openFile():
fileName = listbox_1.get(ACTIVE)
os.system("notepad.exe " + fileName)
Also if you need some if checks before opening the file, feel free to add them. This only shows you how to open the file.
Expanding on FatalError's suggestion with an example.
One additional benefit of using subprocessing rather than os.system is that it uses the same syntax cross-platform (os.system on Windows requires a "start" at the beginning, whereas OS X requires an "open". Not a huge deal, but one less thing to remember).
Opening a file with subprocess.call.
All you need to do to launch a program is call subprocess.call() and pass in a list of arguments where the first is the path to the program, and the rest are additional arguments that you want to supply to the program you're launching.
For instance, to launch Notepad.exe
import subprocess
path_to_notepad = 'C:\\Windows\\System32\\notepad.exe'
path_to_file = 'C:\\Users\\Desktop\\hello.txt'
subprocess.call([path_to_notepad, path_to_file])
Passing multiple arguments and paths is equally as simple. Just add additional items to the list.
Launching with multiple arguments
This, for example, launches a JAR file using a specific copy of the Java runtime environment.
import subprocess
import os
current_path = os.getcwd()
subprocess.call([current_path + '/contents/home/bin/java', # Param 1
'-jar', #Param2
current_path + '/Whoo.jar']) #param3
Argument 1 targets the program I want to launch. Argument2 supplies an argument to that program telling it that it's going to run a JAR, and finally Argument3 tells the target program where to find the file to open.
I want to run subprocess.call (or any other subprocess function) with a cwd that is really long (longer than 260 characters). I am using a recent Windows 10.
I read here that in order to support long paths, you either have to set a registry key or add \\?\ in front of the path. I did both.
It works if the executable I want to run has a long path. But it does not work if the cwd is a long path:
import os, sys
import subprocess
PATH_TO_WRITE_EXE = r"C:\Windows\write.exe"
print(os.path.isfile(PATH_TO_WRITE_EXE))
# error:
my_cwd = "\\\\?\\C:\\a\\really\\long\\path\\a\\really\\long\\path\\a\\really\\long\\path\\a\\really\\long\\path\\a\\really\\long\\path\\a\\really\\long\\path"
print(os.path.isdir(my_cwd))
# no error:
#my_cwd = "\\\\?\\C:\\a\\not\\so\\long\\path"
#print(os.path.isdir(my_cwd))
o = subprocess.call([PATH_TO_WRITE_EXE], timeout=None, cwd=my_cwd)
print(o)
Note that os.path.isdir() returns True on both the short and the long path.
How can I use a long path as cwd on Windows 10?
This is just a proof of concept, and you will would probably want to do something different, but here is "an" answer that will work if you run your script as admin (which is a bad idea... maybe? (depending on the scope)).
import os
import win32file
import subprocess
long_path = '\\\\?\\C:\\Temp\\3d\\RsTYjcEwAA26\\aFmtI0e\\v\\ZZ7\\AWgMBtUP5\\JRGtyZXFj2\\f2rqXnYX3yJ4\\39X11fdRbYEA\\NtPySHqx\\htyDGAtZWv8NDK\\d2VRFFJPuBUVXET\\2QSlBOlMkgO8h\\mES\\sQfPZ1nBAKZNIogOb\\wyGm5Z0RwHV\\n54Si\\2BqDwGnK6TOxjs2P\\p4SnwEre4\\KQzs1NXu5QEZcuZOIct\\YrMfsGq5g5gnMN69ko\\QFIq\\J4IKjZ3vxNrC\\OVDWtz\\Jp1H0M1UclBJqeBuX\\bjN7dA\\lCFmKDg7G1\\OhYtim9AxgX9Bm9\\vrLaaL\\KLvkkJeI0ofdwb\\Es\\ZJi3Q54oIXbQ8NOi10\\VR\\HH3\\O\\5\\zn7\\7EKj96k3BC\\8Q1OqP\\FdX8RLhl1Ce\\mPG\\OtmJWbzFk\\AheYZ8Ypwo\\085dmIvlrg\\Y8tmeJt\\cDYqXPq\\G6EYcqVXaLxv\\XXq6tIfVDhv8WoF\\xM\\PCYkVfFT1Uam9N0e\\G9PfRMOv\\GUWbc6eot4aEuVQIMd\\0NMEq9iDzqgLGOJx09\\HpUN5rBfaq9\\Ve\\Tp0E\\wpXyehjLotcDa4x\\HlBy1LMD83sxzQF0\\1\\NH1be07kdb61aomggou\\D0\\SF\\n0NLPfYTEh\\3k1AooSmx4y2CS6Mrp\\sgAd9N6x1v31jZ\\hof1X6XGdBAU8\\zyzuxVDHuX54PiYW0\\nVJc8\\r\\ukx63N2kY\\6gf8dhUTYad\\L8w4JWwZq\\iixvKOcH13FXljY5D\\zgGuUlXFH1hd\\2Ykw1isPKOKXR4Osv1U\\ncmRIMWf\\i1ioae6pqcsfDsI\\AU7fhnbPCtpaOphXL\\Vxn\\gJFO1o6JAMBmBWP\\8EKwcdps\\JGd\\SgfwKrEd5\\pGSxLp\\DuA8th1\\YRx8u0LF8Cgs6JEfwA\\dIV0Ay\\PEc2\\CSli\\nyRaOzgBtLuM8S09st\\vMd9Ctvc8c6\\2\\H5tpHh\\K6TsNhH\\jXmon6\\BqvEDk\\gsMH20FxEgwlY'
file_name = "test_file"
symlink_name = "C:\\Temp\\long_link"
os.makedirs(long_path)
with open(os.path.join(long_path, file_name), "w") as file:
file.write("I'm some test data in a long path!")
win32file.CreateSymbolicLink(symlink_name, long_path, 0x3)
subprocess.call("type %s" % file_name, shell=True, timeout=None, cwd=symlink_name)
I'm some test data in a long path!0
As #eryksun mentioned in the comments: Creating a symbolic link requires SeCreateSymbolicLinkPrivilege, which by default is only assigned to elevated administrators. (However, it can be explicitly assigned to users and groups.) If os.symlink raises OSError, you can create a junction via _winapi.CreateJunction or CMD's mklink /j command.
Finally here is another answer which should enable the same behavior if you create a junction. I have not tested this answer in conjunction with your question, but it should work.
Edit: If you're running >= Python 3.5 you can use the CreateJunction call to replace the symlink above.
import _winapi
_winapi.CreateJunction(source, target)
Is it possible to send a list of related commands using os.system() in python? I mean if I want to change current directory to a specific directory and then have a list of contents, how can I do it? (I don't want to use dir "path"- I want to do both changing current dir, and listing the directories)
Note : It was just an example, I want to know how I can send multiple commands! (Some related commands in a row)
Note : Python 3.2
os.system uses the local system shell. You can do it as #Rwaing suggests on many unixy shells but not other places like windows. A better option is subprocess.call and the cwd (current working directory) param
import subprocess
subprocess.call('dir', shell=True, cwd='somepath')
As others have mentioned, if all you really want to do is get a list of the files, the existin python api does it quite well.
Edit: sending mutiple commands
One way to send multiple commands is to pump them into the child shell's stdin. Its shell dependant, but here's a windows example:
import os
import subprocess as subp
p=subp.Popen('cmd.exe', shell=True, stdin=subp.PIPE)
p.stdin.write("""dir
cd "\\program files"
dir
""")
p.stdin.write('exit' + os.linesep)
p.wait()
del p
print 'done'
No need for system calls here. os functions chdir and listdir will change your current directory and list the files in a directory respectively.
have a look at os.listdir(path):
https://docs.python.org/3/library/os.html#os.listdir
example:
import os
entries = os.list.dir('/home/foo')
I just started working on Python, and I have been trying to run an outside executable from Python.
I have an executable for a program written in Fortran. Let’s say the name for the executable is flow.exe. And my executable is located in C:\Documents and Settings\flow_model. I tried both os.system and popen commands, but so far I couldn't make it work. The following code seems like it opens the command window, but it wouldn't execute the model.
# Import system modules
import sys, string, os, arcgisscripting
os.system("C:/Documents and Settings/flow_model/flow.exe")
How can I fix this?
If using Python 2.7 or higher (especially prior to Python 3.5) you can use the following:
import subprocess
subprocess.call(args, *, stdin=None, stdout=None, stderr=None, shell=False)
Runs the command described by args. Waits for command to complete, then returns the returncode attribute.
subprocess.check_call(args, *, stdin=None, stdout=None, stderr=None, shell=False)
Runs command with arguments. Waits for command to complete. If the return code was zero then returns, otherwise raises CalledProcessError. The CalledProcessError object will have the return code in the returncode attribute
Example: subprocess.check_call([r"C:\pathToYourProgram\yourProgram.exe", "your", "arguments", "comma", "separated"])
In regular Python strings, the \U character combination signals a
extended Unicode code point escape.
Here is the link to the documentation: http://docs.python.org/3.2/library/subprocess.html
For Python 3.5+ you can now use run() in many cases: https://docs.python.org/3.5/library/subprocess.html#subprocess.run
Those whitespaces can really be a bother. Try os.chdir('C:/Documents\ and\ Settings/') followed by relative paths for os.system, subprocess methods, or whatever...
If best-effort attempts to bypass the whitespaces-in-path hurdle keep failing, then my next best suggestion is to avoid having blanks in your crucial paths. Couldn't you make a blanks-less directory, copy the crucial .exe file there, and try that? Are those havoc-wrecking space absolutely essential to your well-being...?
The simplest way is:
import os
os.startfile("C:\Documents and Settings\flow_model\flow.exe")
It works; I tried it.
I'd try inserting an 'r' in front of your path if I were you, to indicate that it's a raw string - and then you won't have to use forward slashes. For example:
os.system(r"C:\Documents and Settings\flow_model\flow.exe")
Your usage is correct. I bet that your external program, flow.exe, needs to be executed in its directory, because it accesses some external files stored there.
So you might try:
import sys, string, os, arcgisscripting
os.chdir('c:\\documents and settings\\flow_model')
os.system('"C:\\Documents and Settings\\flow_model\\flow.exe"')
(Beware of the double quotes inside the single quotes...)
Use subprocess, it is a smaller module so it runs the .exe quicker.
import subprocess
subprocess.Popen([r"U:\Year 8\kerbal space program\KSP.exe"])
By using os.system:
import os
os.system(r'"C:/Documents and Settings/flow_model/flow.exe"')
Try
import subprocess
subprocess.call(["C:/Documents and Settings/flow_model/flow.exe"])
If it were me, I'd put the EXE file in the root directory (C:) and see if it works like that. If so, it's probably the (already mentioned) spaces in the directory name. If not, it may be some environment variables.
Also, try to check you stderr (using an earlier answer by int3):
import subprocess
process = subprocess.Popen(["C:/Documents and Settings/flow_model/flow.exe"], \
stderr = subprocess.PIPE)
if process.stderr:
print process.stderr.readlines()
The code might not be entirely correct as I usually don't use Popen or Windows, but should give the idea. It might well be that the error message is on the error stream.
in python 2.6 use string enclosed inside quotation " and apostrophe ' marks. Also a change single / to double //.
Your working example will look like this:
import os
os.system("'C://Documents and Settings//flow_model//flow.exe'")
Also You can use any parameters if Your program ingest them.
os.system('C://"Program Files (x86)"//Maxima-gcl-5.37.3//gnuplot//bin//gnuplot -e "plot [-10:10] sin(x),atan(x),cos(atan(x)); pause mouse"')
finally You can use string variable, as an example is plotting using gnuplot directly from python:
this_program='C://"Program Files (x86)"//Maxima-gcl-5.37.3//gnuplot//bin//gnuplot'
this_par='-e "set polar; plot [-2*pi:2*pi] [-3:3] [-3:3] t*sin(t); pause -1"'
os.system(this_program+" "+this_par)
import os
path = "C:/Documents and Settings/flow_model/"
os.chdir(path)
os.system("flow.exe")
Note added by barlop
A commenter asked why this works. Here is why.
The OP's problem is os.system("...") doesn't work properly when there is a space in the path. (Note os.system can work with ('"...."') but anyhow)
Had the OP tried their program from a cmd prompt they'd have seen the error clearly.
C:\carp>type blah.py
import os
os.system(R"C:\Program Files (x86)\Google\Chrome\Application\chrome.exe")
C:\carp>python blah.py
'C:\Program' is not recognized as an internal or external command,
operable program or batch file.
C:\carp>
So it's fine for os.system("calc.exe") (there calc.exe is in the path environment variable). Or for os.system(R"c:\windows\system32\calc.exe"). There's no space in that path.
C:\>md "aa bb cc"
C:\>copy c:\windows\system32\calc.exe "c:\aa bb cc\cccalc.exe"
1 file(s) copied.
This works (Given file "c:\aa bb cc\cccalc.exe" )
import os
os.chdir(R"c:\aa bb cc")
os.system("cccalc.exe")
Other options are subprocess.run and subprocess.popen.
Is that trying to execute C:\Documents with arguments of "and", "Settings/flow_model/flow.exe"?
Also, you might consider subprocess.call().
There are loads of different solutions, and the results will strongly depend on:
the OS you are using: Windows, Cygwin, Linux, MacOS
the python version you are using: Python2 or Python3x
As I have discovered some things that are claimed to work only in Windows, doesn't, probably because I happen to use Cygwin which is outsmarting the OS way to deal with Windows paths. Other things only work in pure *nix based OS's or in Python2 or 3.
Here are my findings:
Generally speaking, os.system() is the most forgiving method.
os.startfile() is the least forgiving. (Windows only && if you're lucky)
subprocess.Popen([...]) not recommended
subprocess.run(winView, shell=True) the recommended way!
Remembering that using subprocess for anything may pose a security risk.
Try these:
import os, subprocess
...
winView = '/cygdrive/c/Windows/explorer.exe %s' % somefile
...
# chose one of these:
os.system(winView)
subprocess.Popen(['/cygdrive/c/Windows/explorer.exe', 'somefile.png'])
subprocess.run(winView, shell=True)
Q: Why would you want to use explorer in Windows?
A: Because if you just want to look at the results of some new file, explorer will automatically open the file with whatever default windows program you have set for that file type. So no need to re-specify the default program to use.
That's the correct usage, but perhaps the spaces in the path name are messing things up for some reason.
You may want to run the program under cmd.exe as well so you can see any output from flow.exe that might be indicating an error.
for the above question this solution works.
just change the path to where your executable file is located.
import sys, string, os
os.chdir('C:\\Downloads\\xpdf-tools-win-4.00\\xpdf-tools-win-4.00\\bin64')
os.system("C:\\Downloads\\xpdf-tools-win-4.00\\xpdf-tools-win-4.00\bin64\\flowwork.exe")
'''import sys, string, os
os.chdir('C:\\Downloads\\xpdf-tools-win-4.00\\xpdf-tools-win-4.00\\bin64')
os.system(r"C:\\Downloads\\xpdf-tools-win-4.00\\xpdf-tools-win-4.00\bin64\\pdftopng.exe test1.pdf rootimage")'''
Here test1.pdf rootimage is for my code .