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Duplicate edit: no, i did that but it doesnt want to launch firefox. I am making a cortana/siri assistant thing, and I want it to lets say open a web browser when I say something. So I have done the if part, but I just need it to launch firefox.exe I have tried different things and I get an error . Here is the code. Please help! It works with opening notepad but not firefox..

#subprocess.Popen(['C:\Program Files\Mozilla Firefox\firefox.exe']) opens the app and continues the script #subprocess.call(['C:\Program Files\Mozilla Firefox\firefox.exe']) this opens it but doesnt continue the script import os import subprocess print "Hello, I am Danbot.. If you are new ask for help!" #intro prompt = ">" #sets the bit that indicates to input to > input = raw_input (prompt) #sets whatever you say to the input so bot can proces raw_input (prompt) #makes an input if input == "help": #if the input is that print "*****************************************************************" #says that print "I am only being created.. more feautrues coming soon!" #says that print "*****************************************************************" #says that print "What is your name talks about names" #says that print "Open (name of program) opens an application" #says that print "sometimes a command is ignored.. restart me then!" print "Also, once you type in a command, press enter a couple of times.." print "*****************************************************************" #says that raw_input (prompt) #makes an input if input == "open notepad": #if the input is that print "opening notepad!!" #says that print os.system('notepad.exe') #starts notepad if input == "open the internet": #if the input is that print "opening firefox!!" #says that subprocess.Popen(['C:\Program Files\Mozilla Firefox\firefox.exe'])
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    Use the absolute path to firefox.exe. Commented May 15, 2016 at 13:06
  • Possible duplicate of How do I execute a program from python? os.system fails due to spaces in path Commented May 15, 2016 at 13:11
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    notepad is normally in system32 folder which is under PATH variable, but firefox is unlikely. Commented May 15, 2016 at 13:12
  • user3549596 What do you mean? this is the path: C:\Program Files\Mozilla Firefox\firefox.exe Commented May 15, 2016 at 14:39
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    ...at a minimum, show the actual error! Is it simply silently failing to open a browser window? Is it throwing an exception? Which exception? Etc, etc. There's no call at all for showing code around input-handling if your bug doesn't have anything to do with that handling; take those parts out. Commented May 15, 2016 at 16:58

3 Answers 3

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The short answer is that os.system doesn't know where to find firefox.exe.

A possible solution would be to use the full path. And it is recommended to use the subprocess module:

import subprocess subprocess.call(['C:\Program Files\Mozilla Firefox\\firefox.exe'])

Mind the \\ before the firefox.exe! If you'd use \f, Python would interpret this as a formfeed:

>>> print('C:\Program Files\Mozilla Firefox\firefox.exe') C:\Program Files\Mozilla Firefox irefox.exe

And of course that path doesn't exist. :-)

So either escape the backslash or use a raw string:

>>> print('C:\Program Files\Mozilla Firefox\\firefox.exe') C:\Program Files\Mozilla Firefox\firefox.exe >>> print(r'C:\Program Files\Mozilla Firefox\firefox.exe') C:\Program Files\Mozilla Firefox\firefox.exe

Note that using os.system or subprocess.call will stop the current application until the program that is started finishes. So you might want to use subprocess.Popen instead. That will launch the external program and then continue the script.

subprocess.Popen(['C:\Program Files\Mozilla Firefox\\firefox.exe', '-new-tab'])

This will open firefox (or create a new tab in a running instance).

A more complete example is my open utility I publish via github. This uses regular expressions to match file extensions to programs to open those files with. Then it uses subprocess.Popen to open those files in an appropriate program. For reference I'm adding the complete code for the current version below.

Note that this program was written with UNIX-like operating systems in mind. On ms-windows you could probably get an application for a filetype from the registry.

"""Opens the file(s) given on the command line in the appropriate program. Some of the programs are X11 programs.""" from os.path import isdir, isfile from re import search, IGNORECASE from subprocess import Popen, check_output, CalledProcessError from sys import argv import argparse import logging __version__ = '1.3.0' # You should adjust the programs called to suit your preferences. filetypes = { '\.(pdf|epub)$': ['mupdf'], '\.html$': ['chrome', '--incognito'], '\.xcf$': ['gimp'], '\.e?ps$': ['gv'], '\.(jpe?g|png|gif|tiff?|p[abgp]m|svg)$': ['gpicview'], '\.(pax|cpio|zip|jar|ar|xar|rpm|7z)$': ['tar', 'tf'], '\.(tar\.|t)(z|gz|bz2?|xz)$': ['tar', 'tf'], '\.(mp4|mkv|avi|flv|mpg|movi?|m4v|webm)$': ['mpv'] } othertypes = {'dir': ['rox'], 'txt': ['gvim', '--nofork']} def main(argv): """Entry point for this script. Arguments: argv: command line arguments; list of strings. """ if argv[0].endswith(('open', 'open.py')): del argv[0] opts = argparse.ArgumentParser(prog='open', description=__doc__) opts.add_argument('-v', '--version', action='version', version=__version__) opts.add_argument('-a', '--application', help='application to use') opts.add_argument('--log', default='warning', choices=['debug', 'info', 'warning', 'error'], help="logging level (defaults to 'warning')") opts.add_argument("files", metavar='file', nargs='*', help="one or more files to process") args = opts.parse_args(argv) logging.basicConfig(level=getattr(logging, args.log.upper(), None), format='%(levelname)s: %(message)s') logging.info('command line arguments = {}'.format(argv)) logging.info('parsed arguments = {}'.format(args)) fail = "opening '{}' failed: {}" for nm in args.files: logging.info("Trying '{}'".format(nm)) if not args.application: if isdir(nm): cmds = othertypes['dir'] + [nm] elif isfile(nm): cmds = matchfile(filetypes, othertypes, nm) else: cmds = None else: cmds = [args.application, nm] if not cmds: logging.warning("do not know how to open '{}'".format(nm)) continue try: Popen(cmds) except OSError as e: logging.error(fail.format(nm, e)) else: # No files named if args.application: try: Popen([args.application]) except OSError as e: logging.error(fail.format(args.application, e)) def matchfile(fdict, odict, fname): """For the given filename, returns the matching program. It uses the `file` utility commonly available on UNIX. Arguments: fdict: Handlers for files. A dictionary of regex:(commands) representing the file type and the action that is to be taken for opening one. odict: Handlers for other types. A dictionary of str:(arguments). fname: A string containing the name of the file to be opened. Returns: A list of commands for subprocess.Popen. """ for k, v in fdict.items(): if search(k, fname, IGNORECASE) is not None: return v + [fname] try: if b'text' in check_output(['file', fname]): return odict['txt'] + [fname] except CalledProcessError: logging.warning("the command 'file {}' failed.".format(fname)) return None if __name__ == '__main__': main(argv)
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7 Comments

(I also copied the exact path of my firefox location) Ok, so I did exactly the subprocces code, and when I try to launch firefox from it, it says:Traceback (most recent call last): File "Danbot.py", line 33, in <module> subprocess.Popen(['C:\Program Files\Mozilla Firefox\firefox.exe']) File "c:\hp\bin\python\lib\subprocess.py", line 593, in init errread, errwrite) File "c:\hp\bin\python\lib\subprocess.py", line 793, in _execute_child startupinfo) WindowsError: [Error 22] The filename, directory name, or volume label syntax is incorrect
@Dan At a guess, that is probably because the '\f' is interpreted by Python as an escape sequence for a formfeed. Change \f to \\f.
What \f ? I did not use anything with \f.. ?
Yes. When Python reads 'C:\Program Files\Mozilla Firefox\firefox.exe' it interprets the \f and replaces it with a formfeed character. So you get C:\Program Files\Mozilla Firefox<formfeed>irefox.exe', which of course doesn't exist. :-)
You need double slashes `\` on all of them, not just the last one...
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You can open any program like this:

import subprocess # This will pause the script until fortnite is closed subprocess.call(['C:\Program Files\Fortnite\fortnite.exe']) # But this will run fortnite and continue the script subprocess.Popen(['C:\Program Files\Fortnite\fortnite.exe', '-new-tab'])

But here's How to find the .exe file of any program (Windows only)

I only know how to find the .exe file in windows

First, click the windows icon then write task manager, open it, then find Fortnite if there's an arrow on it's left click it, if there's another one click it.

Make sure that you press the More Details button at the bottom.

Then scroll to the side until you find Command line written on the top row, here there will be the path of the .exe file of every running task

There might be -(SOME_LETTER) but these are some options for choosing what the exe file will do when it's ran. These are different for every exe file

(Little tips) You can press Ctrl+Shift+Esc to open the task manager. Go in the task manager then press Options (at the top) then press Always on top, which will make the app appear on top of every thing which is pretty useful when an app crashes and you need to close it using the task manager

Thanks.

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If you want to open Google or something on the web just import webbrowser and open the URL. I will give you a quick example.

import webbrowser webbrowser.open("www.google.com")
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The question is not about open something on the web (Firefox is just an example) and not about Python packages which have some external program's functionality. It is about call to external program and there is a generic answer already. So this is not an answer at all.

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