topic Re: Launching .exe in python in Python Questions

Launching .exe in python

DanPajak — Tue, 28 Dec 2010 19:30:26 GMT

I am trying to launch microsoft access.exe and have know errors but also no result with opening access. Does anyone know how to do this. I am using ArcDesktop ten with latest service pack and have python 2.6 installed.

import os
os.spawn("c:\programs files\Microsoft office\office12\msaccess.exe")

Re: Launching .exe in python

LoganPugh — Fri, 10 Dec 2021 20:03:11 GMT

Two things:
1 - Backslashes are an escape character in Python. Either escape your backslashes with another backslash, put an "r" at the front of the string (just before the opening quote) to denote that it is a raw string, or replace the backslashes with forward slashes.

2 - Use the subprocess module instead, as it is intended to replace other methods of executing external programs.

e.g.

import subprocess
subprocess.Popen([r"c:\programs files\Microsoft office\office12\msaccess.exe"])

Re: Launching .exe in python

DanPajak — Wed, 29 Dec 2010 11:52:08 GMT

I used the code as in your example and still I get nothing to launch. Is there anything at the begining or ending of the python script that I could be missing for this not to work. This is a simple command and works if I put it in command line but nothing in python.

Re: Launching .exe in python

LoganPugh — Wed, 29 Dec 2010 16:40:50 GMT

Check the path, I simply copied yours and I see a typo in what should be "program files". Using my own path to MSACCESS.exe works for me. It should be throwing an error if it can't find the file though.

Re: Launching .exe in python

DanPajak — Thu, 30 Dec 2010 11:59:38 GMT

I corrected the typo and still get nothing. Know error is prompted when I run code in python idle. When I run the code as a .py script it flashes but then access doesn't open.

Re: Launching .exe in python

DanPajak — Thu, 30 Dec 2010 12:39:35 GMT

i ran the code in python with in arcmap instead of idle and I get an error.
>>> import subprocess subprocess.Popen([r"C:\Program Files\Microsoft Office\Office12\MSACCESS.EXE])
Parsing error <type 'exceptions.SyntaxError'>: invalid syntax (line 1)
>>>

Any ideas on this one? I know it is the correct path.

Re: Launching .exe in python

LoganPugh — Thu, 30 Dec 2010 14:26:24 GMT

Don't forget the closing quote. Also the two commands should be on separate lines.

Re: Launching .exe in python

DanPajak — Fri, 10 Dec 2021 20:03:14 GMT

>>> import subprocess
... subprocess.Popen([r"c:\programs files\Microsoft office\office12\msaccess.exe"])
... 
Runtime error <type 'exceptions.WindowsError'>: [Error 3] The system cannot find the path specified

I get this error. Which is odd because I know the path is correct.

Re: Launching .exe in python

V_StuartFoote — Mon, 03 Jan 2011 14:08:30 GMT

Jeremy,

You've garbled the path again. And others may know for sure, but wouldn't the raw string be case sensitive?

Needs to be as entered as in your 12/30 6:39 post (but with the extra closing ").

Stuart

Re: Launching .exe in python

LoganPugh — Mon, 03 Jan 2011 14:51:54 GMT

Windows' file system is case-insensitive by default so it does not matter what case you give to Python. On UNIX-like OS's it may be a different story.

Re: Launching .exe in python

DanPajak — Fri, 10 Dec 2021 20:03:17 GMT

Thanks for you help I finally was able to launch the access from python. I realized that I couldn't just copy and past code. It only worked when I typed it out. Now that I get access to open is there a way to launch a macro from python. I tried

>>> import subprocess
>>> subprocess.Popen([r"C:\Program Files\Microsoft Office\Office12\MSACCESS.EXE" "\\vopcc\gis\GISOPS\StreetNetwork\Routing.mdb" /x WSorderMacro])
SyntaxError: invalid syntax
>>>

and also tried

>>> import subprocess
>>> subprocess.Popen([r"C:\Program Files\Microsoft Office\Office12\MSACCESS.EXE" "\\vopcc\gis\GISOPS\StreetNetwork\Routing.mdb""WSorderMacro"])

Traceback (most recent call last):
  File "<pyshell#41>", line 1, in <module>
    subprocess.Popen([r"C:\Program Files\Microsoft Office\Office12\MSACCESS.EXE" "\\vopcc\gis\GISOPS\StreetNetwork\Routing.mdb""WSorderMacro"])
  File "C:\Python26\ArcGIS10.0\lib\subprocess.py", line 633, in __init__
    errread, errwrite)
  File "C:\Python26\ArcGIS10.0\lib\subprocess.py", line 842, in _execute_child
    startupinfo)
WindowsError: [Error 3] The system cannot find the path specified

and get errors. Thanks again for all your help so far it has been much aprreciated:)

Re: Launching .exe in python

V_StuartFoote — Tue, 04 Jan 2011 13:56:33 GMT

Jeremy,

Others may have a better feel for it, but it seems a little ambitious to use UNC ref to the \\vopcc\gis mount while working out syntax and function.

To simplify debugging, I would copy or subset the Access MDB to the local file system along with the macro, or at the least Windows map the share.

Finally, I think that the args provided to the Popen need to be separated by commas, but still raw. So wouldn't that be:

subprocess.Popen([r"C:\Program Files\Microsoft Office\Office12\MSACCESS.EXE", r"\\vopcc\gis\GISOPS\StreetNetwork\Routing.mdb", r"/x WSorderMacro"])

Stuart

Re: Launching .exe in python

LoganPugh — Fri, 10 Dec 2021 20:03:19 GMT

Based on the Python help on subprocess, you can (and on Windows, probably should) use a string instead of a list of arguments to specify the command line. Access seems particularly picky about how you specify its command line. This should work:

import subprocess
subprocess.Popen(r'"C:\Program Files\Microsoft Office\Office12\MSACCESS.EXE" "\\vopcc\gis\GISOPS\StreetNetwork\Routing.mdb" /X WSorderMacro')

Note the raw string indicator, single quotes around the entire string, and the double quotes around paths (which are required if the path contains spaces, but it's good to include them anyways).

Re: Launching .exe in python

ChrisMathers — Fri, 10 Dec 2021 20:03:22 GMT

This is how I open excel and it works fine. I use a list of arguments separated by a comma. Both are rawed strings.

subprocess.Popen([r'C:\Program Files\Microsoft Office\OFFICE11\EXCEL.exe',r'C:\GIS Projects\TaxChange\Export\Change.xls'])

Re: Launching .exe in python

DanPajak — Fri, 10 Dec 2021 20:03:25 GMT

I have finally got it to open and run the macro. Thanks again for all your help.

>>> import subprocess
>>> subprocess.Popen([r'C:\Program Files\Microsoft Office\Office12\MSACCESS.EXE',r'\\vopcc\gis\GISOPS\StreetNetwork\Routing.mdb'])

I am having one small issue when I save the file as a .py file type and then double click on the file nothing happens. How do I make this script into a .py file and run the script off of that.

Re: Launching .exe in python

JasonScheirer — Tue, 04 Jan 2011 20:06:15 GMT

Are the literal >>>s in your script? If so, that's your problem.

Re: Launching .exe in python

DanPajak — Wed, 05 Jan 2011 12:13:30 GMT

Thank You very much. I removed the >>> and am now able to launch this from a .py. Thank you all for your help. Had to redu a whole work process because of ArcMap 10 going away with VB. Hopefully creating all the buttons in Python I will not have to change this again.

Re: Launching .exe in python

JohnHawkins — Fri, 10 Dec 2021 20:03:27 GMT

I also had this problem but didn't see the answer in where vopgis said he got it to work so I am including a script that I got to work. Note: the popen runs the process in parallel and what I wanted was to finished a process then move on. if you use the subprocess.call(program) it will finish the msaccess call then continue.

import subprocess
def MacroAccess(when):
    if when == 'Nightly': #either Nightly or Weekly
        program = r'"C:\Program Files (x86)\Microsoft Office\OFFICE11\msaccess.exe" "c:\ftp\import2.mdb " /x NightlyBatch'
    elif when == 'Weekly':
        program = r'"C:\Program Files (x86)\Microsoft Office\OFFICE11\msaccess.exe" "c:\ftp\import2.mdb " /x WeeklyBatch'
    else:
        return 'no'
    print when + 'batch'
    subprocess.call(program)
    return when + 'Batch has run'
    print 'done'

print MacroAccess('Nightly')

Re: Launching .exe in python

LukeSturtevant — Fri, 10 Dec 2021 20:03:30 GMT

I know this is an old thread but this seems to work for me:

import os 

filepath = r"C:\project_front.accdb"
os.system("start MSACCESS.EXE {}".format(filepath ))

Thanks Wes, I just updated.

Re: Launching .exe in python

WesMiller — Wed, 03 Feb 2016 15:09:15 GMT

Luke shouldn't line 4 be

os.system("start MSACCESS.EXE {}".format(filepath))