Get name of current script in Python

Question

I m trying to get the name of the Python script that is currently running   I have a script called foo py and I d like to do something like this in order to get the script name   print Scriptname

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My fast dirty solution     file   split      -1   0

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The Above answers are good   But I found this method more efficient using above results  This results in actual script file name not a path   import sys     import os     file name    os path basename sys argv 0

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os path abspath   file    will give you an absolute path  relpath   available as well    sys argv -1  will give you a relative path

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Note  If you are using Python 3   then you should use the print   function instead  Assuming that the filename is foo py  the below snippet  import sys print sys argv 0   -3    or  import sys print sys argv 0    -1  3     -1    As for other extentions with more characters  for example the filename foo pypy  import sys print sys argv 0  split      0    If you want to extract from an absolute path  import sys print sys argv 0  split      -1  split      0    will output foo

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For completeness  sake  I thought it would be worthwhile summarizing the various possible outcomes and supplying references for the exact behaviour of each     file   is the currently executing file  as detailed in the official documentation     file   is the pathname of the file from which the module was loaded  if it was loaded from a file  The   file   attribute may be missing for certain types of modules  such as C modules that are statically linked into the interpreter  for extension modules loaded dynamically from a shared library  it is the pathname of the shared library file   From Python3 4 onwards  per issue 18416    file   is always an absolute path  unless the currently executing file is a script that has been executed directly  not via the interpreter with the -m command line option  using a relative path     main     file    requires importing   main    simply accesses the aforementioned   file   attribute of the main module  e g  of the script that was invoked from the command line  From Python3 9 onwards  per issue 20443  the   file   attribute of the   main   module became an absolute path  rather than a relative path   sys argv 0   requires importing sys  is the script name that was invoked from the command line  and might be an absolute path  as detailed in the official documentation   argv 0  is the script name  it is operating system dependent whether this is a full pathname or not   If the command was executed using the -c command line option to the interpreter  argv 0  is set to the string  -c   If no script name was passed to the Python interpreter  argv 0  is the empty string   As mentioned in another answer to this question  Python scripts that were converted into stand-alone executable programs via tools such as py2exe or PyInstaller might not display the desired result when using this approach  i e  sys argv 0  would hold the name of the executable rather than the name of the main Python file within that executable    If none of the aforementioned options seem to work  probably due to an atypical execution process or an irregular import operation  the inspect module might prove useful  In particular  invoking inspect stack   -1  1  should work  although it would raise an exception when running in an implementation without Python stack frame   From Python3 6 onwards  and as detailed in another answer to this question  it s possible to install an external open source library  lib programname  which is tailored to provide a complete solution to this problem  This library iterates through all of the approaches listed above until a valid path is returned  If all of them fail  it raises an exception  It also tries to address various pitfalls  such as invocations via the pytest framework or the pydoc module  import lib programname   this returns the fully resolved path to the launched python program path to program   lib programname get path executed script      type  pathlib Path     Handling relative paths When dealing with an approach that happens to return a relative path  it might be tempting to invoke various path manipulation functions  such as os path abspath      or os path realpath      in order to extract the full or real path  However  these methods rely on the current path in order to derive the full path  Thus  if a program first changes the current working directory  for example via os chdir       and only then invokes these methods  they would return an incorrect path  Handling symbolic links If the current script is a symbolic link  then all of the above would return the path of the symbolic link rather than the path of the real file and os path realpath      should be invoked in order to extract the latter   Further manipulations that extract the actual file name os path basename      may be invoked on any of the above in order to extract the actual file name and os path splitext      may be invoked on the actual file name in order to truncate its suffix  as in os path splitext os path basename        From Python 3 4 onwards  per PEP 428  the PurePath class of the pathlib module may be used as well on any of the above  Specifically  pathlib PurePath      name extracts the actual file name and pathlib PurePath      stem extracts the actual file name without its suffix

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As of Python 3 5 you can simply do   from pathlib import Path Path   file    stem   See more here  https   docs python org 3 5 library pathlib html pathlib PurePath stem  For example  I have a file under my user directory named test py with this inside   from pathlib import Path  print Path   file    stem  print   file      running this outputs    gt  gt  gt  python3 6 test py test test py

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def basename       x   file      y x split          y1 y -1     y2 y1 split         y3 y2 0   return y2 0

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Since the OP asked for the name of the current script file I would prefer  import os os path split sys argv 0   1

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import sys print sys argv 0    This will print foo py for python foo py  dir foo py for python dir foo py  etc  It s the first argument to python   Note that after py2exe it would be foo exe

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If you re doing an unusual import  e g   it s an options file   try   import inspect print  inspect getfile inspect currentframe       Note that this will return the absolute path to the file

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You can use   file   to get the name of the current file  When used in the main module  this is the name of the script that was originally invoked   If you want to omit the directory part  which might be present   you can use os path basename   file

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we can try this to get current script name without extension   import os  script name   os path splitext os path basename   file     0

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Try this   print   file

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For modern Python versions  3 4    Path   file    name should be more idiomatic  Also  Path   file    stem gives you the script name without the  py extension

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Note that   file   will give the file where this code resides  which can be imported and different from the main file being interpreted  To get the main file  the special   main   module can be used   import   main   as main print main   file      Note that   main     file   works in Python 2 7 but not in 3 2  so use the import-as syntax as above to make it portable

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You can do this without importing os or other libs  If you want to get the path of current python script  use    file   If you want to get only the filename without  py extension  use this    file   rsplit  quot   quot   1  1  split      0

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all that answers are great  but have some problems You might not see at the first glance  lets define what we want - we want the name of the script that was executed  not the name of the current module - so   file   will only work if it is used in the executed script  not in an imported module  sys argv is also questionable - what if your program was called by pytest   or pydoc runner   or if it was called by uwsgi   and - there is a third method of getting the script name  I havent seen in the answers - You can inspect the stack  Another problem is  that You  or some other program  can tamper around with sys argv and   main     file   - it might be present  it might be not  It might be valid  or not  At least You can check if the script  the desired result  exists   the library lib programname does exactly that    check if   main   is present check if   main     file   is present does give   main     file   a valid result  does that script exist    if not  check sys argv  is there pytest  docrunner  etc in the sys argv   -- gt  if yes  ignore that can we get a valid result here   if not  inspect the stack and get the result from there possibly if also the stack does not give a valid result  then throw an Exception   by that way  my solution is working so far with setup py test  uwsgi  pytest  pycharm pytest   pycharm docrunner  doctest   dreampie  eclipse there is also a nice blog article about that problem from Dough Hellman   quot Determining the Name of a Process from Python quot  BTW  it will change again in python 3 9   the file attribute of the main module became an absolute path  rather than a relative path  These paths now remain valid after the current directory is changed by os chdir   So I rather want to take care of one small module  instead of skimming my codebase if it should be changed somewere      Disclaimer  I m the author of the lib programname library

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The first argument in sys will be the current file name so this will work import sys print sys argv 0    will print the file name

[python] Get name of current script in Python

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