os independent way of seeing if an executable is on the path?

[Jp Calderone]

On Thu, 26 May 2005 11:53:04 -0700, Don <donald.welch at nospam.hp.com> wrote:
>Steven Bethard wrote:
>
>> This has probably been answered before, but my Google skills have failed
>> me so far...
>>
>> Is there an os independent way of checking to see if a particular
>> executable is on the path?  Basically what I want to do is run code like:
>>      i, o, e = os.popen3(executable_name)
>> but I'd like to give an informative error if 'executable_name' doesn't
>> refer to an executable on the path.
>>
>> [snip]
>>
>> Thanks for the help,
>>
>> STeVe
>
>I wrote this 'which' function for Linux, but I think if you changed the ':'
>character, it would work on Windows (I think its a ';' on Windows, but I
>can't remember):
>
>def which( command ):
>    path = os.getenv( 'PATH' ).split( ':' )
>    found_path = ''
>    for p in path:
>        try:
>            files = os.listdir( p )
>        except:
>            continue
>        else:
>            if command in files:
>                found_path = p
>                break
>
>    return found_path
>

Here's the version that comes with Twisted:

  import os, sys, imp

  def which(name, flags=os.X_OK):
      """Search PATH for executable files with the given name.

      @type name: C{str}
      @param name: The name for which to search.

      @type flags: C{int}
      @param flags: Arguments to L{os.access}.

      @rtype: C{list}
      @param: A list of the full paths to files found, in the
      order in which they were found.
      """
      result = []
      exts = filter(None, os.environ.get('PATHEXT', '').split(os.pathsep))
      for p in os.environ['PATH'].split(os.pathsep):
          p = os.path.join(p, name)
          if os.access(p, flags):
              result.append(p)
          for e in exts:
              pext = p + e
              if os.access(pext, flags):
                  result.append(pext)
      return result

Jp