Finding a File Given an Arbitrary Search Path
Credit: Chui Tey
Problem
Given a search path (a string of directories with a separator in between), you need to find the first file along the path whose name is as requested.
Solution
Basically, you need to loop over the directories in the given search path:
import os, string
def search_file(filename, search_path, pathsep=os.pathsep):
""" Given a search path, find file with requested name """
for path in string.split(search_path, pathsep):
candidate = os.path.join(path, filename)
if os.path.exists(candidate): return os.path.abspath(candidate)
return None
if _ _name_ _ == '_ _ _main_ _':
search_path = '/bin' + os.pathsep + '/usr/bin' # ; on Windows, : on Unix
find_file = search_file('ls',search_path)
if find_file:
print "File found at %s" % find_file
else:
print "File not found"Discussion
This is a reasonably frequent task, and Python makes it extremely
easy. The search loop can be coded in many ways, but returning the
normalized path as soon as a hit is found is simplest as well as
fast. The explicit return None after the loop is
not strictly needed, since None is what Python
returns when a function falls off the end, but having the return
explicit in this case makes the functionality of
search_file much clearer at first sight.
To find files specifically on Python’s own search path, see Recipe 4.22.
See Also
Recipe 4.22; documentation for the module
os in the Library Reference.