Reputation: 171
I want to return the path of a file, If it is found by the program, but I want it to continue to loop(or recursively repeat) the program until all files are checked.
def findAll(fname, path):
for item in os.listdir(path):
n = os.path.join(path, item)
try:
findAll(n, fname)
except:
if item == fname:
print(os.idontknow(item))
So I'm having trouble with calling the path, right now I have
os.idontknow(item)
as a place holder
Input is :
findAll('fileA.txt', 'testpath')
The output is:
['testpat\\fileA.txt', 'testpath\\folder1\\folder11\\fileA.txt','testpath\\folder2\\fileA.txt']
Upvotes: 0
Views: 567
Reputation: 37279
Per my comment above, here is an example that will start at the current directory and search through all sub-directories, looking for files matching fname
:
import os
# path is your starting point - everything under it will be searched
path = os.getcwd()
fname = 'file1.txt'
my_files = []
# Start iterating, and anytime we see a file that matches fname,
# add to our list
for root, dirs, files in os.walk(path):
for name in files:
if name == fname:
# root here is the path to the file
my_files.append(os.path.join(root, name))
print my_files
Or as a function (more appropriate for your case :) ):
import os
def findAll(fname, start_dir=os.getcwd()):
my_files = []
for root, dirs, files in os.walk(start_dir):
for name in files:
if name == fname:
my_files.append(os.path.join(root, name))
return my_files
print findAll('file1.txt')
print findAll('file1.txt', '/some/other/starting/directory')
Upvotes: 2
Reputation: 908
Something like this, maybe?
import os
path = "path/to/your/dir"
for (path, dirs, files) in os.walk(path):
print files
Upvotes: 1