返回文本文件路徑

Question

我想返回文件的路徑，如果程序找到它，但我希望它繼續循環（或遞歸重復）程序，直到檢查完所有文件。

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))

所以我在調用這條路時遇到了麻煩，現在我有了

os.idontknow(item) 

作為占位符

輸入是：

findAll('fileA.txt', 'testpath')

輸出是：

['testpat\\fileA.txt', 'testpath\\folder1\\folder11\\fileA.txt','testpath\\folder2\\fileA.txt']

Answer 1

根據我上面的評論，這是一個示例，它將從當前目錄開始並搜索所有子目錄，查找與fname匹配的文件：

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

或者作為一個函數（更適合你的情況:)）：

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')

Answer 2

也許是這樣的事情？

import os
path = "path/to/your/dir"
for (path, dirs, files) in os.walk(path):
    print files