Traverse directory at URL to root in Python - python-3.8

Question

How can you traverse directory to get to root in Python? I wrote some code using BeautifulSoup, but it says 'module not found'. So I have this:
#
# There is a directory traversal vulnerability in the
# following page http://127.0.0.1:8082/humantechconfig?file=human.conf
# Write a script which will attempt various levels of directory
# traversal to find the right amount that will give access
# to the root directory. Inside will be a human.conf with the flag.
#
# Note: The script can timeout if this occurs try narrowing
# down your search
import urllib.request
import os
req = urllib.request.urlopen("http://127.0.0.1:8082/humantechconfig?file=human.conf")
dirName = "/tmp"
def getListOfFiles(dirName):
listOfFile = os.listdir(dirName)
allFiles = list()
for entry in listOfFile:
# Create full path
fullPath = os.path.join(dirName, entry)
if os.path.isdir(fullPath):
allFiles = allFiles + getListOfFiles(fullPath)
else:
allFiles.append(fullPath)
return allFiles
listOfFiles = getListOfFiles(dirName)
print(listOfFiles)
for file in listOfFiles:
if file.endswith(".conf"):
f = open(file, "r")
print(f.read())
This outputs:
/tmp/level-0/level-1/level-2/human.conf
User : Human 66
Flag: Not-Set (Must be Root Human)
However. If I change the URL to 'http://127.0.0.1:8082/humantechconfig?file=../../../human.conf' it gives me the output:
User : Human 66
Flag: Not-Set (Must be Root Human)
User : Root Human
Flag: Well done the flag is: {}
The level of directory traversal it is at fluctuates wildly, from /tmp/level-2 to /tmp/level-15; if it's at the one I wrote, then it says I'm 'Root Human'. But it won't give me the flag, despite the fact that I am suddenly 'Root Human'. Is there something wrong with the way I am traversing directory?
It doesn't seem to matter at all if I take away the req = urllib.request.urlopen("http://127.0.0.1:8082/humantechconfig?file=human.conf") line. How can I actually send the code to that URL?
Thanks!

Answer 1

cyber discovery moon base challenge?
For this one, you need to keep adding '../' in front of human.conf (for example 'http://127.0.0.1:8082/humantechconfig?file=../human.conf') which becomes your URL. This URL you need to request (using urllib.request.urlopen(URL)).
The main bit of the challenge is to attach the ../ multiple times which shall not be very hard using a simple loop. You don't need to use the OS.
Make sure to break the loop once you find the flag (or it will go into an infinite loop and give you errors).