Scroll through the entire record

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I wonder if there is a way I can go through the entire registry (keys and subkeys, do not need the value of them) of the operating system to find the name of a key in specific?

If so, could someone tell me how to do it in python 3.x+ or at least indicate me a way to do it?

python python-3.x windows

  • If you just want to make sure that a key exists, why go through the whole record? in any case, you can use the function EnumKey module Winreg to enumerate the keys.

  • I want to kind of do a search through the registry without having a predefined path. For example: find the key x across the registry.

  • Got it, do you want to find only the first occurrence or keep looking? maybe recursion is necessary in this case.

  • The key would happen to be the PATH?

1 answer

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Python has a module to access the Windows registry. It comes together in the standard library, no need to import anything external.

import winreg

This module has a function that takes an HKEY and an index, and returns the HKEY key name on that index. If the index is invalid, it releases a Windowserror. Thus it is possible to get the name of all the immediate subchaves of a given HKEY.

def obter_subchaves_imediatas(hkey):
    i = 0

    try:
        while True:
            yield winreg.EnumKey(hkey, i)
            i += 1
    except WindowsError:
        pass

Now, using the above function, it is easy to write a routine that recursively navigates through the entire record tree. It’s basically the pre-order browsing algorithm in some tree.

def navegar_arvore(nome_chave_raiz):
    hkey = winreg.OpenKey(winreg.HKEY_CURRENT_USER, nome_chave_raiz)

    for nome_subchave in obter_subchaves_imediatas(hkey):
        nome_completo = f'{nome_chave_raiz}\\{nome_subchave}'
        yield nome_completo
        yield from navegar_arvore(nome_completo)

A simple loop lets you get all paths from a certain root.

for chave in navegar_arvore('Control Panel'):
    print(chave)

There are several points that can be improved in this code, but what was shown here is enough to solve the problem.

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