移除支架之间的任何东西

Question

因为我现在正在使用windows注册表。。我总是发现以下模式。。注册表包含两个括号{}，中间有数字和字母，您可以看到下面的一些示例：

HKEY_CLASSES_ROOT\Drive\shellex\FolderExtensions\{fbeb8a05-beee-4442-804e-409d6c4515e9}

HKEY_CLASSES_ROOT\CLSID\{00BB2763-6A77-11D0-A535-00C04FD7D062}\InProcServer32

HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\Explorer\MountPoints2\CPC\Volume\{64dcb6fa-03c9-11e6-9e9f-806d6172696f}\

有谁能帮助我理解这些是指什么？它们是随机生成的吗？在

我还试图解析每个注册表项并删除方括号{}之间的任何内容。。首先，我知道这可以用正则表达式实现，但我真的不熟悉它们。。欢迎任何指导。在

Answer 1

当您用Python标记问题时（用2.7测试）：

import re

string = """
HKEY_CLASSES_ROOT\Drive\shellex\FolderExtensions\{fbeb8a05-beee-4442-804e-409d6c4515e9}
HKEY_CLASSES_ROOT\CLSID\{00BB2763-6A77-11D0-A535-00C04FD7D062}\InProcServer32
HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\Explorer\MountPoints2\CPC\Volume\{64dcb6fa-03c9-11e6-9e9f-806d6172696f}
"""

rx = r'\{[^}]+\}\\?'
string = re.sub(rx, '', string)
print string

代码片段提供了所有不带大括号的键。感谢@Liam之前指出了一个错误。
提示：我是一个Mac用户，但是，我很确定大括号是有必要的，不是吗？在

Answer 2

将{.*?}替换为{}（如果一行有两个guid，则使用lazy dot）

demo

字符串是Globally Unique Identifiers。从here：

GUID (or UUID) is an acronym for 'Globally Unique Identifier' (or 'Universally Unique Identifier'). It is a 128-bit integer number used to identify resources. The term GUID is generally used by developers working with Microsoft technologies, while UUID is used everywhere else.

128-bits is big enough and the generation algorithm is unique enough that if 1,000,000,000 GUIDs per second were generated for 1 year the probability of a duplicate would be only 50%. Or if every human on Earth generated 600,000,000 GUIDs there would only be a 50% probability of a duplicate.

GUIDs are used in enterprise software development in C#, Java, and C++ as database keys, component identifiers, or just about anywhere else a truly unique identifier is required. GUIDs are also used to identify all interfaces and objects in COM programming.

Answer 3

在这种情况下，正则表达式是最好的。只需使用：

import re
string = 'HKEY_CLASSES_ROOT\CLSID\{00BB2763-6A77-11D0-A535-00C04FD7D062}\InProcServer32'
string = re.sub('{(.+)}', '', string)