搜索前后的通配符

Question

我正在做一个搜索与字符串匹配的文件名的项目。当前的工作代码在字符串的末尾附加一个*，当后面有文件扩展名和/或数字时，代码会正确地获取字符串并匹配它，这是正确的。我希望它忽略初始字符串之前的内容，并检查文件名的内部。例如CCD_ 2将通过对CCD_；或File B

工作代码：

redeemed = redeemed + "*"
results = [file for file in sounds if re.search(redeemed, file)]

非工作代码：

redeemed = "*" + redeemed + "*"
results = [file for file in sounds if re.search(redeemed, file)]

返回错误：

Task exception was never retrieved
future: <Task finished name='Task-19' coro=<PubSub.__handle_message() done, defined at C:Usersdmvh1AppDataLocalProgramsPythonPython310libsite-packagestwitchAPIpubsub.py:293> exception=error('nothing to repeat at position 0')>
Traceback (most recent call last):
File "C:UsersusernameAppDataLocalProgramsPythonPython310libsite-packagestwitchAPIpubsub.py", line 298, in __handle_message
sub(uuid, msg_data)
File "C:UsersusernameDesktopTwitchSoundsInProgress.py", line 34, in callback_redemptions
results = [file for file in sounds if re.search(redeemed, file)]
File "C:UsersusernameDesktopTwitchSoundsInProgress.py", line 34, in <listcomp>
results = [file for file in sounds if re.search(redeemed, file)]
File "C:UsersusernameAppDataLocalProgramsPythonPython310libre.py", line 200, in search
return _compile(pattern, flags).search(string)
File "C:UsersusernameAppDataLocalProgramsPythonPython310libre.py", line 303, in _compile
p = sre_compile.compile(pattern, flags)
File "C:UsersusernameAppDataLocalProgramsPythonPython310libsre_compile.py", line 788, in compile
p = sre_parse.parse(p, flags)
File "C:UsersusernameAppDataLocalProgramsPythonPython310libsre_parse.py", line 955, in parse
p = _parse_sub(source, state, flags & SRE_FLAG_VERBOSE, 0)
File "C:UsersusernameAppDataLocalProgramsPythonPython310libsre_parse.py", line 444, in _parse_sub
itemsappend(_parse(source, state, verbose, nested + 1,
File "C:UsersusernameAppDataLocalProgramsPythonPython310libsre_parse.py", line 669, in _parse
raise source.error("nothing to repeat",
re.error: nothing to repeat at position 0

我们担心被赎回的代码会捕获所有东西，但它什么也没捕获。关于如何实现此搜索，有什么想法吗？

Answer 1

如果只是纯文本搜索，则不需要regex或glob:

results = [file for file in sounds if redeemed in file]

无需*或.*

Answer 2

如果文件名是作为字符串处理的，那么就不必使用anyString+*只需在字符串中搜索您想要查找的内容

如果我的来源是正确的，pythonsearch()将为你完成任务"search((函数如果在字符串中找到模式的匹配项，则返回Match对象。。。。search((匹配字符串"中的任何位置；

像这个

results = [file for file in sounds if re.search(redeemed, file)]