How do I get string.strip() to strip more than one character

Question

I am attempting to strip an incoming message of all punctuation in order to run an accurate moderation system. However, when I use string.strip(), it only pulls off the ending character of the string, and leaves the rest of the same character in the remainder of the string. What am I misunderstanding? Every resource I can find says that string.strip() is sufficient.

Relevant code:

 async def on_message(message):

    stripString = ".!?,'@#$%^&*()"

    if message.author == bot.user:
        return
    
    message.content = message.content.strip(variables.stripString)

input: This. Is. A. Test.

output: This. Is. A. Test

`.strip()` only removes characters from the start or end of a string — jezza_99, Oct 28 '21 at 20:24
What is the expected output ? You want to remove all punctuations everywhere ? — azro, Oct 28 '21 at 20:24
What is "every resource" to you? How about the python reference manual? https://docs.python.org/3.4/library/stdtypes.html?highlight=strip#str.strip It clearly states what strip does and what it doesn't do. — ypnos, Oct 28 '21 at 20:25
Yeah, I want to remove all punctuation throughout the entire string. Apparently .strip() doesn't work, so what would I do instaead? — Shadrock50, Oct 28 '21 at 20:26
This SO thread might help. https://stackoverflow.com/questions/3939361/remove-specific-characters-from-a-string-in-python — troy, Oct 28 '21 at 20:28

score 4 · Answer 1 · answered Oct 28 '21 at 20:26

The str.strip

Return a copy of the string with the leading and trailing characters removed.

What is you need is a replacement, using the regex [.!?,'@#$%^&*()] and replace by nothing

import re

stripString = ".!?,'@#$%^&*()"
message = "!This. !Is. A. Test.?@"

print(message.strip(stripString))                    # This. !Is. A. Test
print(re.sub("[" + stripString + "]", "", message))  # This Is A Test

How do I get string.strip() to strip more than one character

1 Answers1