How to escape special characters of a string with single backslashes
This is one way to do it (in Python 3.x):
escaped = a_string.translate(str.maketrans({"-": r"\-",
"]": r"\]",
"\\": r"\\",
"^": r"\^",
"$": r"\$",
"*": r"\*",
".": r"\."}))
For reference, for escaping strings to use in regex:
import re
escaped = re.escape(a_string)
Just assuming this is for a regular expression, use re.escape
.
We could use built-in function repr()
or string interpolation fr'{}'
escape all backwardslashs \
in Python 3.7.*
repr('my_string')
or fr'{my_string}'
Check the Link: https://docs.python.org/3/library/functions.html#repr
re.escape
doesn't double escape. It just looks like it does if you run in the repl. The second layer of escaping is caused by outputting to the screen.
When using the repl, try using print
to see what is really in the string.
$ python
>>> import re
>>> re.escape("\^stack\.\*/overflo\\w\$arr=1")
'\\\\\\^stack\\\\\\.\\\\\\*\\/overflo\\\\w\\\\\\$arr\\=1'
>>> print re.escape("\^stack\.\*/overflo\\w\$arr=1")
\\\^stack\\\.\\\*\/overflo\\w\\\$arr\=1
>>>
Simply using re.sub
might also work instead of str.maketrans
. And this would also work in python 2.x
>>> print(re.sub(r'(\-|\]|\^|\$|\*|\.|\\)',lambda m:{'-':'\-',']':'\]','\\':'\\\\','^':'\^','$':'\$','*':'\*','.':'\.'}[m.group()],"^stack.*/overflo\w$arr=1"))
\^stack\.\*/overflo\\w\$arr=1