I want to match space chars or end of string in a text.
import re
uname='abc'
assert re.findall('@%s\s*$' % uname, '@'+uname)
assert re.findall('@%s\s*$' % uname, '@'+uname+' '+'aa')
assert not re.findall('@%s\s*$' % uname, '@'+uname+'aa')
The pattern is not right.
How to use python?
\s*$ is incorrect: this matches "zero or more spaces followed by the end of the string", rather than "one or more spaces or the end of the string".
For this situation, I would use
(?:\s+|$) (inside a raw string, as others have mentioned).
The (?:) part is just about separating that subexpression so that the | operator matches the correct fragment and no more than the correct fragment.
Try this:
assert re.findall('@%s\\s*$' % uname, '@'+uname)
You must escape the \ character if you don't use raw strings.
It's a bit confusing, but stems from the fact that \ is a meta character for both the python interpreter and the re module.
@abc aa doesn't have match the pattern of white spaces followed by the end of the string. See Kampu's answer.Use raw strings.
assert re.findall(r'@%s\s*$' % uname, '@'+uname)
Otherwise the use of \ as a special character in regular strings conflicts with its use as a special character in regular expressions.
But this assertion is impossible to fail. Of course, a string consisting of nothing but "@" plus the contents of the variable uname is going to match a regular expression of "@" plus uname plus optional (always empty) whitespace and then the end of the string. It's a tautology. I suspect you are trying to check for something else?
