[英]python regex: to match space character or end of string
I want to match space chars or end of string in a text. 我想在文本中匹配空格字符或字符串结尾。
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*$' % uname, '@'+uname+'aa')
The pattern is not right. 模式不对。
How to use python? 如何使用python?
\\s*$
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". \\s*$
不正确:这匹配“零或多个空格后跟字符串的结尾”,而不是“一个或多个空格或字符串的结尾”。
For this situation, I would use (?:\\s+|$)
(inside a raw string, as others have mentioned). 对于这种情况,我会使用(?:\\s+|$)
(在原始字符串中,正如其他人提到的那样)。 The (?:)
part is just about separating that subexpression so that the | (?:)
部分只是将子表达式分开,以便| operator matches the correct fragment and no more than the correct fragment. 运算符匹配正确的片段,而不是正确的片段。
Try this: 试试这个:
assert re.findall('@%s\\s*$' % uname, '@'+uname)
You must escape the \\
character if you don't use raw strings. 如果不使用原始字符串,则必须转义\\
字符。
It's a bit confusing, but stems from the fact that \\
is a meta character for both the python interpreter and the re
module. 这有点令人困惑,但源于这样的事实: \\
是python解释器和re
模块的元字符。
Use raw strings . 使用原始字符串 。
assert re.findall(r'@%s\s*$' % uname, '@'+uname)
Otherwise the use of \\
as a special character in regular strings conflicts with its use as a special character in regular expressions. 否则,在常规字符串中使用\\
作为特殊字符会与在正则表达式中用作特殊字符相冲突。
But this assertion is impossible to fail. 但这种说法不可能失败。 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. 当然,除了“@”加上变量uname
的内容之外的字符串将匹配正则表达式“@”加上uname
加上可选(总是空的)空格,然后匹配字符串的结尾。 It's a tautology. 这是一个重言式。 I suspect you are trying to check for something else? 我怀疑你正试图检查别的东西?
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