为什么我不能在grep中使用^ \ s? [英] Why can't I use ^\s with grep?
问题描述
以下两种正则表达式都可以正常工作.
Both of the regexes below work In my case.
grep \s
grep ^[[:space:]]
但是所有下面的失败.我在git bash和腻子中都尝试过.
However all those below fail. I tried both in git bash and putty.
grep ^\s
grep ^\s*
grep -E ^\s
grep -P ^\s
grep ^[\s]
grep ^(\s)
最后一个甚至会产生语法错误.
The last one even produces a syntax error.
如果我在debuggex中尝试 ^ \ s
,它将起作用.
If I try ^\s
in debuggex it works.
如何找到以 grep
开头的空格字符的行?我必须使用 [[:: space:]]
吗?
How do I find lines starting with whitespace characters with grep
? Do I have to use [[:space:]]
?
推荐答案
grep \ s
为您工作,因为您输入的内容包含 s
.在这里,您对 s
进行转义,并且它与 s
匹配,因为它没有被解析为与正则表达式转义符匹配的空格.如果使用 grep ^ \\ s
,则将匹配以空格开头的字符串,因为 \\
将被解析为文字 \
char
grep \s
works for you because your input contains s
. Here, you escape s
and it matches the s
, since it is not parsed as a whitespace matching regex escape. If you use grep ^\\s
, you will match a string starting with whitespace since the \\
will be parsed as a literal \
char.
一个更好的主意是使用 -E
启用POSIX ERE语法并引用该模式:
A better idea is to enable POSIX ERE syntax with -E
and quote the pattern:
grep -E '^\s' <<< "$s"
请参见在线演示:
s=' word'
grep ^\\s <<< "$s"
# => word
grep -E '^\s' <<< "$s"
# => word
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