For example how would I get lines that begin with foo and end with bar?
例如,我如何得到以foo开头并以bar结尾的行?
Example lines:
foo12345abar
fooabcdbar
fooy7ghqqbar
This does not seem to work
这似乎不起作用
grep '^foo[.]*bar$'
What is the correct regex?
什么是正确的正则表达式?
2 个解决方案
#1
2
The reason your expression does not work is because [.] represents a dot, literally. Your expression would match strings that look like this:
你的表达式不起作用的原因是因为[。]代表一个点,字面意思。您的表达式将匹配如下所示的字符串:
foo.......bar
foo...bar
boobar
To make it work remove square brackets [] around the dot meta-character:
要使其工作,请删除点元字符周围的方括号[]:
^foo.*bar$
#1
2
The reason your expression does not work is because [.] represents a dot, literally. Your expression would match strings that look like this:
你的表达式不起作用的原因是因为[。]代表一个点,字面意思。您的表达式将匹配如下所示的字符串:
foo.......bar
foo...bar
boobar
To make it work remove square brackets [] around the dot meta-character:
要使其工作,请删除点元字符周围的方括号[]:
^foo.*bar$