问题描述
使用egrep（（GNU grep）2.5.1），我想知道是否可以选择一部分匹配的文本。 > grep'^（[a-zA-Z .-] +）[0-9] +'./file.txt
因此，我只得到了与括号中相匹配的部分，如
house.com
而不是像我通常得到的那样：
house.com112
假设我在我的file.txt中有一行与house.com112。
（其实这个正则表达式只是一个例子，我只想知道我是否只能打印一个）
我知道在某些语言中，比如PHP，Perl甚至AWK我都可以，但我不知道我是否可以用egrep 。
预先感谢您！ 使用<$ c $
grep'^ sed在grep找到匹配的行后修改结果。 [a-zA-Z .-] + [0-9] +'./file.txt | sed's / [0-9] \ + $ //'
或者如果您想要
grep'^ [a-zA- Z .-] + [0-9] +'./file.txt | grep -o'[a-zA-Z .-] +'
I was wondering if, with egrep ((GNU grep) 2.5.1), I can select a part of the matched text, something like:
grep '^([a-zA-Z.-]+)[0-9]+' ./file.txt
So I get only the part which matched, between the brackets, something like
house.com
Instead of the whole line like I usually get:
house.com112
Assuming I have a line with house.com112 in my file.txt.
(Actually this regular expression is just an example I just want to know if I can print only a part of the whole line.)
I do know in some languages, such as PHP, Perl or even AWK I can, but I do not know if I can with egrep.
Thank you in advance!
Use sed to modify the result after grep has found the lines that match:
grep '^[a-zA-Z.-]+[0-9]+' ./file.txt | sed 's/[0-9]\+$//'
Or if you want to stick with only grep, you can use grep with the -o switch instead of sed:
grep '^[a-zA-Z.-]+[0-9]+' ./file.txt | grep -o '[a-zA-Z.-]+'
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