grep'^([a-zA-Z.-]+)[0-9]+' ./file.txt
So I only get the matching part between the brackets< /p>
house.com
Instead of the whole line like I usually get:
house. com112
Suppose there is a line house.com112 in my file.txt.
(Actually this regular expression is just an example, I just want to know if I can only print the entire line Part of it.)
I know that in some languages, such as PHP, Perl or even AWK, I can do it, but I don’t know if I can use egrep.
Thank you in advance!
< p>
grep'^[a-zA-Z.-]+[0-9]+' ./file.txt | sed's/[0-9]\+$//'
Or if you just want to stick to grep, you can use grep and the -o switch instead of sed:
grep'^[a-zA- Z.-]+[0-9]+' ./file.txt | grep -o'[a-zA-Z.-]+'
I want I know, if I use egrep((GNU grep)2.5.1), I can select a part of the matching text, for example:
grep'^([a-zA -Z.-]+)[0-9]+' ./file.txt
So I only get the matching part between the brackets
house.com
Instead of the whole line like I usually get:
house.com112
Assuming my file. There is a line house.com112 in the txt.
(Actually this regular expression is just an example, I just want to know if I can only print part of the whole line.)
I know In some languages, such as PHP, Perl or even AWK, I can do it, but I don’t know if I can use egrep.
Thank you in advance!
After grep finds the matching line, use sed to modify the result:
grep'^ [a-zA-Z.-]+[0-9]+' ./file.txt | sed's/[0-9]\+$//'
Or if you just want Stick to grep, you can use grep and -o switch instead of sed:
grep'^[a-zA-Z.-]+[0-9]+' ./file.txt | grep -o'[a-zA-Z.-]+'