grep正则表达式以，结束但不包含

Question

I'm working with grep and sed commands in textfiles within Linux. I'm busy playing with the Documents/Data folder and the share/dict/words document. I need to extract all words within the words file that begin with "q", end with "s" but do not contain "a" or "r". I've managed to get this piece of code so far:

grep –n ‘\<q.*[ar]s\>’ /usr/share/dict/words

Which gives me words that shouldn't be included. I've tried adding a "|" and then grep -v to exclude those words like follows:

grep –n ‘\<q.*s\>’ /usr/share/dict/words | grep –nv ‘\<q.*[ar]s\>’ /usr/share/dict/words

This just returns all other words that don't start in "q" or end in "s"

Answer 1

I need to extract all words within the words file that begin with "q", end with "s" but does not contain "a" or "r".

You need to use a negated character class :

grep -n "\<q[^ar]*s\>" /usr/share/dict/words

You may also want to refer to Regular expressions .

Answer 2

The character class [^ar] matches a single charcter which is not a or r. Allow zero or more of these between q and s.

grep '^q[^ar]*s$' /usr/share/dict/words

Answer 3

Through GNU sed ,

sed -n '/^q[^ar]*s$/p' file

It print all the lines that starts with q and ends with s and also it won't contain the the characters a , r inside.

Example:

$ cat cc
qars
qees
qfrs
qoos
jklo
$ sed -n '/^q[^ar]*s$/p' cc
qees
qoos

Through awk ,

$ awk '/^q[^ar]*s$/ {print}' cc
qees
qoos

Answer 4

Thank you, I completed it the long way and worked too. Completely forgot about the "^" this is my method that I then attempted:

grep –in ‘\<q.*s\>’ /usr/share/dict/words | grep –inv "a" | grep -inv "r"