grep only certain expressions involving quotation marks

Question

I have a txt file from which I want to get only the expressions of the type 'USA_word*' where * is whatever ( I don't want the whole line, only the expressions )

I try the command

grep -oP ''USA_word*''

But I get a list :

USA_word USA_word USA_word .....

without the part signified by the *.

Maybe grep -o 'USA_word[^[:blank:]]*'? That should work if by "expressions" you mean any 0+ non-whitespace chars. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Sep 25, 2017 at 12:29

Wiktor Stribiżew · Accepted Answer · 2017-09-25 12:31:43Z

You may use

grep -o 'USA_word[^[:blank:]]*'

The [^[:blank:]]* part matches 0+ non-whitespace chars.

Besides, this does not use -P PCRE option, and uses a pure BRE POSIX regex making it compatible with the majority of grep implementations.

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