[SOLVED] grepping parameters from statement

Issue

I have a statement that looks something like this:

DO prog3 WHILE prog1 arg1 arg2 <= prog2 arg1 END

I would like to extract the parameters prog3, prog1 arg1 arg2, <= (which could be any operator), prog2 arg1 from the statement, using grep-style regex.

My command is:

grep -E 'DO(.*)WHILE(.*)[<=](.*)END' <<< \
'DO prog3 WHILE prog1 arg1 arg2 <= prog2 arg1 END' -o

The regex works on regex101.com, but not in grep, which simply returns the whole statement as a match, ie

DO prog3 WHILE prog1 arg1 arg2 <= prog2 arg1 END

How can I fix this?

Solution

grep doesn't output all capture groups. It would be better to use sed like this:

s='DO prog3 WHILE prog1 arg1 arg2 <= prog2 arg1 END'
sed -E 's/DO (.*) WHILE (.*) ([<>=]+) (.*) END/\1\n\2\n\3\n\4/' <<< "$s"

prog3
prog1 arg1 arg2
<=
prog2 arg1

Here:

• Used [<>=]+ in a separate capture group to grab operator text
• Used spaces around capture groups to handle greediness of .*
• Use \n after each back reference to print each group on a separate line like grep -o

Answered By - anubhava

Answer Checked By - Dawn Plyler (WPSolving Volunteer)