CODEWITHSUNDEEP
bashgreppipe
Vita
Vita

Reputation: 123

Print regex string found with grep

I have this crazy long command-turned-(bash)script that outputs a tiny table with and object and some other data relate to it. I want to single the name of the object out to use it as a variable to further run more commands.

It outputs something like this:

ID          PATH       NAME           VERSION             UPGRADE STATUS
vm-13034    /abc/def   somethingelse  vX.X.X-XXX-XXXXXXX  Up to date

The value I'm interested in is ID, which always is vm-#####. With an online tool I came up with the regex ^vm-\d{5,}$ (, bc the number is incrementing). So I figured vosxls | grep -E ^vm-\d{5,}$ (vosxls is the script's name for the long command) would work but it returns nothing. I tried it with -w, -e, grep -e "vm-\d{5,}" <(vosxls) and a few more adding and remove the enclosing ^ & $ characters when it would throw an error, enclosing the string in soft- and hard quotes or nothing at all, whatever would work.

Everything else that didn't error, returned nothing.

I read some examples with Perl and stuff similar to that but I stayed away and generally have stayed as basic as possible so I can run it in another system without issues. The most I've diverted from the most basic stuff (which are all I know anyway) was egrep which is I believe is the same as grep -e if I'm not mistaken.

How can the string be printed out? Is it because I'm already using a script it somehow affects how grep works? The script has no variables, it is merely a shorthand to avoid typing a lot of difficult to remember syntax+values.

Thanks.

Upvotes: 2

Views: 162

Answers (4)

user1934428
user1934428

Reputation: 22225

You can't use \d with the -E option of grep to denote a digit. -P, respectively -Po, will work. But if you already know that you need the first field of the last line of the output, you don't need grep. A

vosxlx | tail -n 1 | cut -d ' ' -f 1

would do as well.

Upvotes: 0

Taylor G.
Taylor G.

Reputation: 711

Could be as simple as

vosxls | grep -Eo "^vm-\d{5,}"

Specifying -o allows you to return whatever matches the regex only.

-o, --only-matching
    Prints only the matching part of the lines.

Upvotes: 3

Suku
Suku

Reputation: 3880

foo | perl -ne 'print $1 if /^(vm-\d+)/'

Your version of grep might not have extended regex support:

$:/mnt/c/Users/sukuj$ cat /tmp/s.txt
ID          PATH       NAME           VERSION             UPGRADE STATUS
vm-13034    /abc/def   somethingelse  vX.X.X-XXX-XXXXXXX  Up to date
vm-23034    /abc/def   somethingelse  vX.X.X-XXX-XXXXXXX  Up to date

$:/mnt/c/Users/sukuj$ grep -oE 'vm-[0-9]+' /tmp/s.txt
vm-13034
vm-23034

$:/mnt/c/Users/sukuj$ grep -oE 'vm-[0-9]{5,}' /tmp/s.txt
vm-13034
vm-23034

$:/mnt/c/Users/sukuj$ grep -V
grep (GNU grep) 3.4
Copyright (C) 2020 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <https://gnu.org/licenses/gpl.html>.
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.

Written by Mike Haertel and others; see
<https://git.sv.gnu.org/cgit/grep.git/tree/AUTHORS>.
suku@DESKTOP-GID8MQV:/mnt/c/Users/sukuj$

Upvotes: 1

jared_mamrot
jared_mamrot

Reputation: 26505

Perhaps something like vosxlx | sed -n 's/\(vm.*\ \)\(\/.*\)/\1/p'?

Or maybe vosxlx | awk '/vm/ {print $1}'

Upvotes: 1

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