grep and cut a specific pattern

Asked Jul 30 '21 at 09:24

Active Jul 30 '21 at 09:24

Viewed 180 times

I have a a line and I want to get only the hash number :

2   Information example given false  d8a2e503-d8cf-4851-b809-82cc828efbbd   2020-08-23T13:07:22Z

the second field of string could change every time, it is for that I not use awk '{print $1}'

I want to get only the hash number : d8a2e503-d8cf-4851-b809-82cc828efbbd

I can grep with that : grep -e "[0-9a-f]\{8\}-[0-9a-f]\{4\}-[0-9a-f]\{4\}-[0-9a-f]\{4\}-[0-9a-f]\{12\}"

but cannot cut it

asked Jul 30 '21 at 09:24

looock7897

Why not `awk '{print $(NF-1)}'` to print the last but one field? See https://ideone.com/j3KZC4. Or, just use `-o` option with your `grep` – Wiktor Stribiżew Jul 30 '21 at 09:27
1

If you already have the regular expression, you can use sed: `sed -r 's/.*(RE).*/\1/'` where `RE` is your regular expression. – Renaud Pacalet Jul 30 '21 at 09:27
Did you try `grep -o`? – Renaud Pacalet Jul 30 '21 at 09:29

0 Answers0