Store grep result in variable

Question

As above I'm running this grep command which works as it should but how do I store the returned value into a variable?

cat data.txt | grep "" |cut -d\, -f1

I tried this but it didn't work:

rig=$(cat data.txt | grep "" |cut -d\, -f1)

What is grep "" supposed to do? What is the output of echo $rig?
Its ok I'm new to this BASH learning as I go, I believe I need the "" as an argument without it the script wont run? I have found out this works: rig=grep "" data.txt |cut -d\, -f1 and it seems to be storing the value as I expect so this is working for me :)
you don't need cat or grep. grep "" is pass all, a null pipe, so is cat with 1 argument.
No need to escape the comma, either: cut -d, -f1 is sufficient.

RedGrittyBrick · Answer 1 · 2012-09-14 12:29:16Z

up vote 1 down vote

  rig=`grep "" data.txt | cut -d\, -f1`

seems to be working for me so I will go with this unless anyone else has any pointers on how to improve the above?

edited Sep 14 '12 at 12:29

RedGrittyBrick
38k43372

answered Sep 14 '12 at 10:04

twigg
457

Did you make a transcription error? This does not work. The one in question does. – richard Sep 14 '12 at 10:35

2

I think that's the same as rig=$(cut -d\, -f1 data.txt) without the grep - isn't it? – RedGrittyBrick Sep 14 '12 at 10:49

1 Answer