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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)
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What is grep "" supposed to do? What is the output of echo $rig? – Stefan Seidel Sep 14 '12 at 10:00
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 :) – twigg Sep 14 '12 at 10:04
what is the \ for? – richard Sep 14 '12 at 10:29
you don't need cat or grep. grep "" is pass all, a null pipe, so is cat with 1 argument. – richard Sep 14 '12 at 10:31
No need to escape the comma, either: cut -d, -f1 is sufficient. – chepner Sep 14 '12 at 13:39
  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?

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

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