bash: split output of command by columns without specific delimiter

Question

I'm try to extract the contents of a given column, but spaces have been used to format the columns so I don't have a specific delimiter to work with.

The data I'm working with is the result of the following OS X command:

diskutil list | grep 'Apple'

Example output:

   2:                  Apple_HFS Macintosh SSD           511.8 GB   disk0s2
   2:                  Apple_HFS External                79.7 GB    disk1s2

The column I want to extract is the disk name, i.e. "Macintosh SSD" and "External"

tripleee · Accepted Answer · 2013-01-14 06:10:58Z

4

You should use diskutil list -plist and parse the property list instead. The human-readable output is not meant to be machine readable, and might look different than you expect (column boundaries could move to accommodate long outputs or a wide terminal window, etc).

answered Jan 14, 2013 at 6:10

tripleee

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2 Comments

anishsane Over a year ago

Like @ "The human-readable output is not meant to be machine readable"

beej Over a year ago

Thank you. It wasn't the answer I was hoping for, but it's spot on.

Eran Ben-Natan · Accepted Answer · 2013-01-14 09:35:10Z

0

If I understand correctly, you need to print the words from the 3rd to the Nth-3. If so, the following awk will do:

gawk '{disk=""; for (i = 3; i <= NF-3; i++){disk=disk $i " "};print disk}'

answered Jan 14, 2013 at 9:35

Eran Ben-Natan

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Comments

Lri · Accepted Answer · 2013-01-14 11:48:06Z

0

You could also use tr -s ' ' to collapse (squeeze) spaces:

diskutil list | tr -s ' ' | cut -d ' ' -f 3 | grep Apple

Or just regex:

diskutil list | sed -En 's/^[ 0-9]+: +(Apple[^ ]+).+/\1/p'

answered Jan 14, 2013 at 11:48

Lri

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