Print out df command's output using grep in bash script

Question

I'm writing a bash script where I need to print out the content of Available in /home. So in the below example, it would be 458G. I was thinking maybe I can use grep or awk and then print $3 or something like that? I'd appreciate the help here! I'm new to bash script.

**Filesystem          1G-blocks  Used Available Use% Mounted on**

/dev/mapper/cl-root       50G   28G       23G  55% /
devtmpfs                  48G    0G       48G   0% /dev
tmpfs                     48G    1G       48G   1% /dev/shm
tmpfs                     48G    1G       47G   1% /run
tmpfs                     48G    0G       48G   0% /sys/fs/cgroup
/dev/sda2                  1G    1G        1G  45% /boot
/dev/sda1                  1G    1G        1G   5% /boot/efi
/dev/mapper/cl-home      492G   34G      458G   7% /home
tmpfs                     10G    1G       10G   1% /run/user/1000

score 1 · Answer 1 · answered Feb 08 '21 at 15:54

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Specify /home as an argument to df.

df /home

answered Feb 08 '21 at 15:54

Andy Lester

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score 0 · Answer 2 · answered Feb 08 '21 at 16:08

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df -h | grep /home | awk '{print $4}' will give you Available field for /home

answered Feb 08 '21 at 16:08

dejanualex

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I suggest `df -hP /home | awk 'NR==2{print $4}'`. – Cyrus Feb 08 '21 at 16:16

M. Nejat Aydin · Answer 3 · 2021-02-08T16:47:55.763

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This question has nothing to do with bash. Since it has a linux tag, then simply:

df -h --output=avail /home | tail -n1

edited Feb 08 '21 at 16:47

answered Feb 08 '21 at 16:16

M. Nejat Aydin

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Print out df command's output using grep in bash script

3 Answers3