linux - Find OS disk size using df and awk

Question

size=`df / | awk '{print $4}'`

The following code works, however it also prints the word 'Available':

Available
2079328628

How can I take only the size into size variable?

Answer 1

You can skip the first line:

size=$(df / | awk 'NR>1{print $4}')

Here, NR>1 only prints the fourth field value if the line is not the first one.

See an online demo.

Answer 2

Using --output option in gnu df you can pick & chose fields that you want in output. So you can use:

df --output=avail / | sed '1d'

OR:

df --output=avail / | tail -n +2

OR:

df --output=avail / | awk 'NR>1'

Answer 3

Using just one sub-shell for df and parsing with Bash built-in features:

#!/usr/bin/env bash

read -r -d '' _ avail < <(df --output=avail /)

No sub-shell at all version:

#!/usr/bin/env bash

trap 'rm -f "$tmpfile"' EXIT
tmpfile="/tmp/avail_$RANDOM"
df --output=avail / >"$tmpfile"
read -r -d '' _ avail <"$tmpfile"
echo "$avail"

Answer 4

Two more using df and awk:

df | awk '{if ($4 != "Available") print $4}'

and

df | grep -v Available| awk '{print $4}'