How do I remove the header in the df command? - bash

Question

I'm trying to write a bash command that will sort all volumes by the amount of data they have used and tried using
df | awk '{print $1 | "sort -r -k3 -n"}'
Output:
map
devfs
Filesystem
/dev/disk1s5
/dev/disk1s2
/dev/disk1s1
But this also shows the header called Filesystem.
How do I remove that?

Answer 1

For your specific case, i.e. using awk, #codeforester answer (using awk NR (Number of Records) variable) is the best.
In a more general case, in order to remove the first line of any output, you can use the tail -n +N option in order to output starting with line N:
df | tail -n +2 | other_command
This will remove the first line in df output.

Answer 2

Skip the first line, like this:
df | awk 'NR>1 {print $1 | "sort -r -k3 -n"}'

Answer 3

I normally use one of these options, if I have no reason to use awk:
df | sed 1d
The 1d option to sed says delete the first line, then print everything else.
df | tail -n+2
the -n+2 option to tail say start looking at line 2 and print everything until End-of-Input.
I suspect sed is faster than awk or tail, but I can't prove it.
EDIT
If you want to use awk, this will print every line except the first:
df | awk '{if (FNR>1) print}'
FNR is the File Record Number. It is the line number of the input. If it is greater than 1, print the input line.

Answer 4

Count the lines from the output of df with wc and then substract one line to output a headerless df with tail ...
LINES=$(df|wc -l)
LINES=$((${LINES}-1))
df | tail -n ${LINES}
OK - I see oneliner - Here is mine ...
DF_HEADERLESS=$(LINES=$(df|wc -l); LINES=$((${LINES}-1));df | tail -n ${LINES})
And for formated output lets printf loop over it...
printf "%s\t%s\t%s\t%s\t%s\t%s\n" ${DF_HEADERLESS} | awk '{print $1 | "sort -r -k3 -n"}'

Answer 5

This might help with GNU df and GNU sort:
df -P | awk 'NR>1{$1=$1; print}' | sort -r -k3 -n | awk '{print $1}'
With GNU df and GNU awk:
df -P | awk 'NR>1{array[$3]=$1} END{PROCINFO["sorted_in"]="#ind_num_desc"; for(i in array){print array[i]}}'
Documentation: 8.1.6 Using Predefined Array Scanning Orders with gawk

Answer 6

Removing something from a command output can be done very simply, using grep -v, so in your case:
df | grep -v "Filesystem" | ...
(You can do your awk at the ...)
When you're not sure about caps, small caps, you might add -i:
df | grep -i -v "FiLeSyStEm" | ...
(The switching caps/small caps are meant as a clarification joke :-) )