Get only the size of a mounted filesystem

Question

I want to get only the total size of a mounted filesystem. But the catch is that I only know about the mount point. So, I thought of using the df command.

To get the size of this mounted filesystem, I ran the following command:

df --output=target,size | grep -w /mnt/xyz

The result that I got was something like this:

/mnt/xyz             4339044

I know how to use cut but it was of no use here as the space between the string and the integers is unknown to me. Is there a way to just print this size on the terminal?

Solution 1

You can do it without the grep:

df --output=target,size /mnt/xyz | awk ' NR==2 { print $2 } '

df accepts as argument the mount point; you can tell to awk to print both the second line only (NR==2) , and the 2nd argument, $2.

Or better yet, cut the target as you are not outputting it, and it becomes:

df --output=size /mnt/xyz | awk ' NR==2 '

When I was a begginer, I also did manage to get around cut limitations using tr -s " " (squeeze) to cut redundant spaces as in:

df --output=target,size /mnt/xyz | tail -1 | tr -s " " | cut -f2 -d" "

Solution 2

There is also the findmnt command, which can print the number of bytes or a "human" number (powers of 1024 with non-iso abbreviations, sadly):

$ findmnt -no size /mnt/xyz
 9.7G
$ findmnt -bno size /mnt/xyz
10434699264

Comments

[Amoon3234]

Your commands are invalid. Throws error saying that the size column doesn't exist

[meuh]

my version is 2.26.2. Try findmnt --help to see if it lists which columns it allows.

[Manoj Shekhawat]

df -h /mnt/xyz | awk ' NR==3 ' | awk '{print $1}'