使用'df -h'检查特定文件夹的剩余磁盘空间百分比

Question

I am using ' df -h ' command to get disk space details in my directory and it gives me response as below :

Now I want to be able to do this check automatically through some batch or script - so I am wondering, if I will be able to check disk space only for specific folders which I care about, as shown in image - I am only supposed to check for /nas/home that it does not go above 75%.

How can I achieve this ? Any help ?

---

My work till now:

I am using

df -h > DiskData.txt

... this outputs to a text file

grep "/nas/home" "DiskData.txt"

... which gives me the output:

*500G  254G  247G  51% /nas/home*

Now I want to be able to search for the number previous or right nearby '%' sign (51 in this case) to achieve what I want.

Answer 1

This command will give you percentage of /nas/home directory

df /nas/home | awk '{ print $4 }' | tail -n 1| cut -d'%' -f1

So basically you can use store as value in some variable and then apply if else condition.

var=`df /nas/home | awk '{ print $4 }' | tail -n 1| cut -d'%' -f1`
if(var>75){
#send email
}

Answer 2

another variant:

df --output=pcent /nas/home | tail -n 1 | tr -d '[:space:]|%'

output=pcent - show only percent value (for coreutils => 8.21 )

Answer 3

A more concise way without extensive piping could be:

df -h /nas/home | perl -ane 'print substr $F[3],0,-1 if $.==2'

Returns: 51 for your example.