Grep only one word not a whole string

Question

I'm writing a simple script to monitoring the disk space in the / partition and I want to grep only numbers in my output.

tabbi@tabbi:~/scripts$ df -h / 
Filesystem      Size  Used Avail Use% Mounted on
/dev/sda2       234G  7,6G  214G   4% /

Let's say I want to remove the % from the Use column (or remove it everywhere); when I use df -h / | grep -v '%' nothing is printed.

How can I exclude the % symbol from my output?

Answer 1

grep -v removes any line matching the given string; so in your case, any line containing “%” is deleted, in its entirety.

To remove the character only, use tr:

df -h / | tr -d %

Answer 2

You can get the / directory usage, without "%", with awk. Basically, the awk in following command:

• Selects the last line of output, to account for the df header - END;
• deletes "%";

• outputs the fifth field (print $5).

  df -h / | awk '  END { gsub("%",""); print $5 } '
  

Running it:

$ df -h / | awk '  END { gsub("%",""); print $5 } '
4

Or to get all numbers:

$ df -h / | awk '  END { gsub("%",""); gsub("G",""); print $2" "$3" "$4" "$5 } '
234 7,6 214 4