Script to monitor diskspace of /home/ not to exceed 80%

I want to monitor the diskspace of the /home/ directory NOT to exceed 80%.
I want a script_file.sh for this.

I am trying "df -h" command and I am stuck unable to extract the percentage.

1. HOW TO extract the percentage (I believe 5th column)?

2. IF we were able to extract the column, will it return just the value or the value with "%" sign?

3. I want this value to be tested in a IF condition to check its >80%

Please help me quick, its urgent!!
Thanks

You better use df -kP or df -k in a script.
The % in bash/ksh/ash/psh can be chopped off.

percent=`df -kP /home/. | awk 'NR>1 {print $5}'`
percent=${percent%*%}

or

percent=`df -kP /home/. | (read _; read _ _ _ _ percent _; echo  ${percent%*%})`

thanks, but what does NR>1 do? and what is ${percenta%*%}

does it return only one value of the entire /home/ directory ? coz I need that.

will

IF [ $( df -kP /home | awk '{print $5}' ) -gt 80]; then
                  echo "exceeds"
               else
                   echo "don't"
               FI 

this work?

NR > 1 skips the first (= header) line in awk . shell's parameter expansion ${percent%*%} (no "a"!) removes the trailing % sign.

It will not, due to syntactical and semantical errors:

• IF should be lower case
• the closing ] needs a leading space
• the command substitution needs double quotes
• the test ( or [ ) command can't handle multiline arguments, it needs a single item output from the command substitution.
• the -gt comparison can't handle the % sign

${percent%*%} is a variable modifier. So you must store it in a variable first.

percent=`df -kP /home/. | awk 'NR>1 {print $5}'`
if [ ${percent%*%} -gt 80 ]; then
  echo "exceeds"
else
  echo "ok"
done

Or

if [ `df -kP /home/. | (read _; read _ _ _ _ percent _; echo  ${percent%*%})` -gt 80 ]
then
  echo "exceeds"
else
  echo "ok"
fi

The NR>1 or the read _ skip the first line (header).

if [ $( df -kP /home | awk 'NR>1 {print $5}' ) -gt 80 ]; then
                  echo "exceeds"
               else
                   echo "don't"
               fi

this might work if we can remove the % sign..
can't we remove without storing into a variable?
IF not then as per ur first suggestion,
if [ percentage -gt 80 ]; then
echo "exceeds";
else
echo "dont "
fi

this will work anyway right?

How about

df -kP /home | awk 'NR>1 {print ($5+0 > 80)?"exceeds":"doesn''t"}' 

Rudic,
You are taking 6th column? I need 5th as it gives the percentage usage.

No, it's adding 0 to make it numeric (i.e. drop the % sign).
I'm sorry - I made a typo when writing $5+1 - should read $5+0 . Corrected in post#7.