size=`ls -l /var/temp.* | awk '{ print $5}'`
fin_size=0
for row in ${size} ;
do
fin_size=`echo $(( $row + $fin_size )) | bc`;
done
echo $fin_size
is not working !! echo $fin_size is throwing some garbage minus value.
where I'm mistaking?
(my bash is old and I suppose to work in this only Linux kernel: 2.6.39)
du suggestion; just noting that tail -1 is deprecated and you should use tail -n 1 instead.
Because it cannot be stressed enough: Why you shouldn't parse the output of ls(1)
Use e.g. du as suggested by dogbane, or find:
$ find /var -maxdepth 1 -type f -name "temp.*" -printf "%s\n" | awk '{total+=$1}END{print total}'
or stat:
$ stat -c%s /var/temp.* | awk '{total+=$1}END{print total}'
or globbing and stat (unnecessary, slow):
total=0
for file in /var/temp.*; do
[ -f "${file}" ] || continue
size="$(stat -c%s "${file}")"
((total+=size))
done
echo "${total}"
find to awk will fail under the same circumstances that piping the output of ls will fail: when the file name contains a newline character.Below should be enough:
ls -l /var/temp.* | awk '{a+=$5}END{print a}'
No need for you to run the for loop.This means:
size=ls -l /var/temp.* | awk '{ print $5}'`
fin_size=0
for row in ${size} ;
do
fin_size=`echo $(( $row + $fin_size )) | bc`;
done
echo $fin_size
The whole above thing can be replaced with :
fin_size=`ls -l /var/temp.* | awk '{a+=$5}END{printf("%10d",a);}'`
echo $fin_size
ls; it's not guaranteed to work in all cases, and should not be suggested as an answer.
ls(read here why), so this and the accepted answer is the wrong approach to your problem. Usedu,statorfind.