Calculate sum of several sizes of files in Bash
I have list of files in a file, cache_temp
.
In file cache_temp
:
/home/maildir/mydomain.com/een/new/1491397868.M395935P76076.nm1.mydomain.com,S=1740,W=1777
/home/maildir/mydomain.com/een/new/1485873821.M199286P14170.nm1.mydomain.com,S=440734,W=446889
/home/maildir/mydomain.com/td.pr/cur/1491397869.M704928P76257.nm1.mydomain.com,S=1742,W=1779:2,Sb
/home/maildir/mydomain.com/td.pr/cur/1501571359.M552218P73116.nm1.mydomain.com,S=1687,W=1719:2,Sa
/home/maildir/mydomain.com/td.pr/cur/1498562257.M153946P22434.nm1.mydomain.com,S=1684,W=1717:2,Sb
I have a simple script for getting the size of files from cache_temp
:
#!/bin/bash
for i in `grep -v ^# ~/cache_temp | grep -v "dovecot.index.cache"`; do
if [ -f "$i" ]; then
size=$(du -sh "$i" | awk '{print $1}')
echo $size
fi
done
I have a list of sizes of files:
4,0K
4,0K
4,0K
432K
4,0K
How can I calculate the sum of them?
Solution 1:
Use stat
instead of du
:
#!/bin/bash
for i in `grep -v ^# ~/cache_temp | grep -v "dovecot.index.cache"`; do
[ -f "$i" ] && totalsize=$[totalsize + $(stat -c "%s" "$i")]
done
echo totalsize: $totalsize bytes
Solution 2:
According to du(1), there is a -c option whose purpose is to produce the grand total.
% du -chs * /etc/passwd
92K ABOUT-NLS
196K NEWS
12K README
48K THANKS
8,0K TODO
4,0K /etc/passwd
360K total
Solution 3:
If you need to use the file this snippet is hopefully efficient.
xargs -a cache_file stat --format="%s" | paste -sd+ | bc -l
The xargs
is to prevent overflowing the argument limit but getting the max number of files into one invocation of stat
each time.
Solution 4:
If you remove the "-h" flag from your "du" command, you'll get the raw byte sizes. You can then add them with the ((a += b))
syntax:
a=0
for i in $(find . -type f -print0 | xargs -0 du -s | awk {'print $1'})
do
((a += i))
done
echo $a
The -print0
and -0
flags to find/xargs use null-terminated strings to preserve whitespace.
EDIT: turns out I type slower than @HBruijn comments!