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We want to calculate the first numbers that we get from du

du -b /tmp/*
6 /tmp/216c6f99-6671-4865-b8bc-7205f5388752_resources
668669 /tmp/hadoop7887078727316788325.tmp
6 /tmp/hadoop-hdfs
42456 /tmp/hive
32786 /tmp/hsperfdata_hdfs
6 /tmp/hsperfdata_hive
32786 /tmp/hsperfdata_root
262244 /tmp/hsperfdata_yarn

so final sum will be

sum=6+668669+6+42456+32786+6+32786+262244


echo $sum

How we can do it by awk or perl one liners?

In AWK:

 sum += $1 
END print sum 

So

du -b /tmp/* | awk ' sum += $1 END print sum '

Note that the result won’t be correct if the directories under /tmp have subdirectories themselves, because du produces running totals on directories and their children.

du -s will calculate the sum for you correctly (on all subdirectories and files in /tmp, including hidden ones):

du -sb /tmp

and du -c will calculate the sum of the listed directories and files, correctly too:

du -cb /tmp/*

It is simple you can use:

 du -b /tmp/* | awk 'BEGINi=0 i=i+$1 ENDprint i'

If you are not using wildcard, if you are using directory name like /tmp, then you need to avoid the last entry because output of du -b /tmp is like:

size1 file1
size2 file2
size_total .

So now you should avoid this last entry, so use:

du -b /tmp | awk 'BEGINi=0 if( $2 != "." )i=i+$1 ENDprint i'

However you can also use -s option, it will calculate the summary for you then you don't need to add the values, just print the last one, i.e.:

du -s directory

You can also produce a total sum of selected files with du -c. This works even if an argument of du is not a directory, what is not the case of du -s:

$ du -sb file1 file2
17 file1
18 file2

$ du -cb file1 file2
17 file1
18 file2
35 total

BTW, for interactive use I recommend adding -h option instead of -b or any other multiplier of block-size. This will print the size in human readable unit format.

$ du -ch file1 file2
4.0K file1
4.0K file2
8.0K total