How to compare the size of two directories?

问题:

I want to compare the total size of two directories dir1 and dir2 on different file-systems so that if diff -r dir1 dir2 returns 0 then the total sizes will be equal. The du command returns the disk usage, and its option --apparent-size doesn't solve the problem. I now use something like

find dir1 ! -type d |xargs wc -c |tail -1

to know an approximation of dir1's size. Is there a better solution?

edit: for example, I have (diff -r dir1 dir2 returns 0: they are equal):

du -s dir1 --> 540
du -s dir2 --> 166

du -sb dir1 --> 250815 (the -b option is equivalent to --apparent-size -B1)
du -sb dir2 --> 71495

find dir1 ! -type d |xargs wc -c --> 62399
find dir2 ! -type d |xargs wc -c --> 62399 

回答1:

i can't know what you want clearly. Maybe you want this?
diff <(du -sh dir1) <(du -sh dir2)

回答2:

If your version of find has -printf you may find this to be quite a bit faster.

find dir1 ! -type d -printf "%s\n" | awk '{sum += $1} END{print sum}'

There are at least two ways to avoid scientific notation for outputting large numbers in AWK.

END {OFMT = "%.0f"; print sum}

END {printf "%.0f\n", sum}

The .0 truncates the decimal places since we're really dealing with an integer and gawk's %d seems to incorrectly act like %g in version 3.1.5 (but not 3.1.6 and later).

However, from the gawk documentation:

NOTE: When using the integer format-control letters for values that are outside the range of the widest C integer type, 'gawk' switches to the '%g' format specifier.

Beware of exceeding the maximum integer for your system/version of AWK.