How to add mathematically the results of two commands in Linux

Question

In a Debian Linux shell, I am trying to add the results of two separate commands so that I know the sum of both.

I already tried:

echo $(expr $(du -sh /srv/mysql) + $(du -sh /srv/www)) 

and variations of it.

It's returning: expr:

"syntax error: unexpected argument ‘/srv/mysql’

"

of course I also tried to paraphrase the folders, etc.

• Kamil Maciorowski: Not the main issue, but please read: What is wrong with echo $(stuff)?
• Edgar Magallon: Regarding syntax error: unexpected argument ‘/srv/mysql’ that happens because du -sh /srv/mysql returns something like: 500M /srv/mysql and therefore that's not a correct number.
• Stéphane Chazelas: Just use du -csh /srv/www /srv/mysql to get the cumulative disk usage of both directories (which is not the same as the sum of the disk usage of each run in isolation if both directories contain instances of the same file for instance).

Answer 1

Like this:

could be used with any shell (bash, zsh, sh, dash...)

echo "$((
 (
 $(du -s /srv/mysql | awk '{print 1ドル}') +
 $(du -s /srv/www | awk '{print 1ドル}')
 ) / 1024
))MiB"

Or simply:

du -shc /srv/www /srv/mysql | awk 'END{print 1ドル}'

The -c act as the count operator

• Edgar Magallon: Good choice about du -s instead of du -sh. To avoid cofusion: About the MB at the end of echo, it should not be M or MiB?. Because if I'm not wrong Linux works with *bibytes. For example: if du -sh shows 323M it actually means 323MiB or 323 Mibibytes.
• Gilles Quénot: Thanks, edited accordingly

Answer 2

Assuming you are dealing with integers ...

I note that expr is considered archaic and obsolete. It has two main uses, arithmetic and regex matching, both of which are available in most shells.

(By the by, "Debian Linux shell" isn't just one thing; it could be Bash or Dash or indeed whichever shell is chosen when the user is added or changed subsequent.)

For arithmetic, POSIX requires the shell (any shell claiming compliance, including /bin/sh) to evaluate $(( expression )) according to rules similar to those for integer expressions in the C language, so that $((1+2*3)) should give 7.

The main problem with the original question was that other information besides the digits were included in the output. The first thing to try should be to ask the program generating those digits not to output anything else. Only if that fails, then filter the output. There are many many ways to do that, but the simplest is just

tr -dc 0-9

which will remove everything that's not a digit.

If you need decimal fractions ...

(decimal fractions are usually fixed point, like 1.23, but floating point systems can understand fixed point numbers as well, so you can use either.)

Neither expr nor $((...)) can cope with decimal fractions. For that you need bc or to use a separate scripting language that has fixed point or floating point built in, such as awk or perl or python.