Execute a command and put the results into a variable... all in a bash script [duplicate]

I'm working on a bash script that backs up a configuration file before copying over a new file.

Here's what my snippet looks like:

mv ~/myStuff.conf  ~/myStuff.conf.bak
cp ~/new/myStuff.conf ~/myStuff.conf

Every time this script is run, I'd like there the backup to have a unix timestamp in the filename. I tried this

DATEVAR=date +%s
mv ~/myStuff.conf  ~/myStuff.conf.$DATEVAR.bak

But this doesn't work, since the date function doesn't execute and bash sees it as a string, and the resulting file ends up being

myStuff.conf.date+%s.bak

Any ideas on how to get the results of the date function into a variable?

This is possible with command substitution.

DATEVAR=$(date +%s)

--[[z4us|binz--]]

export datevar=`date` # date embedded in backquotes

--[[z4us|binz--]]

echo $datevar

Lun 25 Gen 2016 15:56:14 CET

This does not answer the variable holding the output of a command. That is already answered. As for the rest of your example script;

A little shorter version:

mv ~/myStuff.conf  ~/myStuff.conf.$(date +%s)

No need to set a variable for something you only need or use once. Also, to be compatible with more shells, you could also use this syntax:

mv ~/myStuff.conf  ~/myStuff.conf.`date +%s`

It just seems to me that having the datestamp as an extension eliminates the need for the additional .bak in the filename.

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