Bash - Date command and space

The correct approach is to define your own function inside your Bash script.

function my_date {
  date "+%y-%m-%d %H:%M:%S"
}

Now you can use my_date as if it were an external program.

For example:

echo "It is now $(my_date)."

Or simply:

my_date

Why isn't your approach working?

The first problem is that your assignment is broken.

DATE_COMMAND="date "+%y-%m-%d %H:%M:%S""

This is parsed as an assignment of the string date +%y-%m-%d to the variable DATE_COMMAND. After the blank, the shell starts interpreting the remaining symbols in ways you did not intend.

This could be partially fixed by changing the quotation.

DATE_COMMAND="date '+%y-%m-%d %H:%M:%S'"

However, this doesn't really solve the problem because if we now use

echo $($DATE_COMMAND)

It will not expand the argument correctly. The date program will see the arguments '+%y-%m-%d and %H:%M:%S' (with quotes) instead of a single string. This could be solved by using eval as in

DATE_COMMAND="date '+%y-%m-%d %H:%M:%S'"
echo $(eval $DATE_COMMAND)

where the variable DATE_COMMAND is first expanded to the string date '+%y-%m-%d %H:%M:%S' that is then evaluated as if it were written like so thus invoking date correctly.

Note that I'm only showing this to explain the issue. eval is not a good solution here. Use a function instead.

PS It is better to avoid all-uppercase identifier strings as those are often in conflict with environment variables or even have a magic meaning to the shell.

Escaping the space works for me.

echo `date +%d.%m.%Y\ %R.%S.%3N`