If I run the following command line on Ubuntu:
var="$(date +"%x %r %Z")"
echo var
I can get the desired output. But If I just run
date +%x %r %Z
I will be given errors:
date: extra operand ‘%r’
Try 'date --help' for more information.
So what is the date command corresponding to var="$(date +"%x %r %Z")"? Thanks.
You did not copy exactly the same command. You wrote date +"%x %r %Z" inside parentheses. This just work well outside. Otherwise, date has three arguments (+%x, %r and %Z), rather than one (+%x %r %Z).
From man bash :
When using the $(command) form, all characters between the parentheses make up the command; none are treated specially.
Command substitutions may be nested.
Parentheses are matched before evaluating the content, so the command is not split as $(date +%x, %r, %Z) .
Note :
You forgot the $ of echo $var.
So what is the date command corresponding to var="$(date +"%x %r %Z")"?
Well as already stated in another answer you missed to quote the options right, it should be at least:
date +"%x %r %Z"
This is needed because you are using white-spaces to keep the different information apart, the man page for date suggests using -, or _, with this you could omit the quoting all together in normal terminal, or you escape the white-spaces.
Examples:
using '-' as divider
date +%x-%r%Zgives the following output:
21.09.2018-11:30:15 CESTusing '_' as divider
date +%x_%r%Zgives the following output:
21.09.2018_11:30:15 CESTusing escaped white-spaces as divider
date +%x\ %r%Zgives the following output:
21.09.2018 11:30:15 CESTusing no dividers at all
date +%x%r%Zgives the following output:
21.09.201811:30:15 CEST
Remainder for a special use case. When you try to incorporate date into a crontab command, you need to escape the % sign too. So as undivided information the command would look like this (Before mentioned rules apply for escaping still):
date +\%x\%r\%Z
date +"%x %r %Z"with the quotes on it so that it knows that all the % are part of it. – Terrance Sep 21 '18 at 5:14