Specifying bash command as variable and then executing during runtime

Question

How do I specify the command date as a variable so that it is executed upon calling said variable?

DATE=`date +%H:%M:%S`
echo "$DATE"
sleep 5
echo "$DATE" # Date displays the original time, not the current one

--

EDIT 1: According to anubhava's suggestion:

freshdate[0]=`date +%H:%M:%S`

However, this does not result in:

echo "${freshdate[0]}"  # 19:25:50
sleep 5
echo "${freshdate[0]}"  # 19:25:50

Use a function or array to store command line – anubhava May 18 '17 at 17:08 — anubhava, May 18 '17 at 17:08

score 2 · Accepted Answer · answered May 18 '17 at 17:35

2

Easiest is to wrap this command in a shell function like this:

dt() { date '+%H:%M:%S'; }

Then use it anywhere you want to access it as in:

echo "Time now is: $(dt)"
Time now is: 13:34:30

answered May 18 '17 at 17:35

anubhava

1

Nice explanation. Thanks for your answer! – Michael Gruenstaeudl May 18 '17 at 17:37

score -1 · Answer 2 · answered May 18 '17 at 17:25

-1

You can do

DATE=(date +%H:%M:%S)
${DATE[@]}
sleep 5
${DATE[@]}

and will show different times.

answered May 18 '17 at 17:25

Yes, that prints the new time, but I would like to `echo` the new time (i.e., integrate it into a longer string). Something along the line of `echo "Foo bar baz $DATE[@]"`. How do I do that? – Michael Gruenstaeudl May 18 '17 at 17:27

2 Answers2