CODEWITHSUNDEEP
linuxbashshell
oikonomiyaki
oikonomiyaki

Reputation: 7951

Variable within a Linux shell expression

I have an hour variable in shell that contains UNIX time in seconds.

Then If want to format this hour, I use

date -u -d @$hour +"%Y-%m-%dT%H:%M:%S.000Z"

which works. However, I want to store the result of the above expression to another variable, so when I do:

formatted=$(( date -u -d @$hour +"%Y-%m-%dT%H:%M:%S.000Z" ))

it does not work. I do not know how will I reference the hour variable within an evaluation expression (whatever it is called).

Upvotes: 1

Views: 2269

Answers (3)

Thomas Dickey
Thomas Dickey

Reputation: 54475

The expression

formatted=$(( date -u -d @$hour +"%Y-%m-%dT%H:%M:%S.000Z" ))

use the $(( which works only for arithmetic. Change the double parentheses to single:

formatted=$( date -u -d @$hour +"%Y-%m-%dT%H:%M:%S.000Z" )

The latter works for strings.

For reference (these are POSIX shell features, not bash-specific):

Upvotes: 12

tripleee
tripleee

Reputation: 189317

Your syntax for command substitution is subtly wrong. The use of double parentheses after $ introduces a quite distinct context in Bash, which is known as arithmetic context.

So just drop one pair of parentheses.

formatted=$(date -u -d @$hour +"%Y-%m-%dT%H:%M:%S.000Z")

The use of backticks is syntactically valid, but discouraged.

formatted=`date -u -d @$hour +"%Y-%m-%dT%H:%M:%S.000Z"`

Upvotes: 1

anand
anand

Reputation: 1526

You can do it as following:

hour=1447409296  #Whatever timestamp value you want to set
formatted=`date -u -d @$hour +"%Y-%m-%dT%H:%M:%S.000Z"`
echo $formatted

or:

formatted=`date -u -d @$hour +"%Y-%m-%dT%H:%M:%S.000Z"`

Upvotes: 0

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