CODEWITHSUNDEEP
bashtimestamp
Felix
Felix

Reputation: 51

How to convert this timestamp string into this format in bash script?

I have a csv file I am reading which contains various time stamps of the format "2016-06-13T18:30:31.868Z". I would like to know how ,using bash script, how to convert the string into "18:30:31" leaving only the time component.

Thanks.

Upvotes: 0

Views: 275

Answers (4)

computadoradelcarne
computadoradelcarne

Reputation: 11

In Bash

VAR="2016-06-13T18:30:31.868Z"

FRONT_REMOVED=${VAR%.*}

BACK_REMOVED=${FRONT_REMOVED#T*}

Back Removed should be what you want %.* deletes everything after the period

#T* deletes everything before the T

Upvotes: 1

Andreas Louv
Andreas Louv

Reputation: 47099

Use substring expansion:

$ a="2016-06-13T18:30:31.868Z"
$ echo "${a:11:8}"
18:30:31

Upvotes: 0

glenn jackman
glenn jackman

Reputation: 246774

You can do this just with bash parameter expansion

datetime="2016-06-13T18:30:31.868Z"
no_date=${datetime #*T}               # remove "2016-06-13T"
timestamp=${no_date%.*}               # remove ".868Z"

or with a regular expression:

if [[ $datetime =~ ([0-9][0-9]:[0-9][0-9]:[0-9][0-9]) ]]; then
    timestamp=${BASH_REMATCH[1]}
fi

Upvotes: 2

heemayl
heemayl

Reputation: 42007

It would be better to manipulate date-time with a tool that understands it rather than any text-processing tool.

With GNU date, you can pass the time format and get the output in desired HH:MM:SS format with the %T identifier:

date -d '2016-06-13T18:30:31.868Z' '+%T'

To set a specific timezone, use TZ environment variable:

TZ=EST date -d '2016-06-13T18:30:31.868Z' '+%T'

Upvotes: 1

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