CODEWITHSUNDEEP
groovy
Sri Chakra
Sri Chakra

Reputation: 11

Groovy - Unable to convert date string "2018-07-18T01:40:05-0400" to format required format

Below is my response code from a rest request

"createdDate" : "2012-02-27T18:10:28-0500",
"isDeleted" : "indeterminate",
"modifiedDate" : "2018-07-18T01:40:05-0400"

I am trying to convert a time stamp of format 2018-07-18T01:40:05-0400 to MM/dd/YYYY hh:mm

-0400 indicates time zone

can anyone tell me how to convert this type of time stamp to required format?

Upvotes: 1

Views: 156

Answers (2)

bdkosher
bdkosher

Reputation: 5883

If you're on Java 8 and Groovy 2.5,

java.time.OffsetDateTime.parse('2018-07-18T01:40:05-0400', "yyyy-MM-dd'T'HH:mm:ssXX")
                        .format('MM/dd/yyyy hh:mm')

On Java 8 and an earlier version of Groovy,

import java.time.OffsetDateTime
import java.time.format.DateTimeFormatter

OffsetDateTime.parse('2018-07-18T01:40:05-0400', DateTimeFormatter.ofPattern("yyyy-MM-dd'T'HH:mm:ssXX"))
              .format(DateTimeFormatter.ofPattern('MM/dd/yyyy hh:mm'))

Upvotes: 2

3sky
3sky

Reputation: 890

If i correctly understand problem You should prepare input timestamp and use Date.parse.

def input = '2018-07-18T01:40:05-0400'
pdata = input.replace("T", " ").substring(0, input.replace("T", " ").length() - 5)
def out = Date.parse("yyyy-MM-dd hh:mm:ss", pdata).format("MM/dd/yyyy hh:mm")

/* Output
println "Input date: " + input
println "Parse date: " + out

Input date: 2018-07-18T01:40:05-0400
Parse date: 07/18/2018 01:40
*/

Upvotes: 1

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