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Easy way to get a readable date from a long (timestamp) in scala – Apache

My question is simple

If I do:

var start = System.currentTimeMillis

I get:

start: Long = 1542717303659

How should I do to get a string looking to something readable for a human eye?:

ex: “20/11/2018 13:30:10”

Tags:

3

Answers


  1. 0

    You can use the java.time library like:

      val formatter = DateTimeFormatter.ofPattern("dd/MM/yyyy HH:mm:ss")
  formatter.format(LocalDateTime.now)

    If you only have the timestamp, this solution gets a bit more complex:

    formatter.format(LocalDateTime.ofInstant(Instant.ofEpochMilli(System.currentTimeMillis()), ZoneId.of("UTC")))

    Then I would take java.text.SimpleDateFormat :

    new SimpleDateFormat("dd/MM/yyyy HH:mm:ss").format(System.currentTimeMillis())

    To get back to the Timestamp:

    new SimpleDateFormat("dd/MM/yyyy HH:mm:ss").parse( "02/12/2012 12:23:44" ).getTime
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  2. 0

    Don’t overthink it: nothing wrong with just new Date(start).toString

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  3. 0

    You can use java.time library and get it in readable format as below one-liner.

    scala>  java.time.LocalDateTime.ofEpochSecond(System.currentTimeMillis/1000,0,java.time.ZoneOffset.UTC)
res31: java.time.LocalDateTime = 2018-11-21T18:37:49

scala>

    I’m just diving the Milliseconds by 1000, so that we get EpochSecond.

    For getting it back,

    scala> java.time.LocalDateTime.parse("2018-11-21T18:41:29").toEpochSecond(java.time.ZoneOffset.UTC)
res40: Long = 1542825689

scala>
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