RegExp Date time

[funkey]

Hello,

wisch you a happy new year!

 

Can you help me to find a regexp pattern for replacing timestamps in a csv file please.

 

I have this Format in the file: 21.07.2019;14:12:10 (dd.mm.yyyy;hh:mm:ss)

but I Need this: 2019-07-21 00:00:00 (yyyy-mm-dd hh:mm:ss)

 

Thanks, would be great if you can help me.

P.S.: I want to use StringRegExpReplace for the hole file/string.

 

[Marc]

One possible solution:

$input = "21.07.2019;14:12:10"
$output = StringRegExpReplace($input, "(\d\d)\.(\d\d)\.(\d\d\d\d);", "\3-\2-\1 00:00:00")
ConsoleWrite($output)

best regards,

Marc

[funkey]

Thanks a lot, works great! (without 00:00:00, because I made an error; time has to be the same)

Edited January 7, 2020 by funkey

[Musashi]

 

Local $sInput = "Date Line 01 = 21.07.2019;14:12:10 " & @CRLF & _
                "Date Line 02 = 29.03.2014;03:04:59 " & @CRLF & _
                "Date Line 03 = 03.12.2013;22:41:15 "
Local $sOutput = StringRegExpReplace($sInput, "(\d{2})\.(\d{2})\.(\d{4});(\d{2}):(\d{2}):(\d{2})", "\3-\2-\1 \4:\5:\6")
ConsoleWrite($sInput & @CRLF & @CRLF)
ConsoleWrite($sOutput & @CRLF)