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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.

 


Programming today is a race between software engineers striving to
build bigger and better idiot-proof programs, and the Universe
trying to produce bigger and better idiots.
So far, the Universe is winning.

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Posted (edited)

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

Edited by funkey

Programming today is a race between software engineers striving to
build bigger and better idiot-proof programs, and the Universe
trying to produce bigger and better idiots.
So far, the Universe is winning.

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Share on other sites

 

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)

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"In the beginning the Universe was created. This has made a lot of people very angry and been widely regarded as a bad move."

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