As a more general response, be careful with parsing dates, because as far as I can tell, it is easy to print times (with formatTime) that cannot be parsed (with parseTime). Specifically, if you strip the padding (of zeros or spaces) there is no way to parse it back in. So parseTime . formatTime is not an identity for anything that cuts out padding. (it results in Nothing for those cases). Perhaps this is because too much data is lost (doesn't seem like this would necessarily be the case), but it does seems like an outstanding problem with parseTime. If I'm wrong, and, for example, a non-zero padded month (which occurs quite frequently in the world), which can easily be printed with "%-m", can be parsed in without pre-parsing and inserting zeros as appropriate first, I would be happy to hear how! On Jun 14, 2011, at 5:32 PM, Dmitri O.Kondratiev wrote:

Magnifique, ca marche! Grand merci, Vincent!

On Wed, Jun 15, 2011 at 1:16 AM, Vincent Gerard wrote: Hello Dmitri,

It seems that your format pattern does not match exactly the format of the input, thus the parser returns Nothing.

Try the following format string which seems to work with your date:

parseTime defaultTimeLocale "%m/%d/%Y %l:%M:%S %p" ds :: Maybe UTCTime returns : Just 2009-10-11 19:04:28 UTC

The parsings errors in your format could come from . %D expects a 2 char year . %H expects a 0 padded hour (like 07, not 7)

Regards,

Vincent Gerard

On Wed, 15 Jun 2011 00:33:56 +0400 "Dmitri O.Kondratiev" wrote:

...

I am trying to convert data string to time:

import Data.Time import Data.Time.Format import Locale

ds = "10/11/2009 7:04:28 PM" t = parseTime defaultTimeLocale "%D %H:%M:%S %p" ds :: Maybe UTCTime

and get "Nothing". What is wrong?

Thanks ! Dmitri.

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