Parsing dates

When I do this:

dob=“05/01/1968”
x=Date.parse(dob)
x.month

I get 1 instead of 5.
How do I do this?

When I do this:

dob=“05/01/1968”
x=Date.parse(dob)
x.month

I get 1 instead of 5.
How do I do this?

Well, it’s not a bug or something. The rest of the world uses
day/month/year so that is the default that the Date.parse method uses.

The easiest solution (and the best way to do this IMO) is to just feed
the method “01/05/1968” instead of “05/01/1968”.

If for some reason, you HAVE to have the m/d/y format, you could just
use a regexp. Could be something like this, but note, it’s not really
the best regexp, just something I came up with now (you can use
rubular.com to figure out ruby regexps easily):

dob = “05/01/1968”
dob =~([01]\d)/([0-3]\d)/((19|20)\d\d)

On Fri, Sep 25, 2009 at 1:38 AM, Ehsanul H. [email protected]
wrot>

Well, it’s not a bug or something. The rest of the world uses day/month/year so that is the default that the Date.parse method uses.

The easiest solution (and the best way to do this IMO) is to just feed the method “01/05/1968” instead of “05/01/1968”.

If for some reason, you HAVE to have the m/d/y format, you could just use a regexp. Could be something like this, but note, it’s not really the best regexp, just something I came up with now (you can use rubular.com to figure out ruby regexps easily):

dob = “05/01/1968”
dob =~([01]\d)/([0-3]\d)/((19|20)\d\d)

Or better, IMHO, use strptime

require ‘date’
d = Date.strptime(“05/01/1968”, “%m/%d/%Y”)

d.month #=> 5
d.day #=> 1
d.year #=> 1968

Works for DateTime as well as Date


Rick DeNatale

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