how to get next month string?

[Yinghe Chen]

Hi,
Could someone help on how to use python to output the next month string like
this?

"AUG07", suppose now is July 2007.

I think also need to consider Dec 07 case, it is supposed to output as
below:
"JAN07".

datetime module seems not supporting the arithmatic operations, any hints?

Thanks in advance,

Yinghe Chen

 

[BartlebyScrivener]

Hi,
Could someone help on how to use python to output the next month string like
this?

"AUG07", suppose now is July 2007.

I usually find time and date answers somewhere in here:

http://pleac.sourceforge.net/pleac_python/datesandtimes.html

rd

 

[John Machin]

Hi,
Could someone help on how to use python to output the next month string like
this?

"AUG07", suppose now is July 2007.

I think also need to consider Dec 07 case, it is supposed to output as
below:
"JAN07".

datetime module seems not supporting the arithmatic operations, any hints?

Thanks in advance,

Yinghe Chen

.... mo = d.month
.... yr = d.year
.... mo += 1
.... if mo > 12:
.... mo = 1
.... yr += 1
.... return datetime.date(yr, mo, 1).strftime('%b%y').upper()
....

 

[Carsten Haese]

... mo = d.month
... yr = d.year
... mo += 1
... if mo > 12:
... mo = 1
... yr += 1
... return datetime.date(yr, mo, 1).strftime('%b%y').upper()

A more concise variant:.... mo = d.month
.... yr = d.year
.... nm = datetime.date(yr,mo,1)+datetime.timedelta(days=31)
.... return nm.strftime('%b%y').upper()

Going 31 days from the first of any month will always get us into the
next month. The resulting day of the month will vary, but we're throwing
that away with strftime.

 

[John Machin]

A more concise variant:>>> import datetime

... mo = d.month
... yr = d.year
... nm = datetime.date(yr,mo,1)+datetime.timedelta(days=31)
... return nm.strftime('%b%y').upper()

Going 31 days from the first of any month will always get us into the
next month. The resulting day of the month will vary, but we're throwing
that away with strftime.

+1 for the "+ 31 days" trick

Sorry about the assembly language

Here's an alternative for folks who prefer legibility:
.... yr, mo = divmod(d.year * 12 + d.month, 12)
.... return datetime.date(yr, mo + 1, 1).strftime('%b%y').upper()

<>
And for the other folks, one of these days I'll get around to writing
a PEP for the /% operator.
</>

 

[Silfheed]

Sounds like a job for dateutil (http://labix.org/python-dateutil).
It's not a built in module (it's in the cheeseshop at least), but it
looks like it pretty much does exactly what you want. If you really
dont want to download anything, I suppose you could create something
from datetime's timedeltas which do support arithmetic operations.

 

[Yinghe Chen]

John's method works perfectly, thanks all for your kind help.

 

[holdenweb]

Hello. Is that you? Wondered what you had been up to lately and discovered that your email address hadn't transferred across to my new computer, so I had to resort to Google and found you on camp.lang.python - I hope!

regards
Steve