python日历来计算月份倒退

时间:2010-04-15 06:22:32

标签: python calendar

我们正在尝试在python中创建日历函数。我们创建了一个小型的内容管理系统,要求是,网站的右上角会有一个下拉列表,它会给出选项 - 月 - 1个月,2个月,3个月等等。 ..,如果用户选择8个月,那么它应该显示过去8个月的postscount。问题是我们试图编写一个可以进行月计算的小代码,但问题是它没有考虑当年之后的月份,它只显示当年的几个月的postscount。

例如:如果用户选择3个月,它将显示l 3个月的计数,即当前月份和前2个月,但如果用户选择超过4个月的选项,则不会考虑去年,它仍然只显示今年的月份。

我粘贴下面的代码: -


def __getSpecifiedMailCount__(request, value):

    dbconnector= DBConnector()
    CdateList= "select cdate from mail_records"

    DateNow= datetime.datetime.today()
    DateNow= DateNow.strftime("%Y-%m")
    DateYear= datetime.datetime.today()
    DateYear= DateYear.strftime("%Y")
    DateMonth= datetime.datetime.today()
    DateMonth= DateMonth.strftime("%m")

    #print DateMonth


    def getMonth(value):
        valueDic= {"01": "Jan", "02": "Feb", "03": "Mar", "04": "Apr", "05": "May", "06": "Jun", "07": "Jul", "08": "Aug", "09": "Sep", "10": "Oct", "11": "Nov", "12": "Dec"}
        return valueDic[value]


    def getMonthYearandCount(yearmonth):
        MailCount= "select count(*) as mailcount from mail_records where cdate like '%s%s'" % (yearmonth, "%")
        MailCountResult= MailCount[0]['mailcount']
        return MailCountResult


    MailCountList= []    
    MCOUNT= getMonthYearandCount(DateNow)
    MONTH= getMonth(DateMonth)
    MailCountDict= {}
    MailCountDict['monthyear']= MONTH + ' ' + DateYear
    MailCountDict['mailcount']= MCOUNT
    var_monthyear= MONTH + ' ' + DateYear
    var_mailcount= MCOUNT
    MailCountList.append(MailCountDict)


    i=1
    k= int(value)
    hereMONTH= int(DateMonth)

    while (i < k):
        hereMONTH= int(hereMONTH) - 1
        if (hereMONTH < 10):
            hereMONTH = '0' + str(hereMONTH)
        if (hereMONTH == '00') or (hereMONTH == '0-1'):
            break
        else:
            PMONTH= getMonth(hereMONTH)
            hereDateNow= DateYear + '-' + PMONTH
            hereDateNowNum= DateYear + '-' + hereMONTH
            PMCOUNT= getMonthYearandCount(hereDateNowNum)
            MailCountDict= {}
            MailCountDict['monthyear']= PMONTH + ' ' + DateYear
            MailCountDict['mailcount']= PMCOUNT
            var_monthyear= PMONTH + ' ' + DateYear
            var_mailcount= PMCOUNT
            MailCountList.append(MailCountDict)
        i = i + 1


    #print MailCountList                

    MailCountDict= {'monthmailcount': MailCountList}    
    reportdata = MailCountDict['monthmailcount']
    #print reportdata
    return render_to_response('test.html', locals())

3 个答案:

答案 0 :(得分:7)

python-dateutil包中的relativedelta函数可以解决这个问题:

from dateutil.relativedelta import *
import datetime

five_months_ago = datetime.datetime.now() - relativedelta(months=5)

答案 1 :(得分:1)

这是一个尴尬的问题,因为几个月有不同的长度。什么是7月31日减1个月?我有类似的要求,但我做了一个简化的假设,我总是想要在一个月的第一天,n个月前。如果它对您的要求有帮助,那么它是:

from datetime import date

def add_months(start_date, months):
    return date(
        start_date.year + (start_date.month - 1 + months) / 12,
        (start_date.month - 1 + months) % 12 + 1,
        1)

答案 2 :(得分:-2)

您可以使用timedelta模块中的datetime来减去月数。

from datetime import datetime, timedelta

now = datetime.now()
four_months_ago = now - timedelta(days=(4*365)/12)

这将跟踪在必要时搬回一年......

>>> january_first = datetime(2009, 1,1)
>>> january_first - timedelta(days=(4*365)/12)
datetime.datetime(2008, 9, 2, 0, 0)