python实现的阳历转阴历(农历)算法

2019-10-06 17:15:58王冬梅

    elif dLunar.day==0:   
        tmp2 = '闰'*isLeapMonth + num2GB(dLunar.month+1) +'月'
        return '%s' % tmp2 + ' '*(8-len(tmp2))
    elif dLunar.day<10:
        return '初' + num2GB(dLunar.day+1)
    else:
        return num2GB(dLunar.day+1)


def outputCalendar(year, month):
    dLunar = Date(-1,-1,-1)
    ow ('n     阳历%d年%d月         ' % (year+1900, month+1) )

    for iDay in range( getSolarDaysInMonth(year, month) ):
        dSolar = Date(year, month, iDay)
        dLunar, isLeapMonth = solar2Lunar (dSolar)

        if iDay==0:
            ow ('始于 阴历%s年%s%s月 (%s%s年, 生肖属%s)n' %
                ( num2GB(dLunar.year+1900), '闰'*isLeapMonth,
                  num2GB(dLunar.month+1),
                  ganGB [dLunar.gan], zhiGB[dLunar.zhi], shengXiaoGB[dLunar.zhi]
                ))
            ow ('='*74 + 'n')
            for i in range(7):
                ow ("%3s %2s     " % (weekdayEn[i][:3], weekdayGB[i]) )
            ow('nn')
            for i in range(dLunar.weekday): ow(' '*11)

        elif dLunar.weekday==0: ow('n')

        ow ( "%2d %-8s" %(iDay+1, lunarDate2GB(dLunar, isLeapMonth) ) )
    ow('nn')

 


def checkArgv (argv):
    argc = len(argv)
    if argc==1 or argv[1] in ('-h', '--help'):
        print __doc__; exit(0)

    #in case people input arguments as "4-digit-year month"
    if argc==3 and len(argv[1]) == 4 and len(argv[2]) in (1,2):
        argv[1], argv[2] = argv[2], argv[1]

   
    #Get month
    month=-1
    for iMonth in range(12):