不知在delphi中如何处理闰月问题,要知道闰月好象是没有规律的啊!要找一个合适的算法好象不容易啊...不知那位大侠知道,望告知!!!
解决方案 »
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如没有则为0,每字节存两年: gLunarMonth:array[0..74] of Byte=(
$00, $50, $04, $00, $20, //1910
$60, $05, $00, $20, $70, //1920
$05, $00, $40, $02, $06, //1930
$00, $50, $03, $07, $00, //1940
$60, $04, $00, $20, $70, //1950
$05, $00, $30, $80, $06, //1960
$00, $40, $03, $07, $00, //1970
$50, $04, $08, $00, $60, //1980
$04, $0a, $00, $60, $05, //1990
$00, $30, $80, $05, $00, //2000
$40, $02, $07, $00, $50, //2010
$04, $09, $00, $60, $04, //2020
$00, $20, $60, $05, $00, //2030
$30, $b0, $06, $00, $50, //2040
$02, $07, $00, $50, $03); //2050
)
START_YEAR=1901;
END_YEAR=2050;//返回iYear年iMonth月的天数 1年1月 --- 65535年12月
function MonthDays(iYear,iMonth:Word):Word;//返回阴历iLunarYer年阴历iLunarMonth月的天数,如果iLunarMonth为闰月,
//高字为第二个iLunarMonth月的天数,否则高字为0 1901年1月---2050年12月
function LunarMonthDays(iLunarYear,iLunarMonth:Word):Longword;//返回阴历iLunarYear年的总天数 1901年1月---2050年12月
function LunarYearDays(iLunarYear:Word):Word;//返回阴历iLunarYear年的闰月月份,如没有返回0 1901年1月---2050年12月
function GetLeapMonth(iLunarYear:Word):Word;//把iYear年格式化成天干记年法表示的字符串
procedure FormatLunarYear(iYear:Word;var pBuffer:string);overload;
function FormatLunarYear(iYear:Word):string;overload;//把iMonth格式化成中文字符串
procedure FormatMonth(iMonth:Word;var pBuffer:string;bLunar:Boolean=True);overload;
function FormatMonth(iMonth:Word;bLunar:Boolean=True):string;overload;//把iDay格式化成中文字符串
procedure FormatLunarDay(iDay:Word;var pBuffer:string);overload;
function FormatLunarDay(iDay:Word):string;overload;//计算公历两个日期间相差的天数 1年1月1日 --- 65535年12月31日
function CalcDateDiff(iEndYear,iEndMonth,iEndDay:Word;iStartYear:Word=START_YEAR;iStartMonth:Word=1;iStartDay:Word=1):Longword;overload;
function CalcDateDiff(EndDate,StartDate:TDateTime):Longword;overload;//计算公历iYear年iMonth月iDay日对应的阴历日期,返回对应的阴历节气 0-24
//1901年1月1日---2050年12月31日function GetLunarHolDay(InDate:TDateTime):string;overload;
function GetLunarHolDay(iYear,iMonth,iDay:Word):string;overload;//private function--------------------------------------//计算从1901年1月1日过iSpanDays天后的阴历日期
procedure l_CalcLunarDate(var iYear,iMonth,iDay:Word;iSpanDays:Longword);//计算公历iYear年iMonth月iDay日对应的节气 0-24,0表不是节气
function l_GetLunarHolDay(iYear,iMonth,iDay:Word):Word;implementationvar
//数组gLunarDay存入阴历1901年到2100年每年中的月天数信息,
//阴历每月只能是29或30天,一年用12(或13)个二进制位表示,对应位为1表30天,否则为29天
gLunarMonthDay:array[0..149] of Word=(
//测试数据只有1901.1.1 --2050.12.31
$4ae0, $a570, $5268, $d260, $d950, $6aa8, $56a0, $9ad0, $4ae8, $4ae0, //1910
$a4d8, $a4d0, $d250, $d548, $b550, $56a0, $96d0, $95b0, $49b8, $49b0, //1920
$a4b0, $b258, $6a50, $6d40, $ada8, $2b60, $9570, $4978, $4970, $64b0, //1930
$d4a0, $ea50, $6d48, $5ad0, $2b60, $9370, $92e0, $c968, $c950, $d4a0, //1940
$da50, $b550, $56a0, $aad8, $25d0, $92d0, $c958, $a950, $b4a8, $6ca0, //1950
$b550, $55a8, $4da0, $a5b0, $52b8, $52b0, $a950, $e950, $6aa0, $ad50, //1960
$ab50, $4b60, $a570, $a570, $5260, $e930, $d950, $5aa8, $56a0, $96d0, //1970
$4ae8, $4ad0, $a4d0, $d268, $d250, $d528, $b540, $b6a0, $96d0, $95b0, //1980
$49b0, $a4b8, $a4b0, $b258, $6a50, $6d40, $ada0, $ab60, $9370, $4978, //1990
$4970, $64b0, $6a50, $ea50, $6b28, $5ac0, $ab60, $9368, $92e0, $c960, //2000
$d4a8, $d4a0, $da50, $5aa8, $56a0, $aad8, $25d0, $92d0, $c958, $a950, //2010
$b4a0, $b550, $b550, $55a8, $4ba0, $a5b0, $52b8, $52b0, $a930, $74a8, //2020
$6aa0, $ad50, $4da8, $4b60, $9570, $a4e0, $d260, $e930, $d530, $5aa0, //2030
$6b50, $96d0, $4ae8, $4ad0, $a4d0, $d258, $d250, $d520, $daa0, $b5a0, //2040
$56d0, $4ad8, $49b0, $a4b8, $a4b0, $aa50, $b528, $6d20, $ada0, $55b0); //2050//数组gLanarMonth存放阴历1901年到2050年闰月的月份,如没有则为0,每字节存两年
gLunarMonth:array[0..74] of Byte=(
$00, $50, $04, $00, $20, //1910
$60, $05, $00, $20, $70, //1920
$05, $00, $40, $02, $06, //1930
$00, $50, $03, $07, $00, //1940
$60, $04, $00, $20, $70, //1950
$05, $00, $30, $80, $06, //1960
$00, $40, $03, $07, $00, //1970
$50, $04, $08, $00, $60, //1980
$04, $0a, $00, $60, $05, //1990
$00, $30, $80, $05, $00, //2000
$40, $02, $07, $00, $50, //2010
$04, $09, $00, $60, $04, //2020
$00, $20, $60, $05, $00, //2030
$30, $b0, $06, $00, $50, //2040
$02, $07, $00, $50, $03); //2050
谢谢!