旧的身份证号吗（15位）如何转换为新的（18位）
如题，SQL实现。
[解决办法]
--将15位身份证升级成18位的用户定义函数CREATE FUNCTION ID15TO18 (@id15 char(15)) RETURNS CHAR(18) AS BEGIN DECLARE @ID18 CHAR(18) DECLARE @S1 AS INTEGER DECLARE @S2 AS INTEGER DECLARE @S3 AS INTEGER DECLARE @S4 AS INTEGER DECLARE @S5 AS INTEGER DECLARE @S6 AS INTEGER DECLARE @S7 AS INTEGER DECLARE @S8 AS INTEGER DECLARE @S9 AS INTEGER DECLARE @S10 AS INTEGER DECLARE @S11 AS INTEGER DECLARE @S12 AS INTEGER DECLARE @S13 AS INTEGER DECLARE @S14 AS INTEGER DECLARE @S15 AS INTEGER DECLARE @S16 AS INTEGER DECLARE @S17 AS INTEGER DECLARE @S18 AS INTEGER SET @S1 = SUBSTRING(@ID15,1,1) SET @S2 = SUBSTRING(@ID15,2,1) SET @S3 = SUBSTRING(@ID15,3,1) SET @S4 = SUBSTRING(@ID15,4,1) SET @S5 = SUBSTRING(@ID15,5,1) SET @S6 = SUBSTRING(@ID15,6,1) SET @S7 = 1 SET @S8 = 9 SET @S9 = SUBSTRING(@ID15,7,1) SET @S10 = SUBSTRING(@ID15,8,1) SET @S11 = SUBSTRING(@ID15,9,1) SET @S12 = SUBSTRING(@ID15,10,1) SET @S13 = SUBSTRING(@ID15,11,1) SET @S14 = SUBSTRING(@ID15,12,1) SET @S15 = SUBSTRING(@ID15,13,1) SET @S16 = SUBSTRING(@ID15,14,1) SET @S17 = SUBSTRING(@ID15,15,1) SET @S18 = ( (@S1*7) + (@S2*9) + (@S3*10) + (@S4*5) + (@S5*8) + (@S6*4) + (@S7*2) + (@S8*1) + (@S9*6) + (@S10*3) + (@S11*7) + (@S12*9) + (@S13*10) + (@S14*5) + (@S15*8) + (@S16*4) + (@S17*2) ) % 11 SET @ID18 = SUBSTRING(@ID15,1,6) + '19' + SUBSTRING(@ID15,7,9) + CASE WHEN @S18 = 0 THEN '1' WHEN @S18 = 1 THEN '0' WHEN @S18 = 2 THEN 'X' WHEN @S18 = 3 THEN '9' WHEN @S18 = 4 THEN '8' WHEN @S18 = 5 THEN '7' WHEN @S18 = 6 THEN '6' WHEN @S18 = 7 THEN '5' WHEN @S18 = 8 THEN '4' WHEN @S18 = 9 THEN '3' WHEN @S18 = 10 THEN '2' END RETURN @ID18END GO--调用函数update 表set 身份证号 = dbo.ID15TO18(身份证号)where LEN(身份证号) = 15
[解决办法]
-- 15位身份证号升级为18位，适用于18xx年出生的公民UPDATE 员工表 SET 身份证号= SUBSTRING(身份证号,1,6)+'18'+SUBSTRING(身份证号,7,9)+ SUBSTRING('10X98765432', ( CAST(SUBSTRING(身份证号, 1,1) AS INT)*7 +CAST(SUBSTRING(身份证号, 2,1) AS INT)*9 +CAST(SUBSTRING(身份证号, 3,1) AS INT)*10 +CAST(SUBSTRING(身份证号, 4,1) AS INT)*5 +CAST(SUBSTRING(身份证号, 5,1) AS INT)*8 +CAST(SUBSTRING(身份证号, 6,1) AS INT)*4 +1*2 +8*1 +CAST(SUBSTRING(身份证号, 7,1) AS INT)*6 +CAST(SUBSTRING(身份证号, 8,1) AS INT)*3 +CAST(SUBSTRING(身份证号, 9,1) AS INT)*7 +CAST(SUBSTRING(身份证号,10,1) AS INT)*9 +CAST(SUBSTRING(身份证号,11,1) AS INT)*10 +CAST(SUBSTRING(身份证号,12,1) AS INT)*5 +CAST(SUBSTRING(身份证号,13,1) AS INT)*8 +CAST(SUBSTRING(身份证号,14,1) AS INT)*4 +CAST(SUBSTRING(身份证号,15,1) AS INT)*2 ) % 11 + 1, 1)WHERE LEN(身份证号)=15 AND SUBSTRING(身份证号,13,3) IN ('999','998','997','996')-- 15位身份证号升级为18位，适用于19xx年出生的公民UPDATE 员工表 SET 身份证号= SUBSTRING(身份证号,1,6)+'19'+SUBSTRING(身份证号,7,9)+ SUBSTRING('10X98765432', ( CAST(SUBSTRING(身份证号, 1,1) AS INT)*7 +CAST(SUBSTRING(身份证号, 2,1) AS INT)*9 +CAST(SUBSTRING(身份证号, 3,1) AS INT)*10 +CAST(SUBSTRING(身份证号, 4,1) AS INT)*5 +CAST(SUBSTRING(身份证号, 5,1) AS INT)*8 +CAST(SUBSTRING(身份证号, 6,1) AS INT)*4 +1*2 +9*1 +CAST(SUBSTRING(身份证号, 7,1) AS INT)*6 +CAST(SUBSTRING(身份证号, 8,1) AS INT)*3 +CAST(SUBSTRING(身份证号, 9,1) AS INT)*7 +CAST(SUBSTRING(身份证号,10,1) AS INT)*9 +CAST(SUBSTRING(身份证号,11,1) AS INT)*10 +CAST(SUBSTRING(身份证号,12,1) AS INT)*5 +CAST(SUBSTRING(身份证号,13,1) AS INT)*8 +CAST(SUBSTRING(身份证号,14,1) AS INT)*4 +CAST(SUBSTRING(身份证号,15,1) AS INT)*2 ) % 11 + 1, 1)WHERE LEN(身份证号)=15 AND SUBSTRING(身份证号,13,3) NOT IN ('999','998','997','996')
------解决方案--------------------
create function f_CID15to18 (@sfz char(18)) returns char(18)asbegin declare @osfz varchar(18) declare @i int,@ai int,@wi int,@sum int,@mod int,@result int set @osfz = @sfz set @sum = 0 IF len(@osfz) = 15 begin set @osfz = substring(@osfz,1,6) + '19' + substring(@osfz,7,9) set @i = 2 while @i <= 18 begin set @ai = cast(substring(@osfz,19 - @i,1) as int) set @wi = POWER (2, (@i - 1))% 11 set @sum = @sum + @ai * @wi set @i = @I + 1 end set @mod = @sum % 11 set @result = 12 - @mod IF @result >= 10 IF @result = 10 RETURN @osfz + 'X' ELSE begin set @result = @result - 11 RETURN @osfz + ltrim(@result) end ELSE RETURN @osfz + ltrim(@result) end ELSE RETURN @sfz return @sfzend goselect dbo.f_CID15to18('32108519760502***9')/*------------------ 32108519760502***9（所影响的行数为 1 行）*/select dbo.f_CID15to18('321085760502***')/*------------------ 32108519760502***9（所影响的行数为 1 行）*/drop function f_CID15to18
[解决办法]
CREATE FUNCTION [Helper].[IDCard] ( @Card varchar(18))RETURNS @TCard TABLE ( Input varchar(18) ,IDCard varchar(18) ,Valid bit)ASBEGIN DECLARE @Input as varchar(18) ,@IDCard as varchar(18) ,@Valid as bit DECLARE @Length as smallint ,@TmpCard as varchar(18) ,@IsOld as bit SET @Valid = 0 SET @IDCard = '' SET @Input = '' IF @Card IS NULL GOTO Finish SET @Input = LTRIM(RTRIM(@Card)) /*去空格*/ SET @Length = LEN(@Input) IF NOT @Length IN (15, 18) GOTO Finish /*非15、18位*/ IF @Length = 15 BEGIN IF ISNUMERIC(@Input) = 0 GOTO Finish /*非数字*/ SET @TmpCard = LEFT(@Input, 6) + '19' + RIGHT(@input, 9) /*补充为17位*/ SET @IsOld = 1 END ELSE BEGIN IF ISNUMERIC(LEFT(@Input, 17)) = 0 GOTO Finish /*非数字*/ SET @TmpCard = LEFT(@Input, 17) /*取前17位*/ SET @IsOld = 0 END DECLARE @Birthday varchar(8) SET @Birthday = SUBSTRING(@TmpCard, 7, 8) IF ISDATE(@birthday) = 0 GOTO Finish /*非日期*/ --前17位数与相应加权因子积的和 DECLARE @Sum as smallint ,@WI as tinyint ,@Index as tinyint ,@Num as tinyint SET @Sum = 0 SET @Index = 1 WHILE @Index < 18 BEGIN SET @Num = CAST(SUBSTRING(@TmpCard, @Index, 1) AS tinyint) SELECT @WI = CASE @Index WHEN 1 THEN 7 WHEN 2 THEN 9 WHEN 3 THEN 10 WHEN 4 THEN 5 WHEN 5 THEN 8 WHEN 6 THEN 4 WHEN 7 THEN 2 WHEN 8 THEN 1 WHEN 9 THEN 6 WHEN 10 THEN 3 WHEN 11 THEN 7 WHEN 12 THEN 9 WHEN 13 THEN 10 WHEN 14 THEN 5 WHEN 15 THEN 8 WHEN 16 THEN 4 WHEN 17 THEN 2 END SET @Sum = @Sum + @Num * @WI SET @Index = @Index + 1 END --模11 DECLARE @Mod as tinyint SET @Mod = @Sum % 11 --校验码DECLARE @Parity as varchar(1)SELECT @Parity =CASE @ModWHEN 0 THEN '1'WHEN 1 THEN '0'WHEN 2 THEN 'X'WHEN 3 THEN '9'WHEN 4 THEN '8'WHEN 5 THEN '7'WHEN 6 THEN '6'WHEN 7 THEN '5'WHEN 8 THEN '4'WHEN 9 THEN '3'WHEN 10 THEN '2'END--完整的18位身份证号码SET @TmpCard = @TmpCard + @ParityIF @IsOld = 1SET @Valid = 1ELSE IF @Parity = RIGHT(@Input, 1) /*校验*/SET @Valid = 1--无论对错，都给出有效身份证号码SET @IDCard = @tmpCardFinish: INSERT INTO @TCard VALUES(@Input, @IDCard, @Valid) RETURN END
[解决办法]
★☆★☆ [身份证号] ★☆★☆
18位身份证标准在国家质量技术监督局于1999年7月1日实施的GB11643-1999《公民身份号码》中做了明确的规定。
GB11643-1999《公民身份号码》为GB11643-1989《社会保障号码》的修订版，其中指出将原标准名称“社会保障号码”更名为“公民身份号码”，另外GB11643-1999《公民身份号码》从实施之日起代替GB11643-1989。
GB11643-1999《公民身份号码》主要内容如下：
一、范围
该标准规定了公民身份号码的编码对象、号码的结构和表现形式，使每个编码对象获得一个唯一的、不变的法定号码。
二、编码对象
公民身份号码的编码对象是具有中华人民共和国国籍的公民。
三、号码的结构和表示形式
1、号码的结构
公民身份号码是特征组合码，由十七位数字本体码和一位校验码组成。排列顺序从左至右依次为：六位数字地址码，八位数字出生日期码，三位数字顺序码和一位数字校验码。
2、地址码
表示编码对象常住户口所在县(市、旗、区)的行政区划代码，按GB/T2260的规定执行。
3、出生日期码
表示编码对象出生的年、月、日，按GB/T7408的规定执行，年、月、日代码之间不用分隔符。
4、顺序码
表示在同一地址码所标识的区域范围内，对同年、同月、同日出生的人编定的顺序号，顺序码的奇数分配给男性，偶数分配给女性。
5、校验码
(1)十七位数字本体码加权求和公式
S = Ai * Wi, i = 2, ... , 18
Y = mod(S, 11)
i: 表示号码字符从右至左包括校验码字符在内的位置序号
Ai:表示第i位置上的身份证号码字符值
Wi:表示第i位置上的加权因子
i: 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
Wi: 7 9 10 5 8 4 2 1 6 3 7 9 10 5 8 4 2 1
(2)校验码字符值的计算
Y: 0 1 2 3 4 5 6 7 8 9 10
校验码: 1 0 X 9 8 7 6 5 4 3 2
四、举例如下：
北京市朝阳区: 11010519491231002X
广东省汕头市: 440524188001010014
五、身份证号前6位代表的行政区域: