数据库中有张表(设计的太屁,不过不是我设计的哈哈) 没办法向上面反应过几次 回答都是不能改表 没办法 只能硬着头皮往上顶呢 字段 CustomerUID,BabyName1,BabyName2,BabyName3,Year1,Month1,Day1,Year2,Month2,Day2,Year3,Month3,Day3,
MotherName,FatherName,OtherName,MobilePhone1,MobilePhone2,MobilePhone3,FixPhone1,FixPhone2,FixPhone3
瞧吧 屁吧 想要的结果是 BabyName, BabyBrithday, ParentName, MobilePhone, FixPhone 第一种方案:全部取出来 我到数据访问层去整理数据
这种效率太慢,写的代码也多,没用 第二种方案:写个存储过程,在sql中整理
写sql的时候:举个例子 获取宝宝1的生日: select year1+','+month1+','+day1 from table 出现的情况是 如果某一个值为空 取得的结果就为null 也就是说拼接不成功,不知道这是为什么?
2.想使用union all 那select就超级多 考虑中
3.使用isnull select isnull(year1,0)+','+isnull(month1,0)+','+isnull(day1,0) from table 出现的情况是 year为空时结果就是 0-08-21 这个是不想要的 (也就是说头和尾只要没值就不显示 0-08-09->08-09 2011-0-0->2011) 大家有什么好的建议;
MotherName,FatherName,OtherName,MobilePhone1,MobilePhone2,MobilePhone3,FixPhone1,FixPhone2,FixPhone3
瞧吧 屁吧 想要的结果是 BabyName, BabyBrithday, ParentName, MobilePhone, FixPhone 第一种方案:全部取出来 我到数据访问层去整理数据
这种效率太慢,写的代码也多,没用 第二种方案:写个存储过程,在sql中整理
写sql的时候:举个例子 获取宝宝1的生日: select year1+','+month1+','+day1 from table 出现的情况是 如果某一个值为空 取得的结果就为null 也就是说拼接不成功,不知道这是为什么?
2.想使用union all 那select就超级多 考虑中
3.使用isnull select isnull(year1,0)+','+isnull(month1,0)+','+isnull(day1,0) from table 出现的情况是 year为空时结果就是 0-08-21 这个是不想要的 (也就是说头和尾只要没值就不显示 0-08-09->08-09 2011-0-0->2011) 大家有什么好的建议;
,Year1+'-'+Month1+'-'+Day1 as BabyBrithday
,MotherName+FatherName as ParentName
,MobilePhone1 as MobilePhone
,FixPhone1 as FixPhone
from tab
union all
select BabyName2 as BabyName
,Year2+'-'+Month2+'-'+Day2 as BabyBrithday
,MotherName+FatherName as ParentName
,MobilePhone2 as MobilePhone
,FixPhone2 as FixPhone
from tab
where BabyName2 is not null
union all
select BabyName3 as BabyName
,Year3+'-'+Month3+'-'+Day3 as BabyBrithday
,MotherName+FatherName as ParentName
,MobilePhone3 as MobilePhone
,FixPhone3 as FixPhone
from tab
where BabyName3 is not null
1 'A' 'B' 'C' 0 0 0 2011 0 1 2011 0 0 'D' 'E' 'F' '13455555' '' NULL '027-3333' NULL NULL
2 '' NULL 'C' 2011 0 0 2011 0 1 2011 0 0 'D' 'E' 'F' NULL 'X' NULL '' NULL NULL
3 NULL NULL '' 2011 1 0 2011 0 1 2011 0 0 'D' 'E' 'F' '' '' NULL '' NULL ''
4 'A' '' 'C' 0 0 0 2011 0 1 2011 0 0 'D' 'E' 'F' '13455555' '' NULL '027-3333' NULL '027-3333'
--想要的结果
CustomerID BabyName BabayBrithday ParentName MobilePhone FixPhone
1 'A,B,C' '2011-1,2011' 'D,E,F' '13455555' '027-3333'
2 'C' '2011,2011-0-1' 'D,E,F' 'X' NULL
3 NULL '2011-1,2011-0-1,2011' 'D,E,F' NULL NULL
4 'A,C' '2011-0-1,2011' 'D,E,F' '13455555' '027-3333,027-3333'
select BabyName3 as BabyName
,Year3+'-'+Month3+'-'+Day3 as BabyBrithday
你写的这种不对 Year3,Month3,Day3 这三个字段是int型的可以为null 只要year3为null或'' 得到的结果就是null 我测试过的
,isnull(nullif(BabyName1,''),'') + isnull(nullif(',' + BabyName2,','),'') + isnull(nullif(',' + BabyName3,','),'') as BabyName
,isnull(nullif(Year1,'0'),'') + isnull(nullif('-' + Month1,'-0'),'') + isnull(nullif('-' + Day1,'-0'),'')
+ isnull(nullif(',' + isnull(nullif(Year2,'0'),'') + isnull(nullif('-' + Month2,'-0'),'') + isnull(nullif('-' + Day2,'-0'),''),','),'')
+ isnull(nullif(',' + isnull(nullif(Year3,'0'),'') + isnull(nullif('-' + Month3,'-0'),'') + isnull(nullif('-' + Day3,'-0'),''),','),'')
as BabyBrithday
,isnull(nullif(MotherName,''),'') + isnull(nullif(',' + FatherName,','),'') + isnull(nullif(',' + OtherName,','),'') as ParentName
,isnull(nullif(MobilePhone1,''),'') + isnull(nullif(',' + MobilePhone2,','),'') + isnull(nullif(',' + MobilePhone3,','),'') as MobilePhone
,isnull(nullif(FixPhone1,''),'') + isnull(nullif(',' + FixPhone2,','),'') + isnull(nullif(',' + FixPhone3,','),'') as FixPhone
from tab
哦弥陀佛
--*******************************************************************************************
表结构,数据如下:
id value
----- ------
1 aa
1 bb
2 aaa
2 bbb
2 ccc 需要得到结果:
id values
------ -----------
1 aa,bb
2 aaa,bbb,ccc
即:group by id, 求 value 的和(字符串相加) 1. 旧的解决方法(在sql server 2000中只能用函数解决。)
--=============================================================================
create table tb(id int, value varchar(10))
insert into tb values(1, 'aa')
insert into tb values(1, 'bb')
insert into tb values(2, 'aaa')
insert into tb values(2, 'bbb')
insert into tb values(2, 'ccc')
go
--1. 创建处理函数
CREATE FUNCTION dbo.f_strUnite(@id int)
RETURNS varchar(8000)
AS
BEGIN
DECLARE @str varchar(8000)
SET @str = ''
SELECT @str = @str + ',' + value FROM tb WHERE id=@id
RETURN STUFF(@str, 1, 1, '')
END
GO
-- 调用函数
SELECt id, value = dbo.f_strUnite(id) FROM tb GROUP BY id
drop table tb
drop function dbo.f_strUnite
go
/*
id value
----------- -----------
1 aa,bb
2 aaa,bbb,ccc
(所影响的行数为 2 行)
*/
--===================================================================================
2. 新的解决方法(在sql server 2005中用OUTER APPLY等解决。)
create table tb(id int, value varchar(10))
insert into tb values(1, 'aa')
insert into tb values(1, 'bb')
insert into tb values(2, 'aaa')
insert into tb values(2, 'bbb')
insert into tb values(2, 'ccc')
go
-- 查询处理
SELECT * FROM(SELECT DISTINCT id FROM tb)A OUTER APPLY(
SELECT [values]= STUFF(REPLACE(REPLACE(
(
SELECT value FROM tb N
WHERE id = A.id
FOR XML AUTO
), ' <N value="', ','), '"/>', ''), 1, 1, '')
)N
drop table tb /*
id values
----------- -----------
1 aa,bb
2 aaa,bbb,ccc (2 行受影响)
*/ --SQL2005中的方法2
create table tb(id int, value varchar(10))
insert into tb values(1, 'aa')
insert into tb values(1, 'bb')
insert into tb values(2, 'aaa')
insert into tb values(2, 'bbb')
insert into tb values(2, 'ccc')
go select id, [values]=stuff((select ','+[value] from tb t where id=tb.id for xml path('')), 1, 1, '')
from tb
group by id /*
id values
----------- --------------------
1 aa,bb
2 aaa,bbb,ccc (2 row(s) affected) */ drop table tb /*
标题:分拆列值1
作者:爱新觉罗.毓华(十八年风雨,守得冰山雪莲花开)
时间:2008-11-20
地点:广东深圳
描述有表tb, 如下:
id value
----------- -----------
1 aa,bb
2 aaa,bbb,ccc
欲按id,分拆value列, 分拆后结果如下:
id value
----------- --------
1 aa
1 bb
2 aaa
2 bbb
2 ccc
*/--1. 旧的解决方法(sql server 2000)
SELECT TOP 8000 id = IDENTITY(int, 1, 1) INTO # FROM syscolumns a, syscolumns b SELECT A.id, value = SUBSTRING(A.[value], B.id, CHARINDEX(',', A.[value] + ',', B.id) - B.id)
FROM tb A, # B
WHERE SUBSTRING(',' + A.[value], B.id, 1) = ','DROP TABLE #--2. 新的解决方法(sql server 2005)
create table tb(id int,value varchar(30))
insert into tb values(1,'aa,bb')
insert into tb values(2,'aaa,bbb,ccc')
go
SELECT A.id, B.value
FROM(
SELECT id, [value] = CONVERT(xml,'<root><v>' + REPLACE([value], ',', '</v><v>') + '</v></root>') FROM tb
)A
OUTER APPLY(
SELECT value = N.v.value('.', 'varchar(100)') FROM A.[value].nodes('/root/v') N(v)
)BDROP TABLE tb/*
id value
----------- ------------------------------
1 aa
1 bb
2 aaa
2 bbb
2 ccc(5 行受影响)
*/
select CustomerID
,isnull(nullif(BabyName1,''),'') + isnull(nullif(',' + BabyName2,','),'') + isnull(nullif(',' + BabyName3,','),'') as BabyName
,isnull(nullif(rtrim(Year1),'0'),'') + isnull(nullif('-' + rtrim(Month1),'-0'),'') + isnull(nullif('-' + rtrim(Day1),'-0'),'')
+ isnull(nullif(',' + isnull(nullif(rtrim(Year2),'0'),'') + isnull(nullif('-' + rtrim(Month2),'-0'),'') + isnull(nullif('-' + rtrim(Day2),'-0'),''),','),'')
+ isnull(nullif(',' + isnull(nullif(rtrim(Year3),'0'),'') + isnull(nullif('-' + rtrim(Month3),'-0'),'') + isnull(nullif('-' + rtrim(Day3),'-0'),''),','),'')
as BabyBrithday
,isnull(nullif(MotherName,''),'') + isnull(nullif(',' + FatherName,','),'') + isnull(nullif(',' + OtherName,','),'') as ParentName
,isnull(nullif(MobilePhone1,''),'') + isnull(nullif(',' + MobilePhone2,','),'') + isnull(nullif(',' + MobilePhone3,','),'') as MobilePhone
,isnull(nullif(FixPhone1,''),'') + isnull(nullif(',' + FixPhone2,','),'') + isnull(nullif(',' + FixPhone3,','),'') as FixPhone
from tab
错了也不再改了,有问题楼住自己改
,Year3+'-'+Month3+'-'+Day3 as BabyBrithday from table 测试:
Year3,Month3,Day3 (int) -> 结果 BabyBrithday
2011 1 12 2024
这是么原因呢 难道int与字符相加的时候 int不能自动转换成string ?
declare @a int;
declare @b varchar(10);
declare @c varchar(10);
set @a = 10;
set @b = '20'
set @c = 'HH'
select @a+@b
select @a + @c
1 'A,B,C' '2011-1,2011' 'D,E,F' '13455555' '027-3333'
2 'C' '2011,2011-0-1' 'D,E,F' 'X' NULL
A,B,C 对应的生日是 2011-1,2011,但从上面的数据来看这两个"生日"是B,C的,可是,光看这数据怎么才能知道这种结果?
下面一行更离谱,只有一个C,却有两个生日,哪一个是C的?
如果我来处理,我就把
BabyName, BabyBrithday, ParentName, MobilePhone, FixPhone
处理成一个宝宝一行,这样哪个宝宝没有名,哪个宝宝生日是什么,一清二楚,只是在后面几列有冗余,但不会出逻辑错误啊!