我有表格测试
col_1
----------
aa,ab,ac
ba,bb,bc
ca,cb,cc
我想要输出为
col_1 col_2 col_3
-------------------------
aa ab ac
ba bb bc
ca cb cc
问问题
6377 次
5 回答
0
select substring(col_1,1,CHARINDEX(',',col_1,1)-1) as col_1,
substring(col_1,(CHARINDEX(',',col_1,1)+1),CHARINDEX(',',col_1,CHARINDEX(',',col_1,1)+1)-(CHARINDEX(',',col_1,1)+1)) as col_2,
substring(col_1,CHARINDEX(',',col_1,CHARINDEX(',',col_1,1)+1)+1,LEN(col_1)) as col_3
from test
于 2012-07-10T07:15:57.840 回答
0
嗨，如果没有固定的列，那么我们可以使用下面的查询。
DECLARE @col TABLE(col_1 varchar(255))
INSERT INTO @col
values('aa,ab,ac'),('ba,bb,bc'),('ca,cb,cc')
select * from @col
select LEFT(col_1,CHARINDEX(',',col_1,1)-1) as col_1,
LEFT(RIGHT(col_1,len(col_1)-CHARINDEX(',',col_1,1)),CHARINDEX(',',RIGHT(col_1,len(col_1)-CHARINDEX(',',col_1,1)),1)-1) as col_2,
RIGHT(col_1,2) as col_3
from @col
于 2012-07-10T11:30:21.897 回答
0
即使有超过 2 个用逗号分隔的字符，这也是解决方案。
DECLARE @col TABLE(col_1 varchar(255))
INSERT INTO @col
values('aa,ab,ac'),('ba,bb,bc'),('ca,cb,cc')
select * from @col
select LEFT(col_1,CHARINDEX(',',col_1,1)-1) as col_1,
LEFT(RIGHT(col_1,len(col_1)-CHARINDEX(',',col_1,1)),CHARINDEX(',',RIGHT(col_1,len(col_1)-CHARINDEX(',',col_1,1)),1)-1) as col_2,
RIGHT(RIGHT(col_1,len(col_1)-CHARINDEX(',',col_1,1)),LEN(RIGHT(col_1,len(col_1)-CHARINDEX(',',col_1,1)))-CHARINDEX(',',RIGHT(col_1,len(col_1)-CHARINDEX(',',col_1,1)),1)) as col_3
from @col
于 2012-07-10T11:38:49.253 回答
0
您可以使用函数来拆分字符串，代码如下所示
从测试中选择 dbo.Split(col_1,',')
于 2012-07-10T07:40:23.143 回答
0
declare @T table
(
col_1 varchar(100)
)
insert into @T values
('aa,ab,ac,4,5,6,7,8,9,10'),
('ba,bb,bc,4,5,6,7,8,9,10'),
('ca,cb,cc,4,5,6,7,8,9,10')
select left(T.col_1, C1.Pos-1) as col_1,
substring(T.col_1, C1.Pos+1, C2.Pos-C1.Pos-1) as col_2,
substring(T.col_1, C2.Pos+1, C3.Pos-C2.Pos-1) as col_3,
substring(T.col_1, C3.Pos+1, C4.Pos-C3.Pos-1) as col_4,
substring(T.col_1, C4.Pos+1, C5.Pos-C4.Pos-1) as col_5,
substring(T.col_1, C5.Pos+1, C6.Pos-C5.Pos-1) as col_6,
substring(T.col_1, C6.Pos+1, C7.Pos-C6.Pos-1) as col_7,
substring(T.col_1, C7.Pos+1, C8.Pos-C7.Pos-1) as col_8,
substring(T.col_1, C8.Pos+1, C9.Pos-C8.Pos-1) as col_9,
stuff(T.col_1, 1, C9.Pos, '') as col_10
from @T as T
cross apply (select charindex(',', col_1)) as C1(Pos)
cross apply (select charindex(',', col_1, C1.Pos+1)) as C2(Pos)
cross apply (select charindex(',', col_1, C2.Pos+1)) as C3(Pos)
cross apply (select charindex(',', col_1, C3.Pos+1)) as C4(Pos)
cross apply (select charindex(',', col_1, C4.Pos+1)) as C5(Pos)
cross apply (select charindex(',', col_1, C5.Pos+1)) as C6(Pos)
cross apply (select charindex(',', col_1, C6.Pos+1)) as C7(Pos)
cross apply (select charindex(',', col_1, C7.Pos+1)) as C8(Pos)
cross apply (select charindex(',', col_1, C8.Pos+1)) as C9(Pos)
于 2012-07-10T08:11:00.980 回答