Question

我有一个问题我有一个简单的表，当我选择全部（一列有一些行）时，它看起来像这样。

| a, b, c | - 1st row
| b, d, d | - 2nd row
| d, e, f | - 3rd row

现在尝试用逗号分割这些值，这样每个值都会在单独的行中，如

|a| - 1st row
|b| - 2nd row
|c| - 3rd row
|d| - 4th row
|e| - 5th row
|f| - 6th row

我正在尝试使用类似的东西：

select id,
case when CHARINDEX(', ', [value])>0 
    then SUBSTRING([value] , 1, CHARINDEX(', ',[value])-1) else [value] end firstname, 
CASE WHEN CHARINDEX(', ', [value])>0 
    THEN SUBSTRING([value],CHARINDEX(', ',[value])+1,len([value])) ELSE NULL END as lastname from table

但不是这样。

Answer 1

没有UDF解析/拆分功能

您没有指定表名或列名，因此请将YourTable和YourList替换为您的实际表名和列名。

Select Distinct RetVal
      ,RowNr = Dense_Rank() over (Order by RetVal)
 From  YourTable A
 Cross Apply ( 
                Select RetSeq = Row_Number() over (Order By (Select null))
                      ,RetVal = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)')))
                From  (Select x = Cast('<x>'+ replace((Select A.YourList as [*] For XML Path('')),',','</x><x>')+'</x>' as xml).query('.')) as A 
                Cross Apply x.nodes('x') AS B(i)
       ) B

<强>返回

RetVal  RowNr
a       1
b       2
c       3
d       4
e       5
f       6

使用拆分/解析功能（每个人都应该有一个好的）

Select Distinct RetVal
      ,RowNr = Dense_Rank() over (Order by RetVal)
 From  YourTable A
 Cross Apply (Select * from [dbo].[udf-Str-Parse-8K](A.YourList,',') ) B

UDF - 如果感兴趣

CREATE FUNCTION [dbo].[udf-Str-Parse-8K] (@String varchar(max),@Delimiter varchar(10))
Returns Table 
As
Return (  
    with   cte1(N)   As (Select 1 From (Values(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) N(N)),
           cte2(N)   As (Select Top (IsNull(DataLength(@String),0)) Row_Number() over (Order By (Select NULL)) From (Select N=1 From cte1 a,cte1 b,cte1 c,cte1 d) A ),
           cte3(N)   As (Select 1 Union All Select t.N+DataLength(@Delimiter) From cte2 t Where Substring(@String,t.N,DataLength(@Delimiter)) = @Delimiter),
           cte4(N,L) As (Select S.N,IsNull(NullIf(CharIndex(@Delimiter,@String,s.N),0)-S.N,8000) From cte3 S)

    Select RetSeq = Row_Number() over (Order By A.N)
          ,RetVal = LTrim(RTrim(Substring(@String, A.N, A.L)))
    From   cte4 A
);
--Much faster than str-Parse, but limited to 8K
--Select * from [dbo].[udf-Str-Parse-8K]('Dog,Cat,House,Car',',')
--Select * from [dbo].[udf-Str-Parse-8K]('John||Cappelletti||was||here','||')

Answer 2

一种递归cte解决方案，它可以找到字符串中的所有,，并获得substring在,之间的lead。（假设您使用的是用于with cte as ( select val,charindex(',',','+val+',') as location from t union all select val,charindex(',',','+val+',',location+1) from cte where charindex(',',','+val+',',location+1) > 0 ) ,substrings as (select *, substring(val,location, lead(location,1) over(partition by val order by location)-location-1) as sub from cte) select distinct sub from substrings where sub is not null and sub <> '' order by 1;的sql server版本2012+）

cte

Sample Demo

1）第一个,以递归方式获取字符串中的所有,个位置。 ,附加在字符串的开头和结尾处，以避免错过,之前的第一个子字符串和,之后的最后一个子字符串。

2）对于每个字符串，使用按lead的位置排序的,找到下一个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的位置。

3）最后得到所有那些非空且不是空字符串的子串。

Answer 3

您可以使用交叉应用和XML

来完成此操作

select distinct
    p.a.value('.','varchar(10)') col 
from (
    select 
        cast('<x>' + replace(col,', ','</x><x>') + '</x>' as XML) as x
    from your_table) t
    cross apply x.nodes ('/x') as p(a)
) t

如何用逗号分割结果行

3 个答案: