Question

我需要能够通过定界符*将一个字符串拆分为单独的列，而不必包含* 表x中的y列如下所示：

column y
*1HS*AB*GXX*123*02*PA45*2013-08-10*
*1R1*B*GX*123*02*PA45*2013-08-10*
*1HS*B*GX*13*01*PA45*2013-08-01*
*1P*C*GXX*123*02*PA45*2013-08-10*

STRING_SPLIT不可用

结果应该是这样：

Column1 Column2 Column3 Column4 Column5 Column6 Column7
1HS     AB      GXX     123     2       PA45    10-08-2013
1R1     B       GX      123     2       PA45    10-08-2013
1HS     B       GX      13      1       PA45    01-08-2013
1P      C       GXX     123     2       PA45    10-08-2013

Answer 1

您将使用以下查询。.

select RTRIM (REGEXP_SUBSTR (column y, '[^,]*,', 1, 1), ',')    AS column 1
    ,       RTRIM (REGEXP_SUBSTR (column y, '[^,]*,', 1, 2), ',')    AS column 2
    ,       RTRIM (REGEXP_SUBSTR (column y, '[^,]*,', 1, 3), ',')    AS column 3
    ,       LTRIM (REGEXP_SUBSTR (column y, ',[^,]*', 1, 3), ',')    AS column 4
    from YOUR_TABLE

Answer 2

不幸的是，string_split()不保证保留值的顺序。而且，SQL Server不提供其他有用的字符串函数。

因此，我建议为此使用递归CTE：

with t as (
      select *
      from (values ('*1HS*AB*GXX*123*02*PA45*2013-08-10*'), ('1HSB*GX*13*01*PA45*2013-08-01*')) v(str)
     ),
     cte as (
      select convert(varchar(max), null) as val, 0 as lev, convert(varchar(max), str) as rest,
             row_number() over (order by (select null)) as id
      from t
      union all
      select left(rest, charindex('*', rest) - 1), lev + 1, stuff(rest, 1, charindex('*', rest) + 1, ''), id
      from cte
      where rest <> '' and lev < 10
     )
select max(case when lev = 1 then val end) as col1,
       max(case when lev = 2 then val end) as col2,
       max(case when lev = 3 then val end) as col3,
       max(case when lev = 4 then val end) as col4,
       max(case when lev = 5 then val end) as col5,
       max(case when lev = 6 then val end) as col6,
       max(case when lev = 7 then val end) as col7
from cte
where lev > 0
group by cte.id;

Here是db <>小提琴。

Answer 3

假设您可以向数据库中添加表值函数，那么Jeff Moden的字符串拆分函数是我遇到的最佳方法。它将使您也能保持秩序。

Find details here

带分隔符的SQL子查询

3 个答案: