sql - 如何将字符串值拆分为一列并返回结果表

Question

假设我们有下表：

id name   member
1  jacky  a;b;c
2  jason  e
3  kate   i;j;k
4  alex   null

现在我想使用 sql 或 t-sql 返回下表：

1 jacky a
1 jacky b
1 jacky c
2 jason e
3 kate  i
......

怎么做？我正在使用 MSSQL、MYSQL 和 Oracle 数据库。

Answer 1

这是可以设计的最短且可读的字符串到行拆分器，并且也可以更快。

选择纯 CTE 而不是函数的用例，例如当您不允许在数据库上创建函数时:-)

通过函数创建行生成器（也可以使用循环或通过 CTE 实现）仍需要使用横向连接（DB2 和 Sybase 具有此功能，使用 LATERAL 关键字；在 SQL Server 中，这类似于 CROSS APPLY 和 OUTER APPLY ) 最终将函数生成的拆分行连接到主表。

纯 CTE 方法可能比函数方法更快。速度指标在于分析，如果这确实更快，只需检查它的执行计划与其他解决方案相比：

with Pieces(theId, pn, start, stop) AS
(
      SELECT id, 1, 1, charindex(';', member)
      from tbl

      UNION ALL

      SELECT id, pn + 1, stop + 1, charindex(';', member, stop + 1)
      from tbl 
      join pieces on pieces.theId = tbl.id 
      WHERE stop > 0
)
select 

      t.id, t.name, 

      word = 
         substring(t.member, p.start,             
           case WHEN stop > 0 THEN p.stop - p.start 
           ELSE 512 
           END) 

from tbl t
join pieces p on p.theId = t.id
order by t.id, p.pn 

输出：

ID  NAME    WORD
1   jacky   a
1   jacky   b
1   jacky   c
2   jason   e
3   kate    i
3   kate    j
3   kate    k
4   alex    (null)

此处来源的基本逻辑：T-SQL：与字符串连接相反 - 如何将字符串拆分为多个记录

现场测试：http ://www.sqlfiddle.com/#!3/2355d/1

Answer 2

嗯...首先让我向您介绍亚当·马哈尼奇（Adam Machanic），他教我有关数字表的知识。他还使用这个数字表编写了一个非常快速的拆分函数。

http://dataeducation.com/counting-occurrences-of-a-substring-within-a-string/

实现返回表的拆分函数后，您可以加入它并获得所需的结果。

Answer 3

IF OBJECT_ID('dbo.Users') IS NOT NULL 
    DROP TABLE dbo.Users;

CREATE TABLE dbo.Users
(
  id INT IDENTITY NOT NULL PRIMARY KEY,
  name VARCHAR(50) NOT NULL,
  member VARCHAR(1000)
)
GO

INSERT INTO dbo.Users(name, member) VALUES
  ('jacky', 'a;b;c'),
  ('jason', 'e'),
  ('kate', 'i;j;k'),
  ('alex', NULL);
GO

DECLARE @spliter CHAR(1) = ';';
WITH Base AS
(
    SELECT  1 AS n
    UNION ALL
    SELECT  n + 1
    FROM    Base
    WHERE   n < CEILING(SQRT(1000)) --generate numbers from 1 to 1000, you may change it to a larger value depending on the member column's length.
)
, Nums AS --Numbers Common Table Expression, if your database version doesn't support it, just create a physical table.
(
    SELECT  ROW_NUMBER() OVER(ORDER BY (SELECT  0)) AS n
    FROM    Base AS B1 CROSS JOIN Base AS B2
)
SELECT  id,
        SUBSTRING(member, n, CHARINDEX(@spliter, member + @spliter, n) - n) AS element
FROM    dbo.Users
    JOIN Nums
    ON n <= DATALENGTH(member) + 1
    AND SUBSTRING(@spliter + member, n, 1) = @spliter
ORDER BY id
OPTION (MAXRECURSION 0); --Nums CTE is generated recursively, we don't want to limit recursion count.