Question

我有以下代码清单

Code Meaning 
1    Single 
2    Married/Separate 
3    Divorced 
4    Widowed 
8    Not Applicable 
99   Not known

我正在尝试使用CTE将这些展平为一行。我有一个使用RowNumber函数的CTE解决方案。

WITH Flattened (JoinItem, CodeMeaning) AS
(
    SELECT 1 AS JoinItem, CAST('' AS VARCHAR(255))
    UNION ALL
    SELECT f.JoinItem+1, CAST(f.CodeMeaning + ',' + c.CodeMeaning AS VARCHAR(255))
    FROM
    (
        SELECT JoinItem = ROW_NUMBER() OVER (ORDER BY Code),c.Code + ' - ' + c.Meaning AS CodeMeaning
        FROM Codes c
    ) c
    INNER JOIN Flattened f
    ON f.JoinItem=c.JoinItem
)
SELECT TOP 1 JoinItem,  CodeMeaning 
FROM Flattened 
ORDER BY JoinItem DESC

但是，我想知道我是否可以在不使用RowNumber功能但仍然使用CTE的情况下执行此操作。所以我有以下内容 - 我认为更简单 - Sql

WITH Flattened (JoinItem, CodeMeaning) AS
(
    SELECT 1 AS JoinItem, CAST('' AS VARCHAR(255))
    UNION ALL
    SELECT c.JoinItem, CAST(f.CodeMeaning + ',' + c.CodeMeaning AS VARCHAR(255))
    FROM
    (
        SELECT 1 AS JoinItem,c.Code + ' - ' + c.Meaning AS CodeMeaning
        FROM Codes c            
    ) c
    INNER JOIN Flattened f
    ON f.JoinItem=c.JoinItem
)
SELECT JoinItem, odeMeaning 
FROM Flattened

现在它正在最大化递归并产生类似于笛卡尔联合的东西 - 如果不是更糟糕的话！

我希望每次使用固定的＆＃34; JoinItem＆＃34;

尝试让它加入到锚记录中

因此，如果有解决方案，任何指向我错误的地方都会有所帮助。

编辑SqlFiddle

Answer 1

假设这是SQL Server，您是否考虑过这样的事情：

select stuff((select ',' + c.code + '-' + c.Meaning
              from codes c
              order by code
              for xml path ('')
             ), 1, 1, '')

编辑：

要使用CTE执行此操作，请先定义序列号，然后执行展平：

with c as (
      select row_number() over (order by code) as seqnum, c.code + '-' + c.meaning as CodeMeaning
      from codes c
     ),
     flattened as (
      select CodeMeaning as CodeMeaning
      from c
      where rownum = 1
      union all
      select f.CodeMeaning + ',' + c.CodeMeaning
      from c join
           flattened f
           on c.seqnum = f.seqnum + 1
     )
select *
from flattened;

如果列表太长，则可能必须增加默认递归级别。

Answer 2

如果没有ROW_NUMBER（），您的查询应该是

random_state

您也可以在最后一个选择中使用ORDER BY LEN（CodeMeaning）DESC。在这种情况下，lev（递归级别）列不是必需的。如果您对代码值有额外的了解，那么您可以使用更准确的INNER JOIN谓词来优化查询，例如ON f.Code＆lt; c.Code AND f.Code + 10＆gt; c.Code。

简化CTE字符串连接？

2 个答案: