Question

我有一个表结构如下：

表：

Parent    branchname  branchlevel  level**

tree        a       1.1     1    
tree        b       1.2     1    
tree        c       1.3     1    
tree        d       1.4     1    
tree        e       1.10    1    
b           f       1.1     2    
b           g       1.2     2    
b           h       1.3     2

我需要一个如下所示的结果集：

tree    1    
a       1.1    
e       1.10    
b       1.2    
f       1.2.1    
g       1.2.2    
h       1.2.3    
c       1.3    
d       1.4

到目前为止

我的SQL：

select * 
from ((select distinct parent,'1' as branchlevel 
       from tree 
       where parent not in(select branchname from tree)) 
union all (select branchname,branchlevel 
           from tree 
           where (cast(level as int))<(select max(CAST(level as int)) from tree)) 
union all (select b.branchname,(SUBSTRING(b.branchlevel,1,2)+b.level+'.'+SUBSTRING(b.branchlevel,3‌,2))branchlevel 
           from tree a 
           inner join tree b on a.branchname=b.parent 
           where (cast(a.level as int))<(select max(CAST(level as int)) from tree)))a 
order by branchlevel

Answer 1

您可以使用递归CTE来获得所需的输出：

;WITH TreeRoot AS
( 
   SELECT rootName = (SELECT TOP 1 Parent 
                      FROM #Tree
                      WHERE Parent NOT IN (SELECT branchname FROM #Tree))
),                     
TreeLevels AS (
   -- Recursive CTE Anchor member is simply: (tree, NULL, 0, 1)
   SELECT (SELECT rootName FROM TreeRoot) AS NodeName, 
          CAST(NULL AS VARCHAR(10)) AS ParentName, 
          0 AS level,
          CAST('1' AS VARCHAR(10)) AS branchlevel

   UNION ALL

   SELECT branchname AS NodeName, 
          t.Parent AS ParentName, 
          tl.level + 1 AS level, 
          CAST(CONCAT(tl.branchlevel, 
               '.', 
               SUBSTRING(t.branchlevel, CHARINDEX('.', t.branchlevel)+1, LEN(t.branchlevel) - CHARINDEX('.', t.branchlevel))) AS VARCHAR(10)) AS branchlevel
   FROM #Tree AS t
   INNER JOIN TreeLevels AS tl ON t.Parent = tl.NodeName
)
SELECT NodeName, branchlevel
FROM TreeLevels
ORDER BY branchlevel

SUBSTRING函数用于从branchlevel字段中获取最后一个数字，即得到＆＃39; 2＆＃39;出于＆＃39; 1.2＆＃39;。 CONCAT用于递归地将此数字添加到任何先前获得的分支级别数。

即使分支级别超过2，此查询也会起作用。SQL Fiddle Demo

sql server如何获取此结果集

1 个答案: