创建具有相同的标签名称

Question

表名XML：样品

列名：ID，名称

每一行创建内部独立标签。

显示这样

<Details> 
    <id>1</id> 
    <name>na</name> 
    <Details> 
     <id>2</id> 
     <name>aa</name> 
    </Details> 
    </Details> 

XML值我试过这样，但它不工作

select 
    id  'Details\id' 
    ,name 'Details\name' 
from sample 

如何获取XML输出？

Answer 1

它是硬编码的，但应该工作：

DECLARE @x xml 

SELECT @x = (
    SELECT x+'' 
    FROM (
     SELECT '%details?%id?'+CAST(id as nvarchar(max))+'%/id?%name?'+name+'%/name?' x 
     FROM [sample] s 
     UNION ALL 
     SELECT '%/details?' 
     FROM [sample] s 
     ) as t 
    FOR XML PATH('') 
) 
SELECT CAST(REPLACE(REPLACE((CAST(@x as nvarchar(max))),'%','<'),'?','>') as xml) 

In [sample]表我：

(1,'na'), 
(2,'aa'), 
(3,'sd') 

输出：

<details> 
    <id>1</id> 
    <name>na</name> 
    <details> 
    <id>2</id> 
    <name>aa</name> 
    <details> 
     <id>3</id> 
     <name>sd</name> 
    </details> 
    </details> 
</details> 

编辑

此外，它可以用递归CTE来完成：

DECLARE @x xml 

;WITH rec AS (
SELECT CAST((
      SELECT TOP 1 id, 
          [name] 
      FROM [sample] 
      ORDER BY id DESC 
      FOR XML PATH('details') 
     ) as xml) as d, 
     1 as [Level] 
UNION ALL 
SELECT CAST((
      SELECT id, 
        [name], 
        cast(r.d as xml) 
      FROM [sample] 
      WHERE s.id = id 
      FOR XML PATH('details') 
     ) as xml) as d, 
     r.[Level]+1 
FROM [sample] s 
INNER JOIN rec r 
    ON s.id = CAST(r.d.query('/details/id/text()') as nvarchar(max))-1 
) 

SELECT TOP 1 WITH TIES d 
FROM rec 
ORDER BY [Level] desc 

相同的输出。

Answer 2

您可以使用这样的查询：

SELECT 
    *, 
    (SELECT 
     * 
    FROM #details 
    WHERE id = 2 
    FOR xml PATH ('Details'), TYPE) 
FROM #details 
WHERE id = 1 
FOR xml PATH ('Details') 

对于内环您可以使用CTE

表创建脚本：

CREATE TABLE #details (
    id int, 
    name varchar(10) 
) 

INSERT INTO #details (id, name) 
    VALUES (1, 'test'), (2, 'test2')