XML 到 SQL 的问题 - SQL Server

XML to SQL Issue - SQL Server(XML 到 SQL 的问题 - SQL Server)
本文介绍了XML 到 SQL 的问题 - SQL Server的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

问题描述

我试图在 sql 脚本中获取 xml 特定元素的位置,但我正在获取特定元素的所有位置的详细信息,因为我只是在寻找特定级别.

I am trying to get position of a specific element of xml in sql script , but I am getting details for all the positions of a specific element , where as I am looking for a particular level only.

这是代码:

declare @xmlVar xml ='
<A specVersion="2.09">
  <B id="1" type="Regular">
    <C>
      <D>
        <E actioncode="A" date="06/13/2018 09:20" />
        <E  actioncode="B" date="06/13/2018 09:20" />
      </D>
      <D>
        <E actioncode="C" date="06/13/2018 09:20" />
      </D>
    </C>
  </B>

  <B id="2" type="Regular">
    <C>
      <D>
        <E  actioncode="D" date="06/13/2018 09:20" />
      </D>
    </C>
    <F>
        <D>
            <E  actioncode="F" date="06/13/2018 09:20" />
        </D>
    </F>
  </B>
</A>' ;

WITH Tally(Nmbr) AS
(
    SELECT TOP (SELECT @xmlVar.value(N'count(//D)','int'))
           ROW_NUMBER() OVER(ORDER BY (SELECT NULL)) 
    FROM master..spt_values --just a pre-filled table with many rows
)
SELECT Nmbr
      ,e.value(N'@actioncode[1]','nvarchar(max)') AS Employee
FROM Tally
OUTER APPLY(SELECT  @xmlVar.query(N'//D').query(N'/D[sql:column("Nmbr")]')) AS B(ds)
OUTER APPLY ds.nodes(N'D/E') AS C(e);

以上查询的结果:

1   A
1   B
2   C
3   D
4   F

预期输出:

1 A
1 B
2 C
3 D

我正在寻找路径 A->B->C->D 中所有 D 的位置,而不是其他路径中的位置.

I am looking for positions of all D which comes in path A->B->C->D not the one's which come in other path.

推荐答案

好的,对于您之前的问题,这似乎是一个相当容易的更改.您可以使用相同的逻辑,但将其缩减为您想要查看的路径:

Okay, this seems to be a rather easy change to your previous question. You can use the same logic, but reduce it to the path you want to see:

WITH Tally(Nmbr) AS
(
    SELECT TOP (SELECT @xmlVar.value(N'count(/A/B/C/D)','int'))
           ROW_NUMBER() OVER(ORDER BY (SELECT NULL)) 
    FROM master..spt_values --just a pre-filled table with many rows
)
SELECT Nmbr
      ,e.value(N'@actioncode[1]','nvarchar(max)') AS Employee
FROM Tally
OUTER APPLY(SELECT  @xmlVar.query(N'/A/B/C/D').query(N'/D[sql:column("Nmbr")]')) AS B(ds)
OUTER APPLY ds.nodes(N'D/E') AS C(e);

更新:获取

正如您所发现的,上述方法不能向后移动到 .试试这个:

WITH AllBs AS 
(
    SELECT ROW_NUMBER() OVER(ORDER BY (SELECT NULL)) AS B_position
          ,b.value('@id','int') AS B_id
          ,b.query('.') AS B_node
    FROM @xmlVar.nodes('/A/B') AS A(b)
)
,AllDs As
(
    SELECT AllBs.*
          ,ROW_NUMBER() OVER(ORDER BY (SELECT NULL)) AS D_position
          ,d.query('.') AS D_node 
    FROM AllBs 
    OUTER APPLY B_node.nodes('B/C/D') AS A(d)
)
SELECT AllDs.*
      ,e.value(N'@actioncode[1]','nvarchar(max)') AS Employee
FROM AllDs
OUTER APPLY D_node.nodes('D/E') AS A(e);

正如我在第一个答案中所写的,排序顺序存在很小的风险......但是使用 tally 表 的方法无法获取第三次出现,因为 [sql:column()] 与任何父项重新编号.这意味着:第三个 是第二个 中的第一个.这就是为什么我将 .query() 放在中间......

As written in my first answer, there's a tiny risk about the sort order... But the approach with the tally table cannot fetch the 3rd occurance, because the position within [sql:column()] is re-numbered with any parent. That means: The 3rd <D> is the 1st in the 2nd <B>. That's why I placed the .query() in between...

这篇关于XML 到 SQL 的问题 - SQL Server的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!

本站部分内容来源互联网,如果有图片或者内容侵犯了您的权益,请联系我们,我们会在确认后第一时间进行删除!

相关文档推荐

ibtmp1是非压缩的innodb临时表的独立表空间,通过innodb_temp_data_file_path参数指定文件的路径,文件名和大小,默认配置为ibtmp1:12M:autoextend,也就是说在文件系统磁盘足够的情况下,这个文件大小是可以无限增长的。 为了避免ibtmp1文件无止境的暴涨导致
SQL query to group by day(按天分组的 SQL 查询)
What does SQL clause quot;GROUP BY 1quot; mean?(SQL 子句“GROUP BY 1是什么意思?意思是?)
MySQL groupwise MAX() returns unexpected results(MySQL groupwise MAX() 返回意外结果)
MySQL SELECT most frequent by group(MySQL SELECT 按组最频繁)
Include missing months in Group By query(在 Group By 查询中包含缺失的月份)