Question

我需要以下XML结构

linenumber TEST TEST2 TEST3
1          valu value value

我使用的表

<root>
    <lkz>12345</lkz>
    <styles>
        <style>
            <productNo>
            ..
            </productNo>
        </style>
        <style>
            <productNo>
            ..
            </productNo>
        </style>
    </styles>
</root>

lkz | productNo ------+---------------- 12345 | 123456-789 12345 | 1213456-788始终相同

我的T-SQL代码：

lkz

结果

DECLARE @xmlData XML

SET @xmlData = (SELECT productNo
                FROM dbo.otto
                FOR XML PATH ( 'Style' ), ROOT ('Styles'), ELEMENTS)

SELECT @xmlData

问题（对不起我的英语）：

我如何得到＆＃34;一切＆＃34;
如何将<Styles> <Style> <productNo>123456-789</productNo> </Style> <Style> <productNo>123456-788</productNo> </Style> </Styles>仅作为简单节点的第二行

感谢您的帮助！

Answer 1

在主查询中按lkz分组，并在子查询中获取相应的productNo值。

select O1.lkz,
       (
       select O2.productNo
       from dbo.Otto as O2
       where O1.lkz = O2.lkz
       for xml path('style'), root('styles'), type
       )
from dbo.otto as O1
group by O1.lkz
for xml path(''), root('root');

Answer 2

我的桌子

lkz|stylesNo|ClassOfGoods|ClusterAttribute1
12345|123456-789|test1|Herren
12345|123456-788|test2|Damen

需要xml

<root>
<LKZ>12345</LKZ>
<Styles>
<Style>
<StyleNo>123456-789</StyleNo>
<ClassOfGoods>test1</ClassOfGoods>
<ClusterAttribute1 CountryIsoCode="DE"><![CDATA[Herren]]></ClusterAttribute1>

我的t-sql

SET @xmlData =
    (select O1.lkz,
                (
                select O2.stylesNo
                ,O2.ClassOfGoods

                ,(SELECT
                                1 as Tag,  
                                0 as Parent,
                                'DE' as [ClusterAttribute1!1!CountryIsoCode],
                                ClusterAttribute1 as [ClusterAttribute1!1!!CDATA]
                    FROM
                    dbo.otto as O3
                    --where O1.stylesNo = O3.stylesNo
                    FOR XML EXPLICIT, type
                    )

                from dbo.otto as O2 
                where O1.lkz = O2.lkz
                for xml path('style'), root('styles'), type
                )
    from dbo.otto as O1
    group by O1.lkz
    for xml path(''), root('ottopartner')
    )

SELECT @xmlData

结果

<root>
  <lkz>12345</lkz>
  <styles>
    <style>
      <stylesNo>123456-789</stylesNo>
      <ClassOfGoods>test1</ClassOfGoods>
      <ClusterAttribute1 CountryIsoCode="DE">Herren</ClusterAttribute1>
      <ClusterAttribute1 CountryIsoCode="DE">Damen</ClusterAttribute1>
    </style>
    <style>
      <stylesNo>123456-788</stylesNo>
      <ClassOfGoods>test2</ClassOfGoods>
      <ClusterAttribute1 CountryIsoCode="DE">Herren</ClusterAttribute1>
      <ClusterAttribute1 CountryIsoCode="DE">Damen</ClusterAttribute1>
    </style>
  </styles>
</root>

T-SQL XML节点是交错的

2 个答案: