如何使用 XQuery 根据属性过滤 XML 数据

Question

给定下面的XML结构，我需要过滤掉所有<questionSubType/>值等于ABC，并且<option subType=""/>属性等于001的节点：

<questions>
  <question>
    <text>Some text</text>
    <questionType></questionType>
    <questionSubType>ABC</questionSubType>
    <options>
      <option subType="001">
        <text>Y</text>
        <mappedCodes>
          <code>1</code>
        </mappedCodes>
      </option>
      <option subType="001">
        <text>N</text>
        <mappedCodes>
          <code>2</code>
        </mappedCodes>
      </option>
      <option subType="002">
        <text>Y</text>
        <mappedCodes>
          <code>1</code>
        </mappedCodes>
      </option>
    </options>
  </question>
  <question>
    <text>Some more text</text>
    <questionType></questionType>
    <questionSubType>DEF</questionSubType>    
    <options>
      <option subType="001">
        <text>Single</text>
        <mappedCodes>
          <code>PL0157</code>
        </mappedCodes>
      </option>
      <option subType="001">
        <text>Married</text>
        <mappedCodes>
          <code>PD0241</code>
        </mappedCodes>
      </option>
      <option subType="002">
        <text>Single</text>
        <mappedCodes>
          <code>PL1157</code>
        </mappedCodes>
      </option>
      <option subType="002">
        <text>Married</text>
        <mappedCodes>
          <code>PD1241</code>
        </mappedCodes>
      </option>
    </options>
  </question>
  <question>
    <text>Some last text</text>
    <questionType></questionType>
    <questionSubType>ABC</questionSubType>
    <options>
      <option subType="001">
        <text>T</text>
        <mappedCodes>
          <code>2</code>
        </mappedCodes>
      </option>
      <option subType="002">
        <text>V</text>
        <mappedCodes>
          <code>2</code>
        </mappedCodes>
      </option>
    </options>
  </question>
 </questions>

我尝试了以下操作，但这仅根据 <questionSubType/> 值过滤 XML，因为我不确定如何继续查询 <option/> 节点：

        DECLARE
            @subType varchar(5) = '001'
          , @questionSubType varchar(5) = 'ABC'
        SET @XmlOutput = (
            SELECT
                1 as Tag 
              , null as Parent
              , CONVERT(nvarchar(max), F.N.query('./*')) as [question!1!!XML]
            FROM [MyTable] T
                CROSS APPLY T.[Configuration].nodes('//question') F(N)
            WHERE
                F.N.value('(//questionSubType/text())[1]', 'varchar(100)') = @questionSubType
            FOR XML EXPLICIT, ROOT('questions')
        )

        SELECT @XmlOutput as [Configuration]

所以最后，我的输出应该是这样的：

<questions>
  <question>
    <text>Some text</text>
    <questionType></questionType>
    <questionSubType>ABC</questionSubType>
    <options>
      <option subType="001">
        <text>Y</text>
        <mappedCodes>
          <code>1</code>
        </mappedCodes>
      </option>
      <option subType="001">
        <text>N</text>
        <mappedCodes>
          <code>2</code>
        </mappedCodes>
      </option>
    </options>
  </question>
  <question>
    <text>Some last text</text>
    <questionType></questionType>
    <questionSubType>ABC</questionSubType>
    <options>
      <option subType="001">
        <text>T</text>
        <mappedCodes>
          <code>2</code>
        </mappedCodes>
      </option>
    </options>
  </question>
 </questions>

任何帮助将不胜感激。

Answer 1

这里是 XQuery 来救你：

DECLARE @xml XML=
N'<questions>
  <question>
    <text>Some text</text>
    <questionType></questionType>
    <questionSubType>ABC</questionSubType>
    <options>
      <option subType="001">
        <text>Y</text>
        <mappedCodes>
          <code>1</code>
        </mappedCodes>
      </option>
      <option subType="001">
        <text>N</text>
        <mappedCodes>
          <code>2</code>
        </mappedCodes>
      </option>
      <option subType="002">
        <text>Y</text>
        <mappedCodes>
          <code>1</code>
        </mappedCodes>
      </option>
    </options>
  </question>
  <question>
    <text>Some more text</text>
    <questionType></questionType>
    <questionSubType>DEF</questionSubType>    
    <options>
      <option subType="001">
        <text>Single</text>
        <mappedCodes>
          <code>PL0157</code>
        </mappedCodes>
      </option>
      <option subType="001">
        <text>Married</text>
        <mappedCodes>
          <code>PD0241</code>
        </mappedCodes>
      </option>
      <option subType="002">
        <text>Single</text>
        <mappedCodes>
          <code>PL1157</code>
        </mappedCodes>
      </option>
      <option subType="002">
        <text>Married</text>
        <mappedCodes>
          <code>PD1241</code>
        </mappedCodes>
      </option>
    </options>
  </question>
  <question>
    <text>Some last text</text>
    <questionType></questionType>
    <questionSubType>ABC</questionSubType>
    <options>
      <option subType="001">
        <text>T</text>
        <mappedCodes>
          <code>2</code>
        </mappedCodes>
      </option>
      <option subType="002">
        <text>V</text>
        <mappedCodes>
          <code>2</code>
        </mappedCodes>
      </option>
    </options>
  </question>
 </questions>';

--声明你的变量

    DECLARE @subType varchar(5) = '001'
           ,@questionSubType varchar(5) = 'ABC';

--XQuery 将贯穿您的 XML 并添加具有给定类型的所有问题，然后添加除 <options> 之外的所有内部节点。最后一个节点使用过滤谓词再次添加：

 SELECT @xml.query
 ('<questions>
   {
    for $q in /questions/question[(questionSubType/text())[1]=sql:variable("@questionSubType")]
    return 
        <question>
        {
        $q/*[local-name()!="options"]
        }
        {
        $q/options/option[@subType=sql:variable("@subType")]
        }
        </question>
   } 
   </questions> 
 ');