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Sql 使用节点（）和值（）获取父值

sql sql-server xml sql-server-2008 xquery

Sql 使用节点（）和值（）获取父值,sql,sql-server,xml,sql-server-2008,xquery,Sql,Sql Server,Xml,Sql Server 2008,Xquery,我有一个与此XSD匹配的XML文档片段： <xs:complexType name="QuestionType"> <xs:sequence> <xs:element name="questionId" type="xs:string" minOccurs="1" /> <xs:element name="questionDescription" type="xs:string" minOccurs="1" /> <

我有一个与此XSD匹配的XML文档片段：

<xs:complexType name="QuestionType">
  <xs:sequence>
    <xs:element name="questionId" type="xs:string" minOccurs="1" />
    <xs:element name="questionDescription" type="xs:string" minOccurs="1" />
    <xs:element name="questionHeader" type="xs:string" minOccurs="0" />
    <xs:element name="questionLabel" type="xs:string" minOccurs="0" />
    <xs:element name="version" type="xs:string" minOccurs="1" maxOccurs="1" />
    <xs:element name="SubQuestion" type="QuestionType"
                minOccurs="0" maxOccurs="unbounded" />
  </xs:sequence>
</xs:complexType>
我希望可以使用递归CTE来解决这个问题，但我很难将其组合在一起。

您可以使用CTE：

WITH TopLevel (ID, ParentID, Description, Header, Label,Level)
AS
(
SELECT -- Top-level questions...
  C.value('questionId[1]', 'NVarChar(1000)') Id,
  NULL ParentId,
  C.value('questionDescription[1]', 'NVarChar(1000)') Description,
  NULLIF(C.value('questionHeader[1]', 'NVarChar(1000)'), '') Header,
  NULLIF(C.value('questionLabel[1]', 'NVarChar(1000)'), '') Label,
  C.value('version[1]', 'NVarChar(1000)') Version,
  0 AS Level
FROM @X.nodes('//Question') X(C)
UNION ALL
SELECT -- Sub-questions...
  C.value('questionId[1]', 'NVarChar(1000)') Id,
  C.query('..').value('(Question/questionId)[1]', 'NVarChar(1000)') ParentId,
  C.value('questionDescription[1]', 'NVarChar(1000)') Description,
  NULLIF(C.value('questionHeader[1]', 'NVarChar(1000)'), '') Header,
  NULLIF(C.value('questionLabel[1]', 'NVarChar(1000)'), '') Label,
  C.value('version[1]', 'NVarChar(1000)') Version
  ,Level + 1 AS Level
FROM @X.nodes('//SubQuestion') X(C)
JOIN TopLevel AS t ON C.query('..').value('(Question/questionId)[1]', 'NVarChar(1000)') = t.id )

SELECT * FROM TopLevel
参考资料：

到目前为止，我正在这样做，尽管我仍然希望将查询压缩一点：

WITH Q AS (
  SELECT
    C.value('questionId[1]', 'NVarChar(1000)') Id,
    NULL ParentId,
    C.value('questionDescription[1]', 'NVarChar(1000)') Description,
    NULLIF(C.value('questionHeader[1]', 'NVarChar(1000)'), '') Header,
    NULLIF(C.value('questionLabel[1]', 'NVarChar(1000)'), '') Label,
    C.value('version[1]', 'NVarChar(1000)') Version
  FROM @X.nodes('//Question') X(C)
  UNION ALL
  SELECT
    C.value('questionId[1]', 'NVarChar(1000)') Id,
    COALESCE(
      C.query('..').value('(Question/questionId)[1]', 'NVarChar(1000)'),
      C.query('..').value('(SubQuestion/questionId)[1]', 'NVarChar(1000)')
    ) ParentId,
    C.value('questionDescription[1]', 'NVarChar(1000)') Description,
    NULLIF(C.value('questionHeader[1]', 'NVarChar(1000)'), '') Header,
    NULLIF(C.value('questionLabel[1]', 'NVarChar(1000)'), '') Label,
    C.value('version[1]', 'NVarChar(1000)') Version
  FROM @X.nodes('//SubQuestion') X(C)
)
SELECT Q.Id, Q.ParentId, Q.Description, Q.Header, Q.Label, Q.Version
FROM Q;
这是一个重要的位，因为它将获得第一个非null的值：

COALESCE(
  C.query('..').value('(Question/questionId)[1]', 'NVarChar(1000)'),
  C.query('..').value('(SubQuestion/questionId)[1]', 'NVarChar(1000)')
) ParentId
考虑到您已经用sql-server-2008标记了这个问题，并且IMHO sql server 2008支持XQuery，我想提出一个不同的“角度”：使用XPath表达式选择您感兴趣的节点

.//*[local-name(.) = 'Question' or local-name(.) = 'SubQuestion']
请注意，我正在使用XPath函数local-name（），以防实际XML数据具有名称空间声明

我创建了一个示例XML文件来测试上述表达式：

<?xml version="1.0" encoding="UTF-8"?>
<Questions xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
           xmlns="http://www.acme.com" xsi:schemaLocation="sample.xsd">
    <Question>
        <questionId>1</questionId>
        <questionDescription>Question 1</questionDescription>
        <version>1</version>
        <SubQuestion>
            <questionId>1.1</questionId>
            <questionDescription>Question 1.1</questionDescription>
            <version>1</version>
            <SubQuestion>
                <questionId>1.1.1</questionId>
                <questionDescription>Question 1.1.1</questionDescription>
                <version>1</version>
                <SubQuestion>
                    <questionId>1.1.1.1</questionId>
                    <questionDescription>Question 1.1.1.1</questionDescription>
                    <version>1</version>
                </SubQuestion>
                <SubQuestion>
                    <questionId>1.1.1.2</questionId>
                    <questionDescription>Question 1.1.1.2</questionDescription>
                    <version>1</version>
                </SubQuestion>
            </SubQuestion>
            <SubQuestion>
                <questionId>1.2</questionId>
                <questionDescription>Question 1.2</questionDescription>
            </SubQuestion>
        </SubQuestion>
    </Question>
    <Question>
        <questionId>2</questionId>
        <questionDescription>Question 2</questionDescription>
        <version>1</version>
    </Question>
    <Question>
        <questionId>3</questionId>
        <questionDescription>Question 3</questionDescription>
        <version>1</version>
        <SubQuestion>
            <questionId>3.1</questionId>
            <questionDescription>Question 3.1</questionDescription>
            <version>1</version>
        </SubQuestion>
    </Question>
</Questions>
边栏：如果您的文档非常大，那么XQuery非常慢，您最好使用它OpenXml@kd7：好的，谢谢你指出这一点。我知道XQuery的速度有多慢，但对于我的文档来说应该足够了。@Yuck:我有两个问题：1）您是否可以发布至少一个XML片段（或示例），而不仅仅是XSD？2） 您是在寻找SQL查询还是XQuery查询适合您？文档结构的问题是，
的父级可能有
或
。本CTE仅考虑了前者。我在递归查询“TopLevel”的“ParentID”列中得到了锚和递归部分之间的错误
类型不匹配。
检查答案，我只是添加了一个级别字段来显示问题/子问题的深度。您始终可以在CTE之后的SELECT中将TopLevel连接回自身，以获得附加的子问题关系。+1使用单个查询的解决方案的关键是在回答开始时使用XQuery选择器。我正在编辑我的最终查询，目的是让它在某个地方。 
<?xml version="1.0" encoding="UTF-8"?>
<Questions xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
           xmlns="http://www.acme.com" xsi:schemaLocation="sample.xsd">
    <Question>
        <questionId>1</questionId>
        <questionDescription>Question 1</questionDescription>
        <version>1</version>
        <SubQuestion>
            <questionId>1.1</questionId>
            <questionDescription>Question 1.1</questionDescription>
            <version>1</version>
            <SubQuestion>
                <questionId>1.1.1</questionId>
                <questionDescription>Question 1.1.1</questionDescription>
                <version>1</version>
                <SubQuestion>
                    <questionId>1.1.1.1</questionId>
                    <questionDescription>Question 1.1.1.1</questionDescription>
                    <version>1</version>
                </SubQuestion>
                <SubQuestion>
                    <questionId>1.1.1.2</questionId>
                    <questionDescription>Question 1.1.1.2</questionDescription>
                    <version>1</version>
                </SubQuestion>
            </SubQuestion>
            <SubQuestion>
                <questionId>1.2</questionId>
                <questionDescription>Question 1.2</questionDescription>
            </SubQuestion>
        </SubQuestion>
    </Question>
    <Question>
        <questionId>2</questionId>
        <questionDescription>Question 2</questionDescription>
        <version>1</version>
    </Question>
    <Question>
        <questionId>3</questionId>
        <questionDescription>Question 3</questionDescription>
        <version>1</version>
        <SubQuestion>
            <questionId>3.1</questionId>
            <questionDescription>Question 3.1</questionDescription>
            <version>1</version>
        </SubQuestion>
    </Question>
</Questions>

declare namespace acme = "http://www.acme.com";
<AllQuestions>
 {
   for $question in .//*[local-name(.) = 'Question' or local-name(.) = 'SubQuestion']
   return
        <Question>
            <questionId>{ data($question/acme:questionId) }</questionId>
            <questionDescription>{ data($question/acme:questionDescription) }</questionDescription>
        </Question>
 }
</AllQuestions>

<?xml version="1.0" encoding="UTF-8"?>
<AllQuestions>
   <Question>
      <questionId>1</questionId>
      <questionDescription>Question 1</questionDescription>
   </Question>
   <Question>
      <questionId>1.1</questionId>
      <questionDescription>Question 1.1</questionDescription>
   </Question>
   <Question>
      <questionId>1.1.1</questionId>
      <questionDescription>Question 1.1.1</questionDescription>
   </Question>
   <Question>
      <questionId>1.1.1.1</questionId>
      <questionDescription>Question 1.1.1.1</questionDescription>
   </Question>
   <Question>
      <questionId>1.1.1.2</questionId>
      <questionDescription>Question 1.1.1.2</questionDescription>
   </Question>
   <Question>
      <questionId>1.2</questionId>
      <questionDescription>Question 1.2</questionDescription>
   </Question>
   <Question>
      <questionId>2</questionId>
      <questionDescription>Question 2</questionDescription>
   </Question>
   <Question>
      <questionId>3</questionId>
      <questionDescription>Question 3</questionDescription>
   </Question>
   <Question>
      <questionId>3.1</questionId>
      <questionDescription>Question 3.1</questionDescription>
   </Question>
</AllQuestions>

SELECT
    C.value('questionId[1]', 'NVarChar(1000)') Id,
    COALESCE(
      C.query('..').value('(Question/questionId)[1]', 'NVarChar(1000)'),
      C.query('..').value('(SubQuestion/questionId)[1]', 'NVarChar(1000)')
    ) ParentId,
    C.value('questionDescription[1]', 'NVarChar(1000)') Description,
    NULLIF(C.value('questionHeader[1]', 'NVarChar(1000)'), '') Header,
    NULLIF(C.value('questionLabel[1]', 'NVarChar(1000)'), '') Label,
    C.value('version[1]', 'NVarChar(1000)') Version
FROM
  @X.nodes('.//*[local-name(.)="Question" or local-name(.)="SubQuestion"]') X(C);