Question

我有这两个XML存储在2个表中。

问题XML

<Question>
    <Choice ID="1">
        <Value>Choice A</Value>
    </Choice>
    <Choice ID="2">
        <Value>Choice B</Value>
    </Choice>
    <Choice ID="3">
        <Value>Choice C</Value>
    </Choice>
    <Choice ID="4">
        <Value>Choice D</Value>
    </Choice>
    <Choice ID="5">
        <Value>Choice E</Value>
    </Choice>
</Question>

响应XML

<Response>
    <Question>
        <Value>Choice B</Value>
        <Value>Choice C</Value>
    </Question>
</Response>

我需要为响应XML 值元素添加名为 ID 的新属性 >。可以在问题XML 中找到新 ID 属性的值。

对于实例，如果您看到问题XML ，则值ID和Choice B is 2的正确Choice C is 3

所以我需要的最终 Response XML 应该是这样的

<Response>
    <Question>
        <Value ID="2">Choice B</Value>
        <Value ID="3">Choice C</Value>
    </Question>
</Response>

有人可以告诉我该怎么做吗？

Answer 1

如果要在大多数情况下修改XML中的多个位置，最好是粉碎信息并从头开始重新构建XML：

DECLARE @q XML=
N'<Question>
    <Choice ID="1">
        <Value>Choice A</Value>
    </Choice>
    <Choice ID="2">
        <Value>Choice B</Value>
    </Choice>
    <Choice ID="3">
        <Value>Choice C</Value>
    </Choice>
    <Choice ID="4">
        <Value>Choice D</Value>
    </Choice>
    <Choice ID="5">
        <Value>Choice E</Value>
    </Choice>
</Question>';

DECLARE @r XML=
N'<Response>
    <Question>
        <Value>Choice B</Value>
        <Value>Choice C</Value>
    </Question>
</Response>';

WITH QuestionCTE AS
(
    SELECT c.value('@ID','int') AS qID
          ,c.value('Value[1]','nvarchar(max)') AS qVal
    FROM @q.nodes('Question/Choice') AS A(c)
)
,ResponseCTE AS
(
    SELECT r.value('.','nvarchar(max)') AS rVal
    FROM @r.nodes('Response/Question/Value') AS A(r)
)
SELECT 
(
    SELECT q.qID AS [Value/@ID]
          ,q.qVal AS [Value] 
    FROM ResponseCTE AS r
    LEFT JOIN QuestionCTE AS q ON r.rVal=q.qVal 
    FOR XML PATH(''),TYPE
)
FOR XML PATH('Question'),ROOT('Response')

Answer 2

您可以将XML转换为表格（问题和响应），然后链接值。然后，您可以从问题表中查找ID。例如：

DECLARE @xml_q XML = N'
<Question>
    <Choice ID="1">
        <Value>Choice A</Value>
    </Choice>
    <Choice ID="2">
        <Value>Choice B</Value>
    </Choice>
    <Choice ID="3">
        <Value>Choice C</Value>
    </Choice>
    <Choice ID="4">
        <Value>Choice D</Value>
    </Choice>
    <Choice ID="5">
        <Value>Choice E</Value>
    </Choice>
</Question>';

DECLARE @xml_r XML = N'
<Response>
    <Question>
        <Value>Choice B</Value>
        <Value>Choice C</Value>
    </Question>
</Response>';

;WITH q_id AS (
    SELECT
        n.c.value('(./Value)[1]','NVARCHAR(128)') AS value,
        n.c.value('@ID','INT') AS id 
    FROM
        @xml_q.nodes('//Choice') AS n(c)
),
r_v AS (
    SELECT
        n.c.value('(.)[1]','NVARCHAR(128)') AS value
    FROM
        @xml_r.nodes('//Value') AS n(c)
)
SELECT 
    (
        SELECT
            q_id.id AS "Value/@ID",
            r_v.value AS "Value"
        FROM 
            r_v 
            INNER JOIN q_id ON 
                q_id.value=r_v.value
        FOR
            XML PATH(''), TYPE
    )
FOR 
    XML PATH('Question'), ROOT('Response');

会给出答复：

<Response>
  <Question>
    <Value ID="2">Choice B</Value>
    <Value ID="3">Choice C</Value>
  </Question>
</Response>

使用MS SQL修改XML列

2 个答案: