使用MS SQL修改XML列

Question

I've these 2 XML which is stored in 2 tables.

Question 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>

Response XML

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

I need to add a new attribute called ID , to all the Value elements present in the Response XML . The value of the new ID attribute can be found in the Question XML .

For Instance, If you see the Question XML , the correct ID of the value Choice B is 2 and Choice C is 3

So the final Response XML which i need, should be like this

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

Can someone please tell me how to do this ?

Answer 1

If you want to modify more than one place in an XML in most cases the best is to shredd the information and re-build the XML from scratch:

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

You can convert the XML to tables (question & response) and then link on the value. You then look up the ID from the question table. Example:

---

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');

Will give the response:

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