假设我们有：
CREATE TABLE #Users(id INT PRIMARY KEY, name VARCHAR(100), suggestions XML);
INSERT INTO #Users(id, name, suggestions)
SELECT 1, 'Bob', N'<Products>
<Product id="1" score="1"/>
<Product id="2" score="5"/>
<Product id="3" score="4"/>
</Products>'
UNION ALL
SELECT 2, 'Jimmy', N'<Products>
<Product id="6" score="3"/>
</Products>';
DECLARE @userId INT = 1,
@suggestions XML = N'<Products>
<Product id="2" score="5"/>
<Product id="3" score="2"/>
<Product id="7" score="1" />
</Products>';
现在我想基于id属性合并2个XML：
id = 1的用户的最终结果：
<Products>
<Product id="1" score="1"/> -- nothing changed (but not exists in @suggestions)
<Product id="2" score="5"/> -- nothing changed (but exists in @suggestions)
<Product id="3" score="2"/> -- update score to 2
<Product id="7" score="1"/> -- insert new element
</Products>
请注意，它并没有组合2个XML，而是＆＃34; upsert＆＃34;操作
说明：
.nodes()和.value()函数的解决方案首先解析XML，然后合并并写回来 我正在搜索is XPath/XQuery表达式，它将在一个语句中合并它（没有派生表/ dynamic-sql *）：
*如果绝对需要，可以使用动态SQL，但我想避免使用它。
UPDATE #Users
SET suggestions.modify(... sql:variable("@suggestions") ...); --changes only here
WHERE id = @userId;
/* replace ... for ... where ... with sql:variable */
答案 0 :(得分：4)
经过一段时间的努力，我认为这是不可能的......
这里有类似的问题：XQuery adding or replacing attribute in single SQL update command
.modify(insert Expression1 ... )不允许通过@sql:variable()或sql:column()
在此处阅读：{1}}在Expression1 - ＆gt; “常量XML或独立的sql：column / sql：variable或XQuery（到同一个实例）
DECLARE @xml1 XML= --the existing XML
'<Products>
<Product id="1" score="1" />
<Product id="2" score="5" />
<Product id="3" score="4" />
</Products>';
DECLARE @xml2 XML= --the XML with new or changed data
'<Products>
<Product id="2" score="5" />
<Product id="3" score="2" />
<Product id="7" score="1" />
</Products>';
SET @xml1.modify('insert sql:variable("@xml2") as first into /Products[1]');
SELECT @xml1;
/* The full node is inserted!
Without any kind of preparation there is NO CHANCE to get the inner nodes only
<Products>
<Products>
<Product id="2" score="5" />
<Product id="3" score="2" />
<Product id="7" score="1" />
</Products>
<Product id="1" score="1" />
<Product id="2" score="5" />
<Product id="3" score="4" />
</Products>
*/
您可以声明第二个XML：
DECLARE @xml2 XML= --the XML with new or changed data
'<Product id="2" score="5" />
<Product id="3" score="2" />
<Product id="7" score="1" />';
但是你没有机会将id的值用作XQuery过滤器
SET @xml1.modify('insert sql:variable("@xml2") as first into /Products[**How should one filter here?**]');
最后但并非最不重要的是，我认为在.modify()的一次调用中没有机会合并两个不同的XML_DML语句。
我唯一的想法就是这个，但它不起作用。 IF似乎只能在 表达式中使用，但不能区分两个执行路径
SET @xml1.modify('if (1=1) then
insert sql:variable("@xml2") as first into /Products[1]
else
replace value of /Products[1]/Product[@id=1][1]/@score with 100');
所以我的结论是：不，这是不可能的......
我在这里提供的解决方案https://msdn.microsoft.com/en-us/library/ms175466.aspx在第二部分（“如果你想'合并'两个书籍结构”）将是我解决这个问题的方法。
答案 1 :(得分：0)
尝试这样的事情，很容易理解，但是很冗长。
如果您有任何问题，请告诉我。
declare @Users TABLE(id INT PRIMARY KEY, name VARCHAR(100), suggestions XML);
INSERT INTO @Users(id, name, suggestions)
SELECT 1, 'Bob', N'<Products>
<Product id="1" score="1"/>
<Product id="2" score="5"/>
<Product id="3" score="4"/>
</Products>'
UNION ALL
SELECT 2, 'Jimmy', N'<Products>
<Product id="6" score="3"/>
</Products>';
DECLARE @userId INT = 1,
@suggestions XML = N'<Products>
<Product id="2" score="5"/>
<Product id="3" score="2"/>
<Product id="7" score="1" />
</Products>';
declare @Users1 TABLE(userid INT , productid int,score int);
insert into @Users1
SELECT id userid,
t.c.value('(@id[1])', 'VARCHAR(50)') AS Productid,
t.c.value('(@score[1])', 'VARCHAR(50)') AS score
FROM @Users yt
cross APPLY yt.suggestions.nodes('Products/Product') t(c)
--select * from @Users1
;With CTE1 as
(
select @userId userid,
t.c.value('(@id[1])', 'VARCHAR(50)') AS Productid,
t.c.value('(@score[1])', 'VARCHAR(50)') AS score
from @suggestions.nodes('Products/Product') t(c)
)
Merge @users1 as trg
using cte1 as src
on trg.userid=src.userid and trg.productid=src.productid
when not matched then
insert (userid,productid,score) values(src.userid,src.productid,src.score);
select distinct a.userid
,(select b.productid as '@Productid',b.score as '@Score'
from @users1 b where a.userid=b.userid
for xml path('Product'),root('Products'))
from @users1 a
答案 2 :(得分：0)
这不是你所希望的优雅单线。不过，它有点工作，所以我会分享。 （而且我确信还有改进的余地。）
IF OBJECT_ID('tempdb..#Users') IS NOT NULL DROP TABLE #Users
CREATE TABLE #Users(id INT PRIMARY KEY, name VARCHAR(100), suggestions XML);
INSERT INTO #Users(id, name, suggestions)
SELECT 1, 'Bob', N'<Products>
<Product id="1" score="1"/>
<Product id="2" score="5"/>
<Product id="3" score="4"/>
</Products>'
UNION ALL
SELECT 2, 'Jimmy', N'<Products>
<Product id="6" score="3"/>
</Products>';
--@Suggestions is slightly different (note the "<Products>" parent element is gone).
DECLARE @userId INT = 1,
@suggestions XML = N' <Product id="2" score="5"/>
<Product id="3" score="2"/>
<Product id="7" score="1" />';
--Capture the original suggestions so we can compare later
DECLARE @OldXML As XML = (SELECT suggestions from #Users where id = @userId)
UPDATE #Users --this inserts the new suggestions (we delete old ones as needed, below).
SET suggestions.modify('insert sql:variable("@suggestions") as first into /Products[1]')
WHERE id = @userId;
WHILE (@OldXML.exist('/Products/Product') = 1)
BEGIN --For each of the user's original Products, delete the second instance of it.
DECLARE @prodId int = (SELECT @OldXML.value('(/Products/Product/@id)[1]', 'nvarchar(max)') FROM #Users where id = @userId)
UPDATE #Users SET suggestions.modify('delete /Products/Product[@id= sql:variable("@prodId")][2]') WHERE id = @userId
SET @OldXML.modify('delete /Products/Product[@id=sql:variable("@prodId")][1]')
END
SELECT * FROM #Users where id = @userId
DROP TABLE #Users
答案 3 :(得分：0)
我认为你可以使用这样的查询：
UPDATE #Users
SET suggestions = (
SELECT id, score
FROM
(SELECT
*,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY ord) As seq
FROM (
SELECT
c.value('@id', 'INT') AS id,
c.value('@score', 'INT') AS score,
1 As ord
FROM
@suggestions.nodes('/Products/Product') As t(c)
UNION ALL
SELECT
c.value('@id', 'INT') AS id,
c.value('@score', 'INT') AS score,
2 as ord
FROM
(SELECT suggestions x FROM #Users WHERE id = @userId) As u CROSS APPLY
u.x.nodes('/Products/Product') As t(c)) dt) Product
WHERE seq = 1
FOR XML AUTO, ROOT('Products'))
WHERE
(id = @userId);
SELECT *
FROM #Users;
<强> [Playground]