SQL Server-仅使用.modify()合并两个XML
假设我们有:SQL Server-仅使用.modify()合并两个XML,sql,sql-server,xml,tsql,xpath,Sql,Sql Server,Xml,Tsql,Xpath,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
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>';
请注意,它不是组合2个XML,而是“upsert”操作
备注:
- 我知道这种模式违反了数据库规范化,规范化是一种方法(但在本例中不是)
- 我知道使用派生表的解决方案,
和.nodes()
函数首先解析这两个XML,然后合并并写回.value()
XPath/XQuery
表达式(无派生表/dynamic sql*):
*如果绝对需要,可以使用动态SQL,但我想避免使用它
经过一段时间的尝试,我认为这是不可能的 这里也有类似的问题:
.modify(insert Expression1…
不允许在通过@sql:variable()
或sql:column()传入的XML中获取数据
阅读此处:at Expression1->“常量XML或独立sql:column/sql:variable或XQuery(对同一实例)
最后但并非最不重要的一点是,我认为不可能在一次调用.modify()
中组合两个不同的XML\u DML语句
我唯一的想法是这个,但它不起作用。IF似乎只在表达式中可用,而不是区分两条执行路径
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
我在第二节中提供的解决方案(“如果你想‘合并’两本书的结构”)将是我解决这个问题的方法。试试这样的方法,它很容易理解,但很长
如果你有任何问题,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
declare@Users表(id INT主键,名称VARCHAR(100),XML);
插入@Users(id、名称、建议)
选择1,‘鲍勃’,N'
'
联合所有
选择2,‘吉米’,N'
';
声明@userId INT=1,
@建议XML=N'
';
声明@Users1表(userid INT、productid INT、score INT);
插入到@Users1中
选择id userid,
t、 c.value(“(@id[1])”,“VARCHAR(50)”作为Productid,
t、 c.值(“(@score[1])”,“VARCHAR(50)”作为分数
来自@Users yt
交叉应用yt.建议.节点(“产品/产品”)t(c)
--从@Users1中选择*
;以CTE1作为
(
选择@userId userId,
t、 c.value(“(@id[1])”,“VARCHAR(50)”作为Productid,
t、 c.值(“(@score[1])”,“VARCHAR(50)”作为分数
来自@suggestions.nodes('Products/Product')t(c)
)
将@users1合并为trg
使用cte1作为src
在trg.userid=src.userid和trg.productid=src.productid上
当不匹配时
插入(userid、productid、score)值(src.userid、src.productid、src.score);
选择不同的a.userid
,(选择b.productid为'@productid',b.score为'@score'
来自@users1 b,其中a.userid=b.userid
对于xml路径(“产品”),根(“产品”)
来自@users1a
这并不是你所希望的优雅的一行。不过,它还是挺管用的,所以我将与大家分享。(我相信还有改进的余地。)
如果OBJECT_ID('tempdb..#Users')不是空的DROP TABLE#Users
创建表#Users(id INT主键,name VARCHAR(100),XML);
插入#用户(id、姓名、建议)
选择1,‘鲍勃’,N'
'
联合所有
选择2,‘吉米’,N'
';
--@建议略有不同(请注意,“”父元素已消失)。
声明@userId INT=1,
@建议XML=N'
';
--捕获原始建议,以便我们以后进行比较
将@OldXML声明为XML=(从#Users中选择建议,其中id=@userId)
更新#用户——这将插入新的建议(我们根据需要删除旧建议,如下所示)。
设置建议。修改('insert sql:variable(@suggestions')作为第一个插入到/Products[1]'))
其中id=@userId;
而(@OldXML.exist('/Products/Product')=1)
开始——对于每个用户的原始产品,删除它的第二个实例。
声明@prodId int=(从#Users中选择@OldXML.value('(/Products/Product/@id)[1],'nvarchar(max)'),其中id=@userId)
更新#用户设置建议。修改('delete/Products/Product[@id=sql:variable(@prodId”)][2]”,其中id=@userId
SET@OldXML.modify('delete/Products/Product[@id=sql:variable(@prodId”)][1]”)
结束
选择*FROM#Users,其中id=@userId
删除表#用户
我认为您可以使用如下查询:
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;
我可以放宽要求,所以继续展示动态sql:)无论如何,我会在2天后提供奖励,并在几天后接受。@lad2025即使是动态的,也需要两次修改()。并且没有机会使用特殊sql。实际上,这不是一个选项……谢谢回复,但我知道答案是利用.nodes()/.values()
(请参见备注)。此问题的要点是仅使用.modify()
。并检查结果。1)您的答案未更新建议列2)它未正确处理产品id=3
(请参见其得分为4而非2)。这个问题的原因是:。OP自己在那里写了被接受的答案。看看他的答案、我的答案以及评论……有比你在这个答案中显示的更好的方法(顺便说一句:*没有循环!如果可以避免的话)。这里的问题是:一次调用.modify()
一次性执行条件上插。正如我指出的,我认为这是不可能的。。。
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" />';
SET @xml1.modify('insert sql:variable("@xml2") as first into /Products[**How should one filter here?**]');
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');
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
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
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;