问题描述
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>';
游乐场
现在我想根据 id 属性合并 2 个 XML:
Now I want to merge 2 XMLs based on id attribute:
id = 1 的用户的最终结果:
Final result for user with id = 1:
<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，它不是组合 2 个 XML，而是upsert"操作.
Please note that it is not combining 2 XMLs but "upsert" operation.
备注:
- 我知道这种模式违反了数据库规范化，规范化是可行的方法(但在这种情况下不是)
- 我知道使用派生表的解决方案，
.nodes()和.value()函数首先解析两个 XML，然后合并并写回
- I know that this kind of schema violates database normalization and normalizing it is the way to go (but not in this case)
- I know solution that utilize derived tables,
.nodes()and.value()functions first to parse both XML, then merge and write back
我正在寻找的是 XPath/XQuery 表达式，它将合并到一个语句中(无派生表/动态 sql*):
I am searching for is XPath/XQuery expression that will merge it in one statement (no derived tables/dynamic-sql*):
* 如果绝对需要，可以使用动态 SQL，但我想避免使用.
UPDATE #Users
SET suggestions.modify(... sql:variable("@suggestions") ...); --changes only here
WHERE id = @userId;
/* replace ... for ... where ... with sql:variable */
推荐答案
尝试了一段时间后，我认为这是不可能的...
After trying around a while I think this is not possible...
这里有类似的问题:XQuery 在单个 SQL 更新命令中添加或替换属性
.modify(insert Expression1 ... ) 不允许在通过 @sql:variable() 传入的 XML 内获取数据code> 或 sql:column()
The .modify(insert Expression1 ... ) does not allow to get data within an XML passed in via @sql:variable() or sql:column()
在此处阅读:https://msdn.microsoft.com/en-us/library/ms175466.aspx 在 Expression1 -> "常量 XML 或独立的 sql:column/sql:variable 或 XQuery(到同一实例)
Read here: https://msdn.microsoft.com/en-us/library/ms175466.aspx at Expression1 -> "constant XML or stand alone sql:column / sql:variable or XQuery (to the same instance)
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>
*/
您可以这样声明第二个 XML:
You might declare the second XML as such:
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" />';
但是您将没有机会使用 id 的值作为 XQuery 过滤器
But than you'll have no chance to use the id's value as XQuery filter
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一点，我认为没有机会在 .modify() 的一次调用中组合两个不同的 XML_DML 语句.
And last but not least I think there is no chance to combine two different XML_DML statements within one call of .modify().
我唯一的想法就是这个，但它行不通.IF 似乎只能在内一个表达式中使用，但不能区分两个执行路径
The only idea I had was this, but it doesn't work. IF seems to be usable only within an Expression, but not two distinguish between two execution paths
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...
So my conclusion: No, this is not possible...
我在这里提供的解决方案 https://stackoverflow.com/a/35060150/5089204 在第二部分(如果你想‘合并’两本书结构")将是我解决这个问题的方法.
The solution I provided here https://stackoverflow.com/a/35060150/5089204 in the second section ("If you want to 'merge' two Books-structures") would be my way to solve this.
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