如何将SQL Server SELECT转换为Mule 4中的XML？

Question

如何使用Mule 4中的数据编织将以下SQL输出转换为XML？

SELECT s.RefId
          ,s.LocalId
          ,s.StateProvinceId
          ,s.SchoolName
          ,e.Email
          ,e.EmailType
      FROM SchoolInfo s
      LEFT OUTER JOIN SchoolEmail e
      ON    e.SchoolRefId = s.RefId
      WHERE s.RefId = :ref_id

SQL中的输出是：

RefId                               LocalId StateProvinceId SchoolName      Email               Type
7FDF722B-6BBA-4BF0-8205-A5380B269EF1    1   SA              Steve's School  steven@gmail.com    prm
7FDF722B-6BBA-4BF0-8205-A5380B269EF1    1   SA              Steve's School  test@gmail.com      sec

XML输出应该如下所示：

<ns0:SchoolInfo xmlns:ns0="http://www.sifassociation.org/datamodel/au/3.4" RefId="7FDF722B-6BBA-4BF0-8205-A5380B269EF1">
  <ns0:LocalId>1</ns0:LocalId>
  <ns0:StateProvinceId>SA</ns0:StateProvinceId>
  <ns0:SchoolName>Steve's School</ns0:SchoolName>
  <ns0:SchoolEmailList>
    <ns0:Email Type="prm">steven@gmail.com</ns0:Email>
    <ns0:Email Type="sec">test@gmail.com</ns0:Email>
  </ns0:SchoolEmailList>
</ns0:SchoolInfo>

谢谢，史蒂夫

Answer 1

下面是生成相同XML的DW表达式：

%dw 2.0
output application/xml
ns ns0 http://www.sifassociation.org/datamodel/au/3.4
var rId = payload[0].RefId
var lId = payload[0].LocalId
var sId = payload[0].StateProvinceId
---
ns0#SchoolInfo @(RefId: rId): {
    ns0#LocalId: lId,
    ns0#StateProvinceId: sId,
    ns0#SchoolEmailList: payload reduce (e,acc={}) -> acc ++ {
        ns0#Email @(Type: e.Type): e.Email
    }   
} 

我假设每个查询的RefId、LocalId和StateProvinceId都是相同的。

对reduce：reduce的解释作了详细的解释，并给出了这里的理论基础。这里还有的MuleSoft文档页面。这最后一页很好地解释了reduce

现在用我自己的话来说，reduce接受一个输入(1)一个数组和(2)一个lambda函数。

数组包含reduce将以类似于map函数的方式迭代的元素。map和reduce函数之间的相似之处在这里结束:)。

lambda函数需要两个参数：(1)从数组中迭代的当前元素和(2)累加器。可以将累加器初始化为值(我将其设置为用例中的对象{}，因为XML不喜欢数组)。第一次迭代的lambda函数的结果被设置为下一次迭代的累加器，依此类推。

一旦对数组进行迭代，reduce的结果就是累加器。

因此，如果我要跟踪这个特定的reduce，它将类似于这样，我简化了这些值的表示：

/*
 * 1st iteration: (e=steven@gmail.com, acc={}) -> acc + {Email: steven@gmail.com}
 * 2nd iteration: (e=test@gmail.com, acc={Email: steven@gmail.com} -> acc + {Email: test@gmail.com}
 * result: acc = {Email: steven@gmail.com, Email: test@gmail.com}
 */

Answer 2

只是跟进我上面的评论。通过T的整个解决方案。

SQL

-- DDL and sample data population, start
DECLARE @SchoolInfo TABLE 
(
    RefId VARCHAR(40) PRIMARY KEY,
    LocalId INT,
    StateProvinceId CHAR(2),
    SchoolName VARCHAR(30)
);
DECLARE @SchoolEmail TABLE 
(
    ID INT PRIMARY KEY, 
    RefId VARCHAR(40), 
    Email VARCHAR(30), 
    EmailType CHAR(3)
);
INSERT @SchoolInfo (RefId, LocalId, StateProvinceId, SchoolName) VALUES
('7FDF722B-6BBA-4BF0-8205-A5380B269EF1', 1, 'CA', 'Steve''s School');
INSERT INTO @SchoolEmail (ID, RefId, Email, EmailType) VALUES
(1, '7FDF722B-6BBA-4BF0-8205-A5380B269EF1', 'steven@gmail.com', 'prm')
,(2, '7FDF722B-6BBA-4BF0-8205-A5380B269EF1', 'test@gmail.com ', 'sec');
-- DDL and sample data population, end

DECLARE @ref_id VARCHAR(40) = '7FDF722B-6BBA-4BF0-8205-A5380B269EF1';

;WITH xmlnamespaces ('http://www.sifassociation.org/datamodel/au/3.4' AS ns0)
SELECT s.RefId AS [@RefId]
    , s.LocalId AS [ns0:LocalId]
    , s.StateProvinceId AS [ns0:StateProvinceId]
    , s.SchoolName AS [ns0:SchoolName]
, (
    SELECT e.EmailType AS [ns0:Email/@Type]
        , e.Email AS [ns0:Email]
    FROM @SchoolEmail AS e
    WHERE e.RefId = s.RefId
    FOR XML PATH(''), TYPE, ROOT('ns0:SchoolEmailList')
)
FROM @SchoolInfo AS s
WHERE s.RefId = @ref_id
FOR XML PATH('ns0:SchoolInfo'), TYPE;

输出

<ns0:SchoolInfo xmlns:ns0="http://www.sifassociation.org/datamodel/au/3.4" RefId="7FDF722B-6BBA-4BF0-8205-A5380B269EF1">
  <ns0:LocalId>1</ns0:LocalId>
  <ns0:StateProvinceId>CA</ns0:StateProvinceId>
  <ns0:SchoolName>Steve's School</ns0:SchoolName>
  <ns0:SchoolEmailList xmlns:ns0="http://www.sifassociation.org/datamodel/au/3.4">
    <ns0:Email Type="prm">steven@gmail.com</ns0:Email>
    <ns0:Email Type="sec">test@gmail.com </ns0:Email>
  </ns0:SchoolEmailList>
</ns0:SchoolInfo>