Question

努力编写可以在XML中生成嵌套元素的查询。任何人都可以帮忙

INPUT：

SELECT EmpName AS [Name],
       EmpSalary        AS [WageAmount],
    Add1    AS [Address1],
    Add2    AS [Address2],
    Mobile  AS [Mobile]
FROM   Employee
FOR XML PATH

TABLE DATA（对于具有不同地址的emoployee，存在2条记录）

Conrad Jagger   12345    London   UK    9191919191
Conrad Jagger   12345    Coventry UK    9191919191

预期的输出:(这有附加信息作为嵌套元素）

    <Employee>
    <Name>Conrad</Name>
    <WageAmount>10000</WageAmount>
    <AdditionalInfo>
         <PersonAddress>
                  <Address1>London</Address1>
                  <Address2>UK</Address2>
                 <Mobile>919191919191</Mobile>
        </PersonAddress>
        <PersonAddress>
                  <Address1>Coventry</Address1>  --->Second address
                  <Address2>UK</Address2>
                <Mobile>919191919191</Mobile>
       </PersonAddress>
    </AdditionalInfo>
    </Employee>

如果您有单个记录，此查询有效，我不知道如何使这个工作为多个记录，我得到

SELECT EmpName AS [Name],
       EmpSalary AS [WageAmount],
       Add1 AS [AdditionalInfo/Address1],
       Add2 AS [AdditionalInfo/Address2],
       Mobile AS [AdditionalInfo/Mobile]
FROM   Employee
FOR XML PATH ('Employee')

Answer 1

我希望表格中有一个uniqe id列。由于您还没有提供该列，我正在加入EmpName列。

试试这个

SELECT EmpName AS [Name],
       EmpSalary AS [WageAmount],
       (SELECT 
       Add1 AS [PersonAddress/Address1],
       Add2 AS [PersonAddress/Address2],
       Mobile AS [PersonAddress/Mobile] FROM Employee B WHERE EmpName = A.EmpName
       FOR XML PATH('AdditionalInfo'),type)
FROM   Employee A
GROUP BY EmpName,EmpSalary
FOR XML PATH ('Employee')

SQL Server - 用于XML路径（嵌套元素，多个记录）

1 个答案: