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我一直在使用PATH模式处理T-SQL FOR XML以基于group by字段创建Hierarchy。 以下是我的查询和输出。请帮助我提供宝贵的建议。谢谢。美好的一天!!!T SQL对于XML PATH组通过作为属性或元素
select e.department_id AS [@DepartmentID],
d.DEPARTMENT_NAME AS [@DepartmentName],
e.EMPLOYEE_ID AS [EmployeeInfo/EmployeeID],
e.FIRST_NAME AS [EmployeeInfo/FirstName],
e.LAST_NAME AS [EmployeeInfo/LastName]
from employees e
JOIN departments d
ON e.department_id = d.department_id
GROUP BY e.department_id,d.DEPARTMENT_NAME,
e.EMPLOYEE_ID,e.FIRST_NAME,e.LAST_NAME
FOR XML PATH ('Department'), ROOT ('Departments')
输出:
<Departments>
<Department DepartmentID="10">
<EmployeeInfo>
<EmployeeID>111</EmployeeID>
<FirstName>John</FirstName>
<LastName>Chen</LastName>
</EmployeeInfo>
</Department>
<Department DepartmentID="10">
<EmployeeInfo>
<EmployeeID>201</EmployeeID>
<FirstName>steven</FirstName>
<LastName>Whalen</LastName>
</EmployeeInfo>
</Department>
<Department DepartmentID="30">
<EmployeeInfo>
<EmployeeID>105</EmployeeID>
<FirstName>ANIRUDH</FirstName>
<LastName>RAMESH</LastName>
</EmployeeInfo>
</Department>
<Department DepartmentID="30">
<EmployeeInfo>
<EmployeeID>115</EmployeeID>
<FirstName>Den</FirstName>
<LastName>Raphaely</LastName>
</EmployeeInfo>
</Department>
<Departments>
所需的输出是:
<Departments>
<Department DepartmentID="10">
<EmployeeInfo>
<EmployeeID>111</EmployeeID>
<FirstName>John</FirstName>
<LastName>Chen</LastName>
</EmployeeInfo>
<EmployeeInfo>
<EmployeeID>201</EmployeeID>
<FirstName>steven</FirstName>
<LastName>Whalen</LastName>
</EmployeeInfo>
</Department>
<Department DepartmentID="30">
<EmployeeInfo>
<EmployeeID>105</EmployeeID>
<FirstName>ANIRUDH</FirstName>
<LastName>RAMESH</LastName>
</EmployeeInfo>
<EmployeeInfo>
<EmployeeID>115</EmployeeID>
<FirstName>Den</FirstName>
<LastName>Raphaely</LastName>
</EmployeeInfo>
</Department>
<Departments>
谢谢你:D。现在我能够获得所需的输出。 –
然后将其标记为接受答案:) .... – TriV