测试环境:sql 08, 08 R2, 2010, 2012, 2014 等
declare @agent table ( AgentID int, Fname varchar(5), SSN varchar(11) ) insert into @agent select 1, ‘Vimal‘, ‘123-23-4521‘ union all select 2, ‘Jacob‘, ‘321-52-4562‘ union all select 3, ‘Tom‘, ‘252-52-4563‘ declare @address table ( AddressID int, AddressType varchar(12), Address1 varchar(20), Address2 varchar(20), City varchar(25), AgentID int ) insert into @address select 1, ‘Home‘, ‘abc‘, ‘xyz road‘, ‘RJ‘, 1 union all select 2, ‘Office‘, ‘temp‘, ‘ppp road‘, ‘RJ‘, 1 union all select 3, ‘Home‘, ‘xxx‘, ‘aaa road‘, ‘NY‘, 2 union all select 4, ‘Office‘, ‘ccc‘, ‘oli Com‘, ‘CL‘, 2 union all select 5, ‘Temp‘, ‘eee‘, ‘olkiu road‘, ‘CL‘, 2 union all select 6, ‘Home‘, ‘ttt‘, ‘loik road‘, ‘NY‘, 3 --SELECT -- 1 AS Tag, -- NULL AS Parent, -- 0 AS ‘Agents!1!Sort!hide‘, -- NULL AS ‘Agents!1!‘, -- NULL AS ‘Agent!2!AgentID‘, -- NULL AS ‘Agent!2!Fname!Element‘, -- NULL AS ‘Agent!2!SSN!Element‘, -- NULL AS ‘AddressCollection!3!Element‘, -- NULL AS ‘Address!4!!xml‘, -- NULL AS ‘Address!4!AddressType!Element‘, -- NULL AS ‘Address!4!Address1!Element‘, -- NULL AS ‘Address!4!Address2!Element‘, -- NULL AS ‘Address!4!City!Element‘ --UNION ALL -- SELECT -- 2 AS Tag, -- 1 AS Parent, -- AgentID * 100, -- NULL, AgentID, Fname, SSN, -- NULL, NULL, NULL, NULL, NULL, NULL --FROM @Agent --UNION ALL -- SELECT -- 3 AS Tag, -- 2 AS Parent, -- AgentID * 100 + 1, -- NULL,NULL,NULL, -- NULL, -- NULL, -- NULL, NULL, NULL, NULL, NULL --FROM @Agent --UNION ALL --SELECT -- 4 AS Tag, -- 3 AS Parent, -- AgentID * 100 + 2, -- NULL, NULL, NULL, NULL, NULL, -- ‘<!-- ‘ + AddressType + ‘ Address -->‘, AddressType, -- Address1, Address2, City --FROM @Address --ORDER BY [Agents!1!Sort!hide] --FOR XML EXPLICIT SELECT 1 AS Tag, NULL AS Parent, NULL AS [Agents!1!], NULL AS [Agent!2!AgentID], NULL AS [Agent!2!Fname!Element], NULL AS [Agent!2!SSN!Element], NULL AS [AddressCollection!3!Element], NULL AS [Address!4!AddressType!Element], NULL AS [Address!4!Address1!Element], NULL AS [Address!4!Address2!Element], NULL AS [Address!4!City!Element] UNION ALL SELECT 2 AS Tag, 1 AS Parent, NULL, AgentID, Fname, SSN, NULL,NULL, NULL, NULL, NULL FROM @Agent UNION ALL SELECT 3 AS Tag, 2 AS Parent, NULL,AgentID,NULL, NULL, NULL, NULL, NULL, NULL, NULL FROM @Agent UNION ALL SELECT 4 AS Tag, 3 AS Parent, NULL,AgentID,NULL,NULL,NULL, AddressType, Address1, Address2, City FROM @Address ORDER BY -- all properties of every agent -- (from tag 2 and 4: SSN, fname and adresses) -- will be sorted by agentID and combined into -- separate groups. It is necessary in same cases. [Agent!2!AgentID], [AddressCollection!3!Element], -- optional because NULL everywhere [Address!4!AddressType!Element]-- any ordering by elements of tag 4 FOR XML EXPLICIT
运行结果:
文章来源:http://social.msdn.microsoft.com/Forums/sqlserver/zh-CN/97f79941-324e-479e-ba5b-851cc534ebe5/problem-in-for-xml-explicit-query?forum=sqlxml
sql for xml 另一种写法(采用 tag 与 union all,简洁易懂)
时间: 2024-12-19 10:48:40