我建议使用 XQuery 接口，而不是相当笨拙的老式 OPENXML 方法:

I would recommend using the XQuery interface, rather than the rather clunky old OPENXML approach:

SELECT
    Ppl.Person.value('(@id)[1]', 'int') AS 'ID',
    Ppl.Person.value('(firstname)[1]', 'varchar(20)') AS 'First Name',
    Ppl.Person.value('(surname)[1]', 'varchar(20)') AS 'Last Name'
FROM
    @input.nodes('/people/person') as Ppl(Person)

这是在 SQL Server 2005 及以后版本中执行此操作的首选方式.

This is the preferred way of doing it in SQL Server 2005 and forward.

输出是一样的:

ID  First Name  Last Name
1    John             Doe
2    Mary             Jane

但是，如果您不了解 XML 结构，您就无法真正做到这一点......

You can't really do this however, if you don't know the XML structure.....