我有以下 XML 消息:
DECLARE @XML AS XML
SET @XML =
'<Message>
<Changes>
<Deleted>
<ROW id="1" name="Nicole" surname="Bartlett" city="denver" balance="779.4663" dateOfBirth="1991-12-11T14:05:42.830" maritalStatus="S" />
<ROW id="1" name="Nicole" surname="Bartlett" city="boston" balance="779.4663" dateOfBirth="1991-12-11T14:05:42.830" maritalStatus="S" />
</Deleted>
<Inserted>
<ROW id="1" name="Nicole" surname="Bartlett" city="denver" balance="779.4663" dateOfBirth="1991-12-11T14:05:42.830" maritalStatus="S" />
<ROW id="1" name="Nicole" surname="Bartlett" city="boston" balance="779.4663" dateOfBirth="1991-12-11T14:05:42.830" maritalStatus="S" />
</Inserted>
</Changes>
</Message>'
我需要从此消息中选择数据并在 id 字段上加入另一个表。以下代码不起作用:
SELECT T.c.value('./@id', 'int') as id, t.c.value('./@name', 'varchar(max)') as name
FROM @XML.nodes('/Message/Changes/Deleted/ROW') T(c)
inner join other_table tbl
on tbl.id = id
我怎样才能做到这一点?