您可以通过计算每个节点之前的兄弟节点的数量来模拟 position() 函数:

You can emulate the position() function by counting the number of sibling nodes preceding each node:

SELECT
    code = value.value('@code', 'int'),
    parent_code = value.value('../@code', 'int'),
    ord = value.value('for $i in . return count(../*[. << $i]) + 1', 'int')
FROM @Xml.nodes('//value') AS T(value)

这是结果集:

code   parent_code  ord
----   -----------  ---
1      NULL         1
11     1            1
111    11           1
12     1            2
121    12           1
1211   121          1
1212   121          2

工作原理:

• for $i in . 子句定义了一个名为 $i 的变量，该变量包含当前节点 (.).这基本上是解决 XQuery 缺少类似 XSLT 的 current() 函数的一个技巧.
• ../* 表达式选择当前节点的所有兄弟节点(父节点的子节点).
• [.<<$i] 谓词将兄弟列表过滤到前面的列表 () 当前节点($i).
• 我们count() 前面兄弟的数量，然后加1得到位置.这样，第一个节点(没有前面的兄弟节点)被分配了 1 的位置.

• The for $i in . clause defines a variable named $i that contains the current node (.). This is basically a hack to work around XQuery's lack of an XSLT-like current() function.
• The ../* expression selects all siblings (children of the parent) of the current node.
• The [. << $i] predicate filters the list of siblings to those that precede (<<) the current node ($i).
• We count() the number of preceding siblings and then add 1 to get the position. That way the first node (which has no preceding siblings) is assigned a position of 1.