How to do group capture in xpath - xpath

Question

I am looking for something in xpath can do: //foo[#n="$1"]//bar[#n="$2"]
which can return me $1 and $2. Or at least return me both <foo> and <bar>
Here is more detail, I have a xml document:
<xml>
<foo>
<a n="1">
<b n="1"/>
<b n="2"/>
</a>
</foo>
<a n="2">
<b n="1"/>
</a>
<a n="3">
<b n="1"/>
<foo>
<b n="2"/>
</foo>
<b n="3"/>
</a>
</xml>
I want generate a string base on the n attribute in <a> and <b>
So I have xpath: //a[#n]//b[#n]
Then for every result I get back, I use: ./#n and ./ancestor::a/#n to get the info I want.
This is working fine, but I need something more intelligent, because I have a lot structure like this, and need auto generate the xpath.
So for above example, I am looking for some xpath like: //a[#n="$1"]//b[#n="$2"]
Then return me:
`(1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (3, 3)

Answer 1

Here is one XPath 1.0 expression that selects all wanted n attributes:
//a[.//b]/#n | //a//b/#n
Without optimization the evaluation of the above expression performas at least two complete traversals of the XML document.
This XPath 1.0 expression may be more efficient:
//*[self::a and .//b or self::b and ancestor::a]/#n
Both expressions can be simplified if it is guaranteed that every a has a b descendant.
They become, respectively:
//a/#n | //a//b/#n
And:
//*[self::a or self::b and ancestor::a]/#n
Further simplification is possible if it is guaranteed that every a has a descendant b and every b has an ancestor a.:
//*[self::a or self::b]/#n
It is impossible in a single XPath 1.0 expression to get the string values of all the wanted attributes. One needs to get all the atributes using one of the above expressions, then on each of the selected attributes apply a second XPath expression: string().
In Xpath 2.0 it is possible to get with a single expression all string values of the wanted attributes -- by simply appending each of the expressions with /string(.)
For example, for the simplest one:
//(a|b)/#n/string(.)
Update:
The OP has clarified his question. Now we know that he wants this result to be produced:
(1, 1), (1, 2), (2, 1), (3, 1), (3, 2), (3, 3)
It isn't possible to produce the wanted result with a single XPath 1.0 expression.
The following XPath 2.0 expression produces the wanted result:
for $a in //a[#n and .//b[#n]],
$b in $a//b[#n]
return
concat('(', $a/#n, ',', $b/#n, ') ')