How to select an attribute by a variable in xquery?

Question

I know how to select an attribute like so:

$table/@id

However how do I do this if the attribute name is stored as a variable. For example:

let $x = "id"
$table/@[$x]

score 3 · Accepted Answer · answered Dec 01 '15 at 17:06

3

You can use the functions local-name or node-name to capture the value of the attribute and match it the predicate. local-name will simply return a string that matches the element name, and node-name will return a fully qualified name, which is generally recommended, but practically speaking, is only necessary if you are dealing with namespaces.

let $x = "id"
return $table/@*[local-name(.) = $x]

let $x := xs:QName("id")
return $table/@*[node-name(.) = $x]

answered Dec 01 '15 at 17:06

wst

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Great! How could this be expanded to deal with variable elements? for example `$table/$x` – Yahya Uddin Dec 01 '15 at 17:12
For example: `let $x := "foo" return $table/$x` – Yahya Uddin Dec 01 '15 at 17:18
1

Made it to a separate question here: http://stackoverflow.com/questions/34026352/how-to-select-an-element-by-a-variable-in-xquery – Yahya Uddin Dec 01 '15 at 17:27

How to select an attribute by a variable in xquery?

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