How can I use xsl stylesheet parameters to set a node name in XPATH expression?

Question

I have the following XPATH expression:

select="catalog/product/$category_name = $category_value"

In the given example $category_name and $category_value are the XSL parameters that I receive from my servlet and I want to use them in XSL to filter the XML result based on category and its value.However, for some reason when,say, $category_name parameter equals 'price' attribute of the 'product' parent node and $category_value equals 40, the given expression does not return any result at all! Logically, the expression should be transformed to something like select="catalog/product/price = 40"....I guess there is some problem with specifying the node name which is the category in my case. Can anyone suggest the way to get around this problem?

score 2 · Accepted Answer · answered May 20 '10 at 21:28

2

You probably want:

catalog/product/*[name()=$category_name] [. = $category_value]

answered May 20 '10 at 21:28

Dimitre Novatchev

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score 0 · Answer 2 · edited May 23 '17 at 12:07

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For variable xpath expressions, use dynamic xpath. See Is it possible to use a Dynamic xPath expression in a xslt style sheet?

edited May 23 '17 at 12:07

Community

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answered May 20 '10 at 13:45

mdma

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How can I use xsl stylesheet parameters to set a node name in XPATH expression?

2 Answers2