I have a node like this:
<meta name="og:description" content="Here's the content" />
I want to be able to select this element if the name is "description" whether it's in a namespace or not. I need to be able to select the meta tag if it's name is "og:description", "description", "blah:description", etc.
I've seen resources for xpath that show how to select within a namespace, but not irrespective of a namespace.
Use:
//meta[@*[local-name() = 'description']]
This selects all meta elements in the XML document that have an attribute with local-name "description".
By definition, the standard XPath function local-name() produces the name of the node from which the namespace prefix (if any) is stripped off.
If, on the other side you need to select a <meta> element the string value of whose name attribute is either "description" or any string that ends with ":description", use this XPath 1.0 expression:
//meta
[@name = 'description'
or substring(@name, string-length(@name) - 11) = ':description']
Do note: Always avoid using the // pseudo operator if the structure of the XML document is statically known. Often using // causes slow execution.
Using XPath 2 you could do:
/meta[ends-with(@name, 'description')]
For XPath 1 we need:
/meta['description' = substring(@name, string-length(@name) - string-length('description') + 1)]
meta. Also the meta top element will be selected even when the attribute name is Mydescription, yourdescription, ..., whateverdescription, Just noting these facts for the readers. Not downvoting!
og:name="description"is namespace, butname="og:description"is just namespace-looking-like content, isn't it?[ends-with(@name, 'description')]), apparently OP wanted to filter on "namespace-looking content". Based on the highest-voted answer with[local-name()="description"], you're right, other people want to filter on namespaced attribute name itself!