Find all namespace declarations in an XML document - xPath 1.0 vs xPath 2.0

Question

As part of a Java 6 application, I want to find all namespace declarations in an XML document, including any duplicates. Edit: Per Martin's request, here's the Java code I am using: ``` XPathFactory xPathFactory = XPathFactory.newInstance(); XPath xPath = xPathFactory.newXPath(); XPathExpression xPathExpression = xPathExpression = xPath.compile("//namespace::*"); NodeList nodeList = (NodeList) xPathExpression.evaluate(xmlDomDocument, XPathConstants.NODESET); ``` Suppose I have this XML document: ``` a e txt:f ``` To find all namespace declarations, I applied this xPath statement to the XML document using xPath 1.0: ``` //namespace::* ``` It finds 4 namespace declarations, which is what I expect (and desire): ``` /root[1]/@xmlns:att - attribute.com /root[1]/@xmlns:ele - element.com /root[1]/@xmlns:txt - t…

Accepted Answer

I think this will get all namespaces, without any duplicates:

```
for $i in 1 to count(//namespace::*) return 
if (empty(index-of((//namespace::*)[position() = (1 to ($i - 1))][name() = name((//namespace::*)[$i])], (//namespace::*)[$i]))) 
then (//namespace::*)[$i] 
else ()
```

Find all namespace declarations in an XML document - xPath 1.0 vs xPath 2.0

Solution

Problem