Any chance you can show an example of what your xml file looks like?
I was goign to suggest that you try:
countryName = XPath.selectNodes(country, ./name);
or
countryName = XPath.selectNodes(country, ./country/name);
But I can't be sure without looking at the xml structure.
On Sun, Apr 30, 2006 at 10:39:23AM -0400, Howard Nager wrote:
Any chance you can show an example of what your xml file looks like?
I was goign to suggest that you try:
countryName = XPath.selectNodes(country, ./name);
or
countryName = XPath.selectNodes(country, ./country/name);
But
That's because you're searching from the country-node, that's the
structure you're searching in. Forget about all its parentnodes, we're
talking about a new xml-document here.
var countriesList:Array = XPath.selectNodes(xmlData, /map/country);
for (var i=0; icountriesList; i++) {
var
Bart,
Thanks for the explanation, I understand it a better now.
Cheers,
- Kevin
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That's because you're searching from the country-node, that's the
structure you're searching in. Forget about all its parentnodes, we're
talking about a new xml-document here.
var countriesList:Array = XPath.selectNodes(xmlData, /map/country);
for (var i=0; icountriesList; i++) {
var
OK, I get it.
It's definitively a problem of reference.
In order to filter an already filtered XMLNode, you've got to clone it
before.
XMLNode has such a function, cloneNode.
So your example should work like this :
function parseXML(xmlData) {
countriesList =
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