How to filter xml nodes by nested children values in ActionScript 3?

Question

I have a very simple scenario where I have a structure similar to this

<tours>
    <tour>
        <name>Italy 1</name>
        <destinations>
            <destination>Rome</destination>
            <destination>Milan</destination>
        <destinations>
    </tour>
    <tour>
        <name>Italy 2</name>
        <destinations>
            <destination>Rome</destination>
            <destination>Venice</destination>
        <destinations>
    </tour>
</tours>

Now I want to query all the tours that go to Milan.

Below is the logical format that I can think of based on EX4 style

XmlData.tour.(destinations.destination.(name == "Milan"))

But of course this doesn't work.

What is the correct way of pulling this data without using any extra logic?

Answer 1

XmlData.tour.destinations.destination is XMLList. It contains XMLs, which looks like <destination>Milan</destination> Each these XML have name = destination, so you get nothing as result.

Try this:

XmlData.tour.(destinations.destination.children().contains("Milan"));

Answer 2

1)Badly formatted xml. (missing '/' on closing destinations)

2)You seem to have run into some kind of bug in AS3 E4X-finder. This was really weird, here's a workaround though...

var foo:XMLList = data.tour.destinations.(destination == "Milan");
trace("direct check:  " + foo); //fails - 0 matches
trace("------");
for each(var child:XML in data.tour.destinations.destination) {
    if (child == "Milan") {
        trace("found match in foreach Milan");
    }
}

Further investigating, it seems like the E4X-engine screws up since you have multiple children inside a tag with the same identifier (<destination>).

typing following makes the "filter function" behave as expected:

<destinations>
<destination2>Rome</destination2>
<destination>Milan</destination>
</destinations>

.... that's really weird... Anyone who can elaborate on this? Because according to the xml-standards of E4X it should be possible to do it as done in the question.