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ActionScript - Count matching children in XML

Posted on 2010-11-30
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Last Modified: 2012-05-10
I have the XML below come back from a server and put it into _historicalAlarms which is an XML object.

I am wanting to build a pie chart that shows % red alerts vs yellow alerts

I build out my Pie chart XML by calling this function for each distinct Alarm name:

                  private function getNameCount(name:String):XML {
                        var nameCount:int = _historicalAlarms.Alarm.(Name == name).length();
                        trace("Name count for " + name + " = "  + nameCount);
                        return <NameCount name={name} count={nameCount}/>
                  }

nameCount is always 0.  Can anyone help me diagnose why this is not returning a true count.  In the example XML it would send the name "Red Alert" and should get 3 back for a count.
<Results>
  <Alarm id="2266">
    <Name>Red Alert</Name>
  </Alarm>
  <Alarm id="2286">
    <Name>Red Alert</Name>
  </Alarm>
  <Alarm id="2306">
    <Name>Yellow Alert</Name>
  </Alarm>
  <Alarm id="2486">
    <Name>Red Alert</Name>
  </Alarm>
</Results>

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Question by:trippy1976
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6 Comments
 
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Accepted Solution

by:
Gvoid earned 250 total points
Comment Utility
You can't do it like that, it's a good try.

whenever you call
var nameCount:int = _historicalAlarms.Alarm.(Name == name).length();

the code will not know which 'Alarm' node to use, also there is only ever one Name node embeded, so you will only ever get back 0 from something like this.

you need to itterate through the list and count the occurances of each item you are interested in.

e.g.
count 

private function getNameCount():void {
 var redCount : int = 0;
 var yellowCount : int = 0;
 for (var i in _historicalAlarms.Alarm) {
   if (_historicalAlarms.Alarm[i].Name == "Red Alert"){
     redCount++;
   } else {
     yellowCount++;
   }
  }
  trace ("redCount: " + redCount);
  trace ("yellowCount: " + yellowCount);
}

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by:trippy1976
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The example was probably too convenient, but I won't really know what's coming in that list.  I can see how to adapt what you have, but I'd rather do it with some more "elegant" finder if that's feasible.

If I do just

_historicalAlarms.Alarm.length()

I actually get the right count of total alarms.
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Expert Comment

by:Gvoid
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oh right, sorry I miss understood try this

var names : XMLList = historicalAlarms.descendants("Name");
trace(names.length())
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Assisted Solution

by:tomaugerdotcom
tomaugerdotcom earned 250 total points
Comment Utility
Use an associative array to count the occurrences of like Names
var alertsXML:XML = 

	<Results>

	  <Alarm id="2266">

		<Name>Red Alert</Name>

	  </Alarm>

	  <Alarm id="2286">

		<Name>Red Alert</Name>

	  </Alarm>

	  <Alarm id="2306">

		<Name>Yellow Alert</Name>

	  </Alarm>

	  <Alarm id="2486">

		<Name>Red Alert</Name>

	  </Alarm>

	</Results>;

	



// create an associative array to hold the various alarm types

var alarmsCount:Object = new Object();



// iterate through the XML

for each (var alarm:XML in alertsXML.Alarm){

	// increment the corresponding associative array key

	alarmsCount[alarm.Name] = alarmsCount[alarm.Name] ? alarmsCount[alarm.Name] + 1 : 1;

}



// now, list em out

for (var alarmType:String in alarmsCount){

	trace(alarmType + ":" + alarmsCount[alarmType]);

}

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Author Closing Comment

by:trippy1976
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I believe either way will work.  I hoped there was a more elegant way to do it similar to xpath, but based on two replies I and failure to find it myself I'm guessing that this is the way.  You have to parse through the XML one way or another and keep track.

Thanks guys.

FWIW - here is what I did...

- Put the XML into a string variable
- Split the string on the name I was looking for with the element text around it
- Took the count as array.length() - 1

This is also brute, but it works.
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Expert Comment

by:tomaugerdotcom
Comment Utility
The challenge here was that you didn't want to hard-code the names. We could have done something that looked more elegant if we knew the Names beforehand (Alarm.(Name=="Red Alert").length()); but that didn't meet your requirements. Not sure how XPath would have been any more use in this scenario.

-- T
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