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Rebuild XML data

Posted on 2004-10-10
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Last Modified: 2010-04-03
Hi,

I am reading a XML file into Flash.

Creating Arrays from the Data.

Then, I am modifying these arrays.

And then, I want to recreate the XML again which will be sent to a server side script to update the origional XML on the server.

Can anyone suggest how to i do this? I dont know how to recreate the XML after the Arrays have been modified.
Any suggestion is welcome.

thanks.
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Question by:ssdesign
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6 Comments
 
LVL 27

Accepted Solution

by:
negatyve earned 300 total points
ID: 12270203
ok, follow me:

• create a new xml file, named "test.xml"
• write inside it:

<?xml version="1.0" encoding="UTF-8" ?>
<data>
      <item>10</item>
      <item>45</item>
      <item>12</item>
      <item>34</item>
</data>

• create a flash movie in the same folder
• test this code:

data_xml = new XML();
data_xml.ignoreWhite = true;
data_xml.onLoad = function(success)
{
      if(success){
            // edit xml file
            editXML(this);
      } else {
            // xml file missing
            trace("XML FILE NOT FOUND");
      }
};
function editXML(xml)
{
      // show source xml file in the output windo
      trace(xml);
      // create an empty array
      var temp_array = new Array();
      // select the items node sequence in the xml file
      var nodes = xml.firstChild.childNodes;
      for(var i = 0; i < nodes.length; i++){
            // push inside the array the numeric value of each item
            temp_array.push(Number(nodes[i].firstChild.nodeValue));
      }
      // show in the output window the filled array
      trace(temp_array);
      // add a value to each element (modify)
      for(var i = 0; i < temp_array.length; i++){
            temp_array[i] = temp_array[i] * 2;
      }
      // show in the output window the modified array
      trace(temp_array);
      // create the new xml
      var new_xml = new XML('<?xml version="1.0" encoding="UTF-8" ?><data></data>');
      // add nodes
      for(var i = 0; i < temp_array.length; i++){
            // create the node
            var new_node = new_xml.createElement("item");
            // append inside it the corresponding array element as text
            new_node.appendChild(new_xml.createTextNode(temp_array[i]));
            // insert it in the new xml
            new_xml.firstChild.appendChild(new_node)
      }
      // trace the new xml
      trace(new_xml)
}
data_xml.load("test.xml");
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LVL 7

Author Comment

by:ssdesign
ID: 12284401
ok thats looks quiet ok, i will need to try it.

I will get back soon with my feedback wether it worked or not.

Just need to try it with my application.
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LVL 27

Expert Comment

by:negatyve
ID: 12295537
That's ok, take your time.. :)
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LVL 7

Author Comment

by:ssdesign
ID: 12320910
I have not forgotten, give me just one more day. Ok?
0
 
LVL 27

Expert Comment

by:negatyve
ID: 12346439
>I have not forgotten, give me just one more day. Ok?

hehe, that's right, no worries...
0
 

Expert Comment

by:subhailc
ID: 12472116
i know this is closed and all but that was a great answer.  to the original poster: feel lucky :D  information like that can be (is) expensive and rare.
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