Roey Zada
Roey Zada

Reputation: 683

NodeJS xml-stream, parsing unknown nodes

I need a powerful package to parse large XML files, and I came across xml-stream. It is great, but for some reason I can't seem to find how to parse unknown nodes, it expects me to send the expected field as a parameter.

That's the XML file

<Offers>
<Offer>
    <Model>
        694547
    </Model>
    <RegularPrice>
        219.00
    </RegularPrice>
    <ParentSKU>
        654987
    </ParentSKU>
</Offer>
<Offer>
    <Model>
        694548
    </Model>
    <RegularPrice>
        235.00
    </RegularPrice>
    <ParentSKU>
        123456
    </ParentSKU>
</Offer>

And this is the code I use

var XmlStream = require('xml-stream'); 
var fs = require('fs');

var stream = fs.createReadStream(filePath);
var xml = new XmlStream(stream);

xml.on('endElement: Offer', function (item) { //<<<- Here I need to add 'Offer' as an expected node
    console.log('loaded', JSON.stringify(item));
});

xml.on('end', function () {
    console.log('END XML');
});

The response is great, as expected

loaded {"Model":"694547","RegularPrice":"219.00","ParentSKU":"654987"}
loaded {"Model":"694548","RegularPrice":"235.00","ParentSKU":"123456"}
END XML

How can I get the same list without me having to set 'Offer' as an expected node? The XML's I'm expecting to get can have different nodes and values.

Upvotes: 1

Views: 2695

Answers (1)

Roey Zada
Roey Zada

Reputation: 683

ok, in the end - I've override the xml-stream lib. (located in '/xml-stream/lib/xml-stream.js')

xml.on('endElement', function(name) {

  //xml-stream code

  if (self._bufferLevel === 0 && self._emitData) {
    emitEnd.call(self, name);
  }

  //my override <<-----
  if (prev.path.split('/').length == 2) { //the split length check give me the first child
    self.emit('endItem', val);
  }
  //my override end

  curr = prev;
  this._collect = curr.collect;
});

to implement

  xml.on('endItem', function (item) {
      //single item 
  });

Upvotes: 2

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