CODEWITHSUNDEEP
javascriptdomscala.js
bbarker
bbarker

Reputation: 13078

How to serialize XML without xmlns attributes?

My code is in Scala.js, but I think the gist of it should be easy to understand from a JavaScript perspective:

  def htmlToXHTML(input: String)
  (implicit parser: DOMParser, serializer: XMLSerializer): String = {
    val doc = parser.parseFromString(input, "text/html")
    val body = getElementByXpath("/html/body", doc).singleNodeValue
    val bodyXmlString = serializer.serializeToString(body)
    val xmldoc = parser.parseFromString(bodyXmlString, "application/xml")
    val xmlDocElems: NodeList = xmldoc.getElementsByTagName("*")
    xmlDocElems.foreach{
      case elem: Element =>
        elem.removeAttribute("xmlns")
        println(s"Found element $elem with html: ${elem.outerHTML}")
      case node => println(s"Warning: found unexpected non-element node: $node.")
    }
    xmldoc.firstElementChild.innerHTML
  }

This is used above, so including it for completeness (https://stackoverflow.com/a/14284815/3096687):

  def getElementByXpath(xpath: String, doc: Document): XPathResult =
    doc.evaluate(
      xpath, doc, null.asInstanceOf[XPathNSResolver],
      XPathResult.FIRST_ORDERED_NODE_TYPE, null
    )

In short, this function reads an HTML string, converts it to an HTML document, serializes to XML, reparses as XML, and finds all the elements in the doc and loops over them (foreach), and then removes the xmlns attribute. It seems that the resulting innerHTML, however, still has the xmlns attributes on elements, even though the first println (aka console.log) indicates we are finding the elements in question, but not removing the xmlns attributes.

The problem may derive from default values specified in a DTD:

If a default value for the attribute is defined in a DTD, a new attribute immediately appears with the default value

Upvotes: 2

Views: 2087

Answers (3)

Ben Bucksch
Ben Bucksch

Reputation: 660

If you only want to remove the xmlns="" from <html>, you can use this specific regexp:

xmls.serializeToString(domNode)
  .replace(/^<html xmlns="[^"]+">/, "<html>");

Upvotes: 0

Andre Elrico
Andre Elrico

Reputation: 11480

As mentioned, the easiest is to remove it from the result string:

xmls.serializeToString(domNode).replace(/xmlns="[^"]+"/, '')

Upvotes: 5

Nickolay
Nickolay

Reputation: 32063

I probably would cheat and remove the xmlns from the resulting string, as it's a huge pain to make the elements lose their namespace.

If you insist on doing that, you could try building a document from scratch while walking the original DOM -- pedantically copying everything but the namespaces (i.e. using createElementNS with an empty namespace?)

Upvotes: 3

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