Parse SOAP XML Response

Code to parse a SOAP XML response.

This is the SOAP response parsed by the code below:

<?xml version="1.0"?>
<soap:Envelope
xmlns:soap="http://www.w3.org/2001/12/soap-envelope"
soap:encodingStyle="http://www.w3.org/2001/12/soap-encoding">

<soap:Body xmlns:m="http://www.example.org/stock">
  <m:GetStockPriceResponse>
    <m:Price>34.5</m:Price>
  </m:GetStockPriceResponse>
</soap:Body>

Downloads for Windows/Linux and Install Instructions

require 'rubygems'
require 'chilkat'



xml = Chilkat::CkXml.new()

#  Load the XML object with the SOAP response.
#  The XML may be loaded from a file by calling
#  LoadXmlFile, or directly from a string by calling
#  LoadXml.
success = xml.LoadXmlFile("soapResponse.xml")
if (success != true)
    print xml.lastErrorText() + "\n";
    exit
end

#  The quickest way to get a piece of data is w/ the
#  ChilkatPath method.
#  To get the content of the m:Price node:

strPrice = xml.chilkatPath("soap:Body|m:GetStockPriceResponse|m:Price|*")
print "Price = " + strPrice + "\n";

#  ------------------------------------
#  Another way of doing it:

#  Another way to get it is to navigate to the node.
#  Navigate to soap:Body
xml.FirstChild2()

#  Navigate to m:GetStockPriceResponse
#  You may call FirstChild2, or you may call GetChildWithTag2
#  to get the Nth direct child having a given tag.
#  In this case, get the 1st direct child having the tag
#  "m:GetStockPriceResponse"
#  (Indexing begins at 0, so the 1st occurance is at index 0.)
bFound = xml.GetNthChildWithTag2("m:GetStockPriceResponse",0)
if (bFound != true)
    print "Did not find m:GetStockPriceResponse node." + "\n";
    exit
end

#  One more to go..
bFound = xml.GetNthChildWithTag2("m:Price",0)
if (bFound != true)
    print "Did not find m:Price node." + "\n";
    exit
end

#  The content of the "m:Price" node is "34.5"
strPrice = xml.content()
print "Price = " + strPrice + "\n";

#  Restore the xml object's internal reference to the root
#  of the XML document:
xml.GetRoot2()

#  ------------------------------------
#  Another way of doing it:

#  If you know that the node w/ the content you want to
#  extract is uniquely tagged (i.e. no other node has the same
#  tag), then search for the node having the tag:
bFound = xml.SearchForTag2(xml,"m:Price")
if (bFound != true)
    #  Not found...
    exit
end

#  The content of the "m:Price" node is "34.5"
strPrice = xml.content()
print "Price = " + strPrice + "\n";

#  Restore the xml object's internal reference to the root
#  of the XML document:
xml.GetRoot2()

#  ------------------------------------
#  Another way of doing it:

#  Again, assuming there is only a single node with the
#  tag to be fetched:
strPrice = xml.accumulateTagContent("m:Price","")
print "Price = " + strPrice + "\n";

#  There are many other possible ways of doing the same thing...