title: How to find out the content of a XML file using Unix Sed/Awk?
body: <p>I have a XML file(MyXML.xml) like this :</p>
  
  <pre><code>&lt;?xml version="1.0" encoding="UTF-8"?&gt;
  &lt;S:Envelope xmlns:S="http://schemas.xmlsoap.org/soap/envelope/"&gt;
      &lt;S:Body&gt;
          &lt;ns3:GetAllInfoFromRest xmlns:ns2="http://com.lanuk.cfe/b2_7/service/objects" xmlns:ns3="http://com.lanuk.cfe/b2_7/service/operations"&gt;
              1111,GH43567,Hamburger,GET,278598655,\n000001,                    ,Kunal,Bhyuo,Ramond,856 K. 98 Rd,                                        ,                                        ,Tripura,AGT,INDIA,856987,                    ,S,S,S,8956,\666666
          &lt;/ns3:GetAllInfoFromRest&gt;
      &lt;/S:Body&gt;
  &lt;/S:Envelope&gt;
  </code></pre>
  
  <p>Now i need to strip out the SOAP content and all the tag attributes from this xml and get only the string response <code>1111,GH43567,Hamburger,GET,278598655,\n000001,                    ,Kunal,Bhyuo,Ramond,856 K. 98 Rd,                                        ,                                        ,Tripura,AGT,INDIA,856987,                    ,S,S,S,8956,\666666</code>.</p>
  
  <p>How can i do it with <code>awk</code> or  <code>sed</code> ?</p>
  
  <p>I tried it in this  way :</p>
  
  <pre><code>$ xgawk -lxml 'XMLATTR["xmlns:ns3"]=="http://com.lanuk.cfe/b2_7/service/operations"{print $2}' MyXML.xml
  </code></pre>
  
  <p>But obviously I am making some mistake  due to which it is not working. 
  Can some one suggest any other way around this ?</p>