XmlSlurper.parse(uri) with HTTP basic authentication

Question

I need to grab a data from XML-RPC web-service.

new XmlSlurper().parse("http://host/service") works fine, but now I have a particular service that requires basic HTTP authentication.

How can I set username and password for parse() method, or modify HTTP headers of the request?

Using http://username:password@host/service doesn't help - I still get java.io.IOException: Server returned HTTP response code: 401 for URL exception.

Thanks

Hylan

Answer 1

I found this code over here which might help?

Editing this code to your situation, we get:

def addr       = "http://host/service"
def authString = "username:password".getBytes().encodeBase64().toString()

def conn = addr.toURL().openConnection()
conn.setRequestProperty( "Authorization", "Basic ${authString}" )
if( conn.responseCode == 200 ) {
  def feed = new XmlSlurper().parseText( conn.content.text )

  // Work with the xml document

} else {
  println "Something bad happened."
  println "${conn.responseCode}: ${conn.responseMessage}" 
}

Clamworm

Answer 2

This will work for you

Please remember to use this instead of the 'def authString' mentioned above:

def authString  = "${usr}:${pwd}".getBytes().encodeBase64().toString()

Torruella

I defined my parameters as usr and pwd. Cheers! – Torruella