groovy - problem parsing xml

Question

I am new to Groovy and I am trying to parse both a valid rest resource and an invalid one. For example:

this code works fine -

def entity = new XmlSlurper().parse('http://api.twitter.com/1/users/show/slashdot.xml')
println entity.name()
println entity.screen_name.text()

when I run it, I get output:

user
slashdot

but when I pass an invalid url to xmlSlurper, like this

def entity = new XmlSlurper().parse('http://api.twitter.com/1/users/show/slashdotabc.xml')
println entity.name()
println entity.screen_name.text(

)

I get this error message:

Caught: java.io.FileNotFoundException: http://api.twitter.com/1/users/show/slashdotabc.xml
    at xmltest.run(xmltest.groovy:1)

Although the url returns an hash code (like below) with an error message which I would like to parse and display it.

<hash>
<request>/1/users/show/slashdotabc.xml</request>
<error>Not found</error>
</hash>

How can I parse a url which returns a 404 but with error information?

Any help will be appreciated.

-- Thanks & Regards, Frank Covert

score 4 · Accepted Answer · answered Apr 13 '11 at 17:17

The response you want to see is available on the URL connection's errorStream instead of the inputStream. Fortunately, given the InputStream, XmlSlurper.parse can read an InputStream as well.

Here's a sample to switch to reading the errorStream when you don't get a 200 status:

def con = new URL("http://api.twitter.com/1/users/show/slashdotaxxxbc.xml").openConnection()
def xml = new XmlSlurper().parse(con.responseCode == 200 ? con.inputStream : con.errorStream)

thanks, that works. Do I need to close the con in the end? like con.Close() or something like that? — frank.C, Apr 13 '11 at 19:21

groovy - problem parsing xml

when I run it, I get output:

but when I pass an invalid url to xmlSlurper, like this

)

1 Answers1