Android - Read an XML file with HTTP GET

Question

I need to explore for my project use of web services on Android. I know that there is no official library for XML - RPC web service.

But there is for REST XML and i need to test it.

I would like to read XML on my web page (Where i have to pass username and Password) from Android with HTTP GET.

OR

Suppose, i follow This link, then where do i pass username and password?

Can anybody help me on this.

How are the username and password required? If it's HTTP Basic Auth then the HttpClient library supports that and you can add to Gianni's code. — Nick H
– Nick H, Commented Aug 4, 2010 at 21:00

Rajat Singhal · Accepted Answer · 2014-03-01 14:41:58Z

12

HttpGet uri = new HttpGet("http://example.com");    

DefaultHttpClient client = new DefaultHttpClient();
HttpResponse resp = client.execute(uri);

StatusLine status = resp.getStatusLine();
if (status.getStatusCode() != 200) {
    Log.d(tag, "HTTP error, invalid server status code: " + resp.getStatusLine());  
}

DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
Document doc = builder.parse(resp.getEntity().getContent());

edited Mar 1, 2014 at 14:41

Rajat Singhal

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answered Aug 3, 2010 at 15:09

Gianni

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10 Comments

Nick H Over a year ago

Don't forget to close up your HTTP request! resp.getEntity().consumeContent() - only need to do this in the event that the XML parsing doesn't fully read the file though, so in a finally block in some form.

Paresh Mayani Over a year ago

just add this line to pass "Username" and "Password" : httpclient.getCredentialsProvider().setCredentials( new AuthScope(null, -1), new UsernamePasswordCredentials("username", "password"));

Łukasz Siwiński Over a year ago

Should be? => if (status.getStatusCode() != 200) {

Pieter Over a year ago

For future reference: that uri object can be constructed as follows: new HttpGet("http://example.com")

Czechnology Over a year ago

StatusLine.getStatusCode() returns int, shouldn't be compared to String.

|

iTom · Accepted Answer · 2010-08-03 09:29:40Z

2

This link helped me to get started understanding how to HTTP GET XML and Parse using the SAX Parser.

http://www.anddev.org/parsing_xml_from_the_net_-_using_the_saxparser-t353.html

Hope this helps,

iTom

answered Aug 3, 2010 at 9:29

iTom

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4 Comments

Paresh Mayani Over a year ago

@iTom...I already read this article....but then how do i pass "Username" and "Password" value.....btw thanx

iTom Over a year ago

With a GET request the form values are usually appended to the URI request instead of posted i.e. example.com/… Hope that helps, iTom

iTom Over a year ago

Oh also if your after a function that builds the URI for a GET request then this article may also help.. anddev.org/doing_http_post_with_android-t492.html

iTom Over a year ago

Hey Paresh please explain what's happening are you getting an error message? or is nothing being returned?

Ryan · Accepted Answer · 2010-08-03 14:51:01Z

1

A few lines of code for HTTP Basic Auth, if you mean this.

String auth = Base64Converter.encode(String.format("%s:%s", user, pass));
URL u = new URL(url);
conn = (HttpsURLConnection) u.openConnection();
conn.addRequestProperty("Authorization", "Basic " + auth);

Where "Base64Converter" is a utility class converts a string to its Base64 compiled form. Do this before the openConnection() call in parsingxml.java, line 36.

answered Aug 3, 2010 at 14:51

Ryan

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3 Comments

Paresh Mayani Over a year ago

i think this helps me...let me try...give u reply if running..thanx for the support

Paresh Mayani Over a year ago

Base64Converter is not supported in Android..showing Red line under(error) the Base64Converter

Ryan Over a year ago

@Paresh yes Base64Converter is a public domain utility class that I found somewhere else. A Google search might get you the right place. Android meant to have provided an Base64 encoder / decoder class but is actually unavailable to app developers.

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