Server returned HTTP response code: 400

Question

I am trying to get an InputStream from a URL. The URL can be a opened from Firefox. It returns a json and I have installed an addon for viewing json in Firefox so I can view it there.

So I tried to get it from Java by:

URL url = new URL(urlString);
URLConnection urlConnection = url.openConnection();
BufferedReader reader = new BufferedReader(new InputStreamReader(urlConnection.getInputStream()));

But it is throwing an IOException in urlConnection.getInputStream().

I also tried:

HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
InputStream inputStream = url.openStream();

But no luck.

Any information is appreciable. Thanks in advance.

Answer 1

Thank you everybody. This is a weird problem but at last I solved it.

The URL I am requesting is

http://api.themoviedb.org/2.1/Movie.search/en/json/api_key/a nightmare on elm street 

Now browser replaces the spaces between "a nightmare on elm street" by "%20" internally and parses. That is why the requested server can response by that request. But From Java I didn't replaced that spaces by "%20", so it turns into Bad Request, source.

Now it is working.

BufferedReader reader = new BufferedReader(new InputStreamReader(((HttpURLConnection) (new URL(urlString)).openConnection()).getInputStream(), Charset.forName("UTF-8")));

Answer 2

I had a similar issue and my url was:

http://www.itmat.upenn.edu/assets/user-content/documents/ITMAT17. October 10 2017_.pdf

which obviously contained spaces.

These caused java.io.IOException Server returned HTTP response code: 400 in the following code:

java.net.URL url = new URL(urlString);  
java.io.InputStream in = url.openStream();

If you copy the above url and paste in browser, you will realize that browser adds '%20' for the spaces. So I did it manually with the following code and the problem is solved.

if(urlString.contains(" "))
    urlString = urlString.replace(" ", "%20");

Answer 3

are you setting up the connection correctly? here's some code that illustrates how to do this. Note that I am being lazy about exception handling here, this is not production quality code.

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URL;


public class URLFetcher {

    public static void main(String[] args) throws Exception {
        URL myURL = new URL("http://www.paulsanwald.com/");
        HttpURLConnection connection = (HttpURLConnection) myURL.openConnection();
        connection.setRequestMethod("GET");
        connection.setDoOutput(true);
        connection.connect();
        BufferedReader reader = new BufferedReader(new InputStreamReader(connection.getInputStream()));
        StringBuilder results = new StringBuilder();
        String line;
        while ((line = reader.readLine()) != null) {
            results.append(line);
        }

        connection.disconnect();
        System.out.println(results.toString());
    }
}

Answer 4

encode the parameters in the URL as follows:

String messageText = URLEncoder.encode(messageText, "UTF-8");