CODEWITHSUNDEEP
javaurlioexceptionhttp-status-code-400
Tapas Bose
Tapas Bose

Reputation: 29806

Server returned HTTP response code: 400

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.

Upvotes: 19

Views: 146344

Answers (5)

Aneesh
Aneesh

Reputation: 1

I a encountered same error. In my case, it was because the sizjwt token in the header was larger than acceptable size by mule soft proxy. One option is to increase the size of acceptable header size in mule soft, or reduce the size of token by removing some of the permissions assigned to the user id

Upvotes: 0

harshainfo
harshainfo

Reputation: 534

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");

Complete code/answer should be:

if(urlString.contains(" "))
    urlString = urlString.replace(" ", "%20");
java.net.URL url = new URL(urlString);  
java.io.InputStream in = url.openStream();

Upvotes: 8

Mahmoud Saleh
Mahmoud Saleh

Reputation: 33605

encode the parameters in the URL as follows:

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

Upvotes: 2

Tapas Bose
Tapas Bose

Reputation: 29806

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")));

Upvotes: 39

Paul Sanwald
Paul Sanwald

Reputation: 11329

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());
    }
}

Upvotes: 3

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