如何使用Java客户端使用liferay编写的liferay Web服务？

Question

I have web service written in Liferay. I want to get data in Java Client that is running on a separate machine.how can i do this? I have tried this but i am getting following error

{"message":"Authenticated access required","exception":"java.lang.SecurityException"}

My Code is below

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

public class JavaClient {

    // http://localhost:8080/RESTfulExample/json/product/get
    public static void main(String[] args)
    {

        try
        {

            // URL url = new
            // URL("http://localhost:8080/api/jsonws/us-pharmacy-ui-portlet.ph_fax/get-documents-to-fax
            // \-u test@liferay.com:test");
            HttpURLConnection conn = (HttpURLConnection) getURL().openConnection();

            conn.setRequestMethod("GET");
            conn.setRequestProperty("Accept", "application/json");

            if(conn.getResponseCode() != 200){ throw new RuntimeException("Failed : HTTP error code : " + conn.getResponseCode()); }

            BufferedReader br = new BufferedReader(new InputStreamReader((conn.getInputStream())));

            String output;
            System.out.println("Output from Server .... \n");
            while ((output = br.readLine()) != null)
            {
                System.out.println(output);
            }

            conn.disconnect();

        }
        catch (MalformedURLException e)
        {

            e.printStackTrace();

        }
        catch (IOException e)
        {

            e.printStackTrace();

        }

    }

    private static URL getURL() throws MalformedURLException
    {
        String url = "http://localhost:8080";
        String screenName = "shahid.rana";
        String password = "123";

        int pos = url.indexOf("://");
        String protocol = url.substring(0, pos + 3);
        String host = url.substring(pos + 3, url.length());

        StringBuilder sb = new StringBuilder();
        sb.append(protocol);
        sb.append(screenName);
        sb.append(":");
        sb.append(password);
        sb.append("@");
        sb.append(host);
        sb.append("/api/jsonws/us-pharmacy-ui-portlet.ph_fax/get-documents-to-fax/");
        // sb.append(serviceName);
        System.out.println("sb.toString()" + sb.toString());
        return new URL(sb.toString());
    }

}

Answer 1

Yes, you will get this error because Liferay Web services needs basic Auth. you need to set test@liferay.com & whatever-password in base64 & pass it along.

Example code using HttpClient

HttpHost targetHost = new HttpHost("localhost", 8080, "http");
           DefaultHttpClient httpclient = new DefaultHttpClient();
           BasicHttpContext ctx = new BasicHttpContext();
           // Plugin Context Use for Liferay 6.1
           HttpPost post = new HttpPost("/api/jsonws/country/get-countries");
           Base64 b = new Base64();
        String encoding = b.encodeAsString(new String("test@liferay.com:test").getBytes());
        post.setHeader("Authorization", "Basic " + encoding);
           List<NameValuePair> params = new ArrayList<NameValuePair>();
           //params.add(new BasicNameValuePair("emplyeeId", "30722"));
           UrlEncodedFormEntity entity = new UrlEncodedFormEntity(params, "UTF-8");
           post.setEntity(entity);
           HttpResponse resp = httpclient.execute(targetHost, post, ctx);
           resp.getEntity().writeTo(System.out);
           httpclient.getConnectionManager().shutdown();

     }

Please note

String encoding = b.encodeAsString(new String("test@liferay.com:test").getBytes());
        post.setHeader("Authorization", "Basic " + encoding);