Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
262 views
in Technique[技术] by (71.8m points)

java - Httpclient 4, error 302. How to redirect?

I want to access one site that first requires an (tomcat server) authentication and then log in with a POST request and keep that user to see the site's pages. I use Httpclient 4.0.1

The first authentication works fine but not the logon that always complains about this error: "302 Moved Temporarily"

I keep cookies & I keep a context and yet nothing. Actually, it seems that the logon works, because if I write incorrect parameters or user||password, I see the login page. So I guess what doesn't work is the automatic redirection.

Following my code, which always throws the IOException, 302:

    DefaultHttpClient httpclient = new DefaultHttpClient();
    CookieStore cookieStore = new BasicCookieStore();
    httpclient.getParams().setParameter(
      ClientPNames.COOKIE_POLICY, CookiePolicy.BROWSER_COMPATIBILITY); 
    HttpContext context = new BasicHttpContext();
    context.setAttribute(ClientContext.COOKIE_STORE, cookieStore);
    //ResponseHandler<String> responseHandler = new BasicResponseHandler();

    Credentials testsystemCreds = new UsernamePasswordCredentials(TESTSYSTEM_USER,  TESTSYSTEM_PASS);
    httpclient.getCredentialsProvider().setCredentials(
            new AuthScope(AuthScope.ANY_HOST, AuthScope.ANY_PORT),
            testsystemCreds);

    HttpPost postRequest = new HttpPost(cms + "/login");
    List<NameValuePair> formparams = new ArrayList<NameValuePair>();
    formparams.add(new BasicNameValuePair("pUserId", user));
    formparams.add(new BasicNameValuePair("pPassword", pass));
    postRequest.setEntity(new UrlEncodedFormEntity(formparams, "UTF-8"));
    HttpResponse response = httpclient.execute(postRequest, context);
    System.out.println(response);

    if (response.getStatusLine().getStatusCode() != HttpStatus.SC_OK)
        throw new IOException(response.getStatusLine().toString());

    HttpUriRequest currentReq = (HttpUriRequest) context.getAttribute( 
            ExecutionContext.HTTP_REQUEST);
    HttpHost currentHost = (HttpHost)  context.getAttribute( 
            ExecutionContext.HTTP_TARGET_HOST);
    String currentUrl = currentHost.toURI() + currentReq.getURI();        
    System.out.println(currentUrl);

    HttpEntity entity = response.getEntity();
    if (entity != null) {
        long len = entity.getContentLength();
        if (len != -1 && len < 2048) {
            System.out.println(EntityUtils.toString(entity));
        } else {
            // Stream content out
        }
    }
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

For 4.1 version:

DefaultHttpClient  httpclient = new DefaultHttpClient();
    httpclient.setRedirectStrategy(new DefaultRedirectStrategy() {                
        public boolean isRedirected(HttpRequest request, HttpResponse response, HttpContext context)  {
            boolean isRedirect=false;
            try {
                isRedirect = super.isRedirected(request, response, context);
            } catch (ProtocolException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
            if (!isRedirect) {
                int responseCode = response.getStatusLine().getStatusCode();
                if (responseCode == 301 || responseCode == 302) {
                    return true;
                }
            }
            return isRedirect;
        }
    });

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...