HTTP/1.1 400錯誤的請求Apache

Question

我試圖用我寫的下面的代碼登錄到twitter。問題出現在每次執行時，我收到一個400 Bad Request作爲回覆。我嘗試過無數次嘗試使這個工作無濟於事。

public void login(String url) throws ClientProtocolException, IOException{ 
     HttpClient client = HttpClientBuilder.create().build(); 
     HttpGet request = new HttpGet(url); 

     // add request header 
     request.addHeader("User-Agent", USER_AGENT); 
     HttpResponse response = client.execute(request); 

     System.out.println("Response Code : " 
        + response.getStatusLine().getStatusCode()); 

     BufferedReader rd = new BufferedReader(
      new InputStreamReader(response.getEntity().getContent())); 

     StringBuffer result = new StringBuffer(); 
     String line = ""; 
     while ((line = rd.readLine()) != null) { 
      result.append(line); 
     } 

     // set cookies 
     setCookies(response.getFirstHeader("Set-Cookie") == null ? "" : response.getFirstHeader("Set-Cookie").toString()); 

     Document doc = Jsoup.parse(result.toString()); 
     System.out.println(doc); 

     // Get input elements 
     Elements loginform = doc.select("div.clearfix input[type=hidden][name=authenticity_token]"); 
     String auth_token = loginform.attr("value"); 
     System.out.println("Login: "+auth_token); 

     List<NameValuePair> paramList = new ArrayList<NameValuePair>(); 
     paramList.add(new BasicNameValuePair("authenticity_token", auth_token)); 
     paramList.add(new BasicNameValuePair("session[username_or_email]", "twitter_username")); 
     paramList.add(new BasicNameValuePair("session[password]", "twitter_password")); 

     System.out.println(paramList); 

     HttpPost post = new HttpPost(url); 

     // add header 
     post.setHeader("Host", "twitter.com"); 
     post.setHeader("User-Agent", USER_AGENT); 
     post.setHeader("Accept", "text/html,application/xhtml;q=0.9,*/*;q=0.8"); 
     post.setHeader("Accept-Language", "en-US,en;q=0.5"); 
     post.setHeader("Keep-Alive", "115"); 
     post.setHeader("Cookie", getCookies()); 
     post.setHeader("Connection", "keep-alive"); 
     post.setHeader("Referer", "https://twitter.com/"); 
     post.setHeader("Content-Type", "application/x-www-form-urlencoded"); 
     post.setEntity(new UrlEncodedFormEntity(paramList)); 
     // Execute POST data 
     HttpResponse res = client.execute(post); 

     int responseCode = res.getStatusLine().getStatusCode(); 

     System.out.println("\nSending 'POST' request to URL : " + url); 
     System.out.println("Post parameters : " + paramList); 
     System.out.println("Response Code : " + responseCode); 
     System.out.println("Headers: "+res.getAllHeaders().toString()); 
     System.out.println("Response: "+res.getStatusLine()); 

     BufferedReader rd1 = new BufferedReader(
        new InputStreamReader(res.getEntity().getContent())); 

     StringBuffer resul = new StringBuffer(); 
     String line1 = ""; 
     while ((line1 = rd1.readLine()) != null) { 
      resul.append(line1); 
     } 
     Document doc2 = Jsoup.parse(res.toString()); 
     System.out.println(doc2); 
    } 
      public static void main(String[] args) throws ClientProtocolException, IOException{ 
      Browser b = new Browser(); 

      b.login("https://twitter.com/login"); 
} 

我相信所有需要POST'd的東西都是存在的，比如用戶名，密碼以及真實性標記。

Answer 1

原來我在POST請求中發送了錯誤的會話信息！如果其他人有類似的問題，我建議使用Chrome開發人員工具檢查發送/接收的標題。