Java HttpURLConnection响应不工作

Java HttpURLConnection响应不工作,java,android,httpurlconnection,Java,Android,Httpurlconnection,在我的项目中,我有一个url 比如: 它包含Json值。要访问此字段,需要将access key作为标头 现在,我所做的是: private String doHttpUrlConnectionAction(String desiredUrl) throws Exception { URL url = null; BufferedReader reader = null; StringBuilder stringBuilder; try {

在我的项目中,我有一个url 比如:

它包含Json值。要访问此字段,需要将access key作为标头

现在,我所做的是:

private String doHttpUrlConnectionAction(String desiredUrl)
  throws Exception
  {
    URL url = null;
    BufferedReader reader = null;
    StringBuilder stringBuilder;

    try
    {
      // create the HttpURLConnection
      url = new URL(desiredUrl);
      HttpURLConnection connection = (HttpURLConnection) url.openConnection();

      // just want to do an HTTP GET here
      connection.setRequestMethod("GET");
     // connection.setRequestProperty("Content-Type","application/json");
      connection.setRequestProperty("API-KEY", "value");

      // uncomment this if you want to write output to this url
      connection.setDoOutput(true);

      // give it 15 seconds to respond
      connection.setReadTimeout(15*1000);
      connection.connect();

      // read the output from the server
     // reader = new BufferedReader(new InputStreamReader(connection.getInputStream()));


      StringBuilder sb = new StringBuilder();
      BufferedReader rd = new BufferedReader(new InputStreamReader(connection.getInputStream()));
      String line;
      while ((line = rd.readLine()) != null) {
          sb.append(line);
      }
      return sb.toString();

    }
    catch (Exception e)
    {
      e.printStackTrace();
      throw e;
    }
    finally
    {
      // close the reader; this can throw an exception too, so
      // wrap it in another try/catch block.
      if (reader != null)
      {
        try
        {
          reader.close();
        }
        catch (IOException ioe)
        {
          ioe.printStackTrace();
        }
      }
    }
  }
此代码返回输出:

Response Code : 200
<table border="0" cellpadding="4" cellspacing="0"><thead><tr><th>status</th><th>statusCode</th><th>data</th></tr></thead><tbody><tr><td statusCode="200" status="1">Array</td></tr></tbody></table>
在这里,我得到了正确的输出

public String ReadHttpResponse(String url){
      StringBuilder sb= new StringBuilder();
      HttpClient client= new DefaultHttpClient();     
      HttpGet httpget = new HttpGet(url);  
      httpget.addHeader("API-KEY", "value");
      try {
          HttpResponse response = client.execute(httpget);
          StatusLine sl = response.getStatusLine();
          int sc = sl.getStatusCode();
          if (sc==200)
          {
              HttpEntity ent = response.getEntity();
              InputStream inpst = ent.getContent();
              BufferedReader rd= new BufferedReader(new InputStreamReader(inpst));
              String line;
              while ((line=rd.readLine())!=null)
              {
                  sb.append(line);
              }
             // System.out.println(sb.toString());

          }
          else
          {
              System.out.println("I didn't  get the response!");


          }
      } catch (ClientProtocolException e) {
          e.printStackTrace();
      } catch (IOException e) {
          e.printStackTrace();
      }
      return sb.toString();
  }

HttpUrlConnection
中的问题在哪里??我在这里做错了什么??我必须使用
HttpUrlConnection
。请大家帮帮我。

您可以为请求POST和GET创建这样的代码逻辑。它有助于降低代码复杂性。您可以为此创建一个方法,并根据需要为GET和POST方法传递参数

  HttpURLConnection urlConnection = null;
    try {
        // http client

        murl=new URL(url);
        urlConnection = (HttpURLConnection) murl.openConnection();

        urlConnection.setRequestProperty("Content-Type", "application/json;odata=verbose");
        urlConnection.setRequestProperty("Accept", "application/json;odata=verbose");

        // Checking http request method type
        if (method == POST) {
            urlConnection.setRequestMethod("POST");
            urlConnection.setDoInput(true);
            urlConnection.setDoOutput(true);



            if(!jsondata.equals("null")) {
                OutputStream os = urlConnection.getOutputStream();
                BufferedWriter writer = new BufferedWriter(
                        new OutputStreamWriter(os, "UTF-8"));
                writer.write(jsondata);
                writer.flush();
                writer.close();
                os.close();
            }


        } else if (method == GET) {
            // appending params to url
            urlConnection.setRequestMethod("GET");

        }
        resCode = urlConnection.getResponseCode();
        Log.i("TAG", "response code=>" + resCode);

URL是否在浏览器中工作?您必须在请求中添加此标题:urlConnection.setRequestProperty(“内容类型”,“应用程序/json;odata=verbose”);setRequestProperty(“Accept”、“application/json;odata=verbose”);您的代码中没有问题,因为您从服务器端获得了200。服务器端出现问题,所以您可以从服务器端获取html代码。请检查服务器端以解决此问题。@ρц∑ρєK,URL正在工作。我在使用httpclient时也获得了价值。@Android Developer,谢谢。。现在,我得到了值。。。
  HttpURLConnection urlConnection = null;
    try {
        // http client

        murl=new URL(url);
        urlConnection = (HttpURLConnection) murl.openConnection();

        urlConnection.setRequestProperty("Content-Type", "application/json;odata=verbose");
        urlConnection.setRequestProperty("Accept", "application/json;odata=verbose");

        // Checking http request method type
        if (method == POST) {
            urlConnection.setRequestMethod("POST");
            urlConnection.setDoInput(true);
            urlConnection.setDoOutput(true);



            if(!jsondata.equals("null")) {
                OutputStream os = urlConnection.getOutputStream();
                BufferedWriter writer = new BufferedWriter(
                        new OutputStreamWriter(os, "UTF-8"));
                writer.write(jsondata);
                writer.flush();
                writer.close();
                os.close();
            }


        } else if (method == GET) {
            // appending params to url
            urlConnection.setRequestMethod("GET");

        }
        resCode = urlConnection.getResponseCode();
        Log.i("TAG", "response code=>" + resCode);