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我正在尝试 json 解析，但是当解析时给出错误 httppost 方法不受此 url 支持我在这里写下我的代码

http://ajax.googleapis.com/ajax/services/search/local?v=1.0&q=restaurants&rsz=8&sll=-27.5595451,-48.6206452&radius=1000&output=json

SearchListActivity.java

 public class SearchlistActivity extends Activity {

    private static String url="http://ajax.googleapis.com/ajax/services/search/local?v=1.0&q=restaurants&rsz=8&sll=-27.5595451,-48.6206452&radius=1000&output=json/";   private static final String TAG_RESULTS = "results";    private static final String TAG_RESPONSEDATA="responseData";    

        @Override
    public void onCreate(Bundle savedInstanceState)

    {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        Log.e("json ----","urlll---->"+url);       
        //JSONArray results = null;       
         JSONArray responseData=null;
     // Hashmap for ListView
        ArrayList<HashMap<String, String>> contactList = new ArrayList<HashMap<String, String>>();

        // Creating JSON Parser instance
        JsonParserSearch jParser = new JsonParserSearch();

        // getting JSON string from URL
        JSONObject json = jParser.getJSONFromUrl(url);

        try {
            // Getting Array of Contacts

             responseData = json.getJSONArray(TAG_RESPONSEDATA);
             Log.e("jsonnn data","sfsssss00000-------->"+responseData);

            }
        catch (JSONException e) {
            e.printStackTrace();
        }

    } }

>

Jsonparsersearch.java

公共类 JsonParserSearch { 静态 InputStream is = null; 静态 JSONObject jObj = null; 静态字符串 json = "";

// constructor
public JsonParserSearch() {

}

public JSONObject getJSONFromUrl(String url) {

    // Making HTTP request
    try {
        // defaultHttpClient
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(url);

        HttpResponse httpResponse = httpClient.execute(httpPost);
        HttpEntity httpEntity = httpResponse.getEntity();
        is = httpEntity.getContent();          

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    try {

        BufferedReader reader = new BufferedReader(new InputStreamReader(
                is, "iso-8859-1"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
        is.close();
        json = sb.toString();

       Log.e("json object url","display"+json);


    } catch (Exception e) {
        Log.e("Buffer Error", "Error converting result " + e.toString());
    }

    // try parse the string to a JSON object
    try {
        jObj = new JSONObject(json);
    } catch (JSONException e) {
        Log.e("JSON Parser finaal", "Error parsing data " + e.toString());
    }

    // return JSON String
    return jObj;

}      }

 >     > > lOGCAT ERROR
    >     > > 
    >     > > 11-09 12:23:17.700: E/json ----(481): urlll---->http://ajax.googleapis.com/ajax/services/search/local?v=1.0&q=restaurants&rsz=8&sll=-27.5595451,-48.6206452&radius=1000&output=json 11-09 12:23:18.260: E/json object url(481): display<HTML> 11-09
    >     > 12:23:18.260: E/json object url(481): <HEAD> 11-09 12:23:18.260:
    >     > E/json object url(481): <TITLE>HTTP method POST is not supported by
    >     > this URL</TITLE> 11-09 12:23:18.260: E/json object url(481): </HEAD>
    >     > 11-09 12:23:18.260: E/json object url(481): <BODY BGCOLOR="#FFFFFF"
    >     > TEXT="#000000"> 11-09 12:23:18.260: E/json object url(481): <H1>HTTP
    >     > method POST is not supported by this URL</H1> 11-09 12:23:18.260:
    >     > E/json object url(481): <H2>Error 405</H2> 11-09 12:23:18.260: E/json
    >     > object url(481): </BODY> 11-09 12:23:18.260: E/json object url(481):
    >     > </HTML> 11-09 12:23:18.260: E/JSON Parser finaal(481): Error parsing
    >     > data org.json.JSONException: Value <HTML> of type java.lang.String
    >     > cannot be converted to JSONObject 11-09 12:23:18.270:
    >     > D/AndroidRuntime(481): Shutting down VM 11-09 12:23:18.270:
    >     > W/dalvikvm(481): threadid=1: thread exiting with uncaught exception
    >     > (group=0x4001d800) 11-09 12:23:18.290: E/AndroidRuntime(481): FATAL
    >     > EXCEPTION: main 11-09 12:23:18.290: E/AndroidRuntime(481):
    >     > java.lang.RuntimeException: Unable to start activity
    >     > ComponentInfo{com.example.ssss/com.example.ssss.SearchlistActivity}:
    >     > java.lang.NullPointerException 11-09 12:23:18.290:
    >     > E/AndroidRuntime(481):  at
    >     > android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2663)
    >     > 11-09 12:23:18.290: E/AndroidRuntime(481):  at
    >     > android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2679)
    >     > 11-09 12:23:18.290: E/AndroidRuntime(481):  at
    >     > android.app.ActivityThread.access$2300(ActivityThread.java:125)
4

4 回答 4

2

您正在尝试使用POST仅支持GET.

尝试使用 aHttpGet而不是HttpPost.

于 2012-11-09T07:43:08.193 回答
1

您应该使用Get请求而不是POST请求。

public class MainActivity extends Activity {

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    //make async request 
    new myAsynctask().execute();
}

//********************************************
//get json string
public String getJSONString() {
    StringBuilder builder = new StringBuilder();
    HttpClient client = new DefaultHttpClient();

    String urlString="http://www.ajax.googleapis.com/ajax/services/search/local?v=1.0&q=restaurants&rsz=8&sll=-27.5595451,-48.6206452&radius=1000&output=json";

    HttpGet httpGet = new HttpGet(urlString);
    try {
        HttpResponse response = client.execute(httpGet);
        StatusLine statusLine = response.getStatusLine();
        int statusCode = statusLine.getStatusCode();
        if (statusCode == 200) {
            HttpEntity entity = response.getEntity();
            InputStream content = entity.getContent();
            BufferedReader reader = new BufferedReader(new InputStreamReader(content));
            String line;
            while ((line = reader.readLine()) != null) {
                builder.append(line);
            }
        } else {
            Log.e(getClass().getSimpleName(), "Failed to download json response");
        }
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
    return builder.toString();
}
//********************************************
private class myAsynctask extends AsyncTask<Void, Void, String>{

    @Override
    protected String doInBackground(Void... params) {

        String jsonString=getJSONString();
        return jsonString;
    }
    @Override
    protected void onPostExecute(String jsonString) {
        super.onPostExecute(jsonString);
        Log.e(getClass().getSimpleName(), jsonString);
    }
}

}

于 2012-11-09T07:57:28.950 回答
0

jParser.getJSONFromUrl(url) 使用 POST 方法，但你应该使用 GET

于 2012-11-09T07:46:39.280 回答
0

您收到的错误来自服务器，因为它不支持该实际 URL 的 POST。

在处理结果之前检查 HTTP 状态代码是一种很好的做法。通过调用来做到这一点int statusCode = httpResponse.getStatusLine().getStatusCode();。状态码 2XX 表示成功（此处列出的代码）。

您的代码的另一个问题是，尽管您正在创建一个 POST 请求，但所有信息都在请求的查询参数中提供。请求的 POST 部分（正文部分）实际上是空的。

MultipartEntity entity = new MultipartEntity();
entity.addPart("alma", new StringBody("beka", Charset.forName("UTF-8")));
entity.addPart("apple", new StringBody("frog", Charset.forName("UTF-8")));
httpPost.setEntity(entity);

这是将参数作为 POST 参数添加到请求的示例...

于 2012-11-09T07:51:01.297 回答