解析JSONobject時出錯

Question

我有一個android的登錄應用程序，它使用JSON將數據從數據庫解析到應用程序。接受HTTP請求通過標籤識別的PHP API，要麼是「登錄」或「註冊」，像這樣：

if (isset($_POST['tag']) && $_POST['tag'] != '') { 
    "Do stuf 
    } else { 
     echo "access denied"; 

的應用程序已經工作正常，但現在我只是得到響應「訪問拒絕」。

public JSONObject loginUser(String email, String password){ 
    // Building Parameters 
    List<NameValuePair> params = new ArrayList<NameValuePair>(); 
    params.add(new BasicNameValuePair("tag", "login")); 
    params.add(new BasicNameValuePair("email", email)); 
    params.add(new BasicNameValuePair("password", password)); 
    JSONObject json = jsonParser.getJSONFromUrl(loginURL, params); 
    // return json 
    // Log.e("JSON", json.toString()); 
    return json; 
} 

這是發送請求的JSON對象，並且我懷疑它沒有正確發送標記。有沒有人知道發生了什麼？

更新：新增JSONparser.class

public class JSONParser { 

static InputStream is = null; 
static JSONObject jObj = null; 
static String json = ""; 

// constructor 
public JSONParser() { 

} 

public JSONObject getJSONFromUrl(String url, List<NameValuePair> params) { 

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

     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, "UTF-8"), 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", 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", "Error parsing data " + e.toString()); 
    } 

    // return JSON String 
    return jObj; 

} 

}

Answer 1

其實在看它，它看起來像你使用JSON查詢錯誤反正unles有更多的代碼，你不向我們展示。你錯過了jsonParser類

public class JSONParser {

static InputStream is = null; 
static JSONObject jObj = null; 
static String json = ""; 

// constructor 
public JSONParser() { 

} 

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(); 
    } 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", "Error parsing data " + e.toString()); 
    } 

    // return JSON String 
    return jObj; 

} } 

創建瀏覽此網站了解更多信息http://www.androidhive.info/2012/01/android-json-parsing-tutorial/