org.json.JSONException: Value <!DOCTYPE of type java.lang.String cannot be converted to JSONObject

问题:

Here I want to display the JSON content using API key. But I am unable to get the authentication.

I am getting the error in JsonObject:

org.json.JSONException: Value Authorization of type java.lang.String cannot be converted to JSONObject 

In my android application, I just pass the API key and URL id to get the JSON response in the following URL. I display the JSON content using JSON array.

But if I:

public class AndroidAPiActivity extends Activity {  /*  * FlickrQuery = FlickrQuery_url  * + FlickrQuery_per_page  * + FlickrQuery_nojsoncallback  * + FlickrQuery_format  * + FlickrQuery_tag + q  * + FlickrQuery_key + FlickrApiKey  */  String FlickrQuery_url = "http://192.138.11.9/api/interests/"; String FlickrQuery_per_page = "&per_page=1"; String FlickrQuery_nojsoncallback = "&nojsoncallback=1"; String FlickrQuery_format = "&format=json"; String FlickrQuery_tag = "&tags="; String FlickrQuery_key = "&api_key=";  // Apply your Flickr API: // www.flickr.com/services/apps/create/apply/?    String FlickrApiKey = "f65215602df8f8af";     EditText searchText;    Button searchButton;    TextView textQueryResult, textJsonResult;    ImageView imageFlickrPhoto;    Bitmap bmFlickr;     /** Called when the activity is first created. */    @Override    public void onCreate(Bundle savedInstanceState) {        super.onCreate(savedInstanceState);        setContentView(R.layout.main);         searchText = (EditText)findViewById(R.id.searchtext);        searchButton = (Button)findViewById(R.id.searchbutton);        textQueryResult = (TextView)findViewById(R.id.queryresult);        textJsonResult = (TextView)findViewById(R.id.jsonresult);        imageFlickrPhoto = (ImageView)findViewById(R.id.flickrPhoto);        searchButton.setOnClickListener(searchButtonOnClickListener);    }     private Button.OnClickListener searchButtonOnClickListener    = new Button.OnClickListener(){   public void onClick(View arg0) {   // TODO Auto-generated method stub   String searchQ = searchText.getText().toString();   String searchResult = QueryFlickr(searchQ);   textQueryResult.setText(searchResult);   String jsonResult = ParseJSON(searchResult);   textJsonResult.setText(jsonResult);    if (bmFlickr != null){    imageFlickrPhoto.setImageBitmap(bmFlickr);   }  }};     private String QueryFlickr(String q){      String qResult = null;      String qString =       FlickrQuery_url       + FlickrQuery_per_page       + FlickrQuery_nojsoncallback       + FlickrQuery_format       + FlickrQuery_tag + q        + FlickrQuery_key + FlickrApiKey;      HttpClient httpClient = new DefaultHttpClient();        HttpGet httpGet = new HttpGet(qString);         try {   HttpEntity httpEntity = httpClient.execute(httpGet).getEntity();    if (httpEntity != null){    InputStream inputStream = httpEntity.getContent();    Reader in = new InputStreamReader(inputStream);    BufferedReader bufferedreader = new BufferedReader(in);    StringBuilder stringBuilder = new StringBuilder();     String stringReadLine = null;     while ((stringReadLine = bufferedreader.readLine()) != null) {     stringBuilder.append(stringReadLine + "\n");     }     qResult = stringBuilder.toString();    }   } catch (ClientProtocolException e) {   // TODO Auto-generated catch block   e.printStackTrace();  } catch (IOException e) {     // TODO Auto-generated catch block   e.printStackTrace();  }         return qResult;    }     private String ParseJSON(String json){      String jResult = null;     bmFlickr = null;     String key_id;     String category;     String subcategory;     String title;     String icon_image;      try      {   JSONObject JsonObject = new JSONObject(json);   JSONObject Json_photos = JsonObject.getJSONObject("interests");   JSONArray JsonArray_photo = Json_photos.getJSONArray("interest");    //We have only one photo in this exercise   JSONObject FlickrPhoto = JsonArray_photo.getJSONObject(0);    key_id = FlickrPhoto.getString("row_key");   category = FlickrPhoto.getString("category");   subcategory = FlickrPhoto.getString("subcategory");    title = FlickrPhoto.getString("title");    jResult = "\n key_id: " + key_id + "\n"     + "category: " + category + "\n"     + "subcategory: " + subcategory + "\n"     + "title: " + title + "\n";    bmFlickr = LoadPhotoFromFlickr(key_id, category, subcategory,title);   } catch (JSONException e) {   // TODO Auto-generated catch block   e.printStackTrace();  }      return jResult;    }     private Bitmap LoadPhotoFromFlickr(      String key_id, String category, String subcategory,      String title){     Bitmap bm= null;      String icon_image = null;  //   String FlickrPhotoPath ="";    String FlickrPhotoPath ="http://182.72.180.34/media/"+icon_image+".jpg";      URL FlickrPhotoUrl = null;      try {    FlickrPhotoUrl = new URL(FlickrPhotoPath);    HttpURLConnection httpConnection = (HttpURLConnection) FlickrPhotoUrl.openConnection();   httpConnection.setDoInput(true);   httpConnection.connect();   InputStream inputStream = httpConnection.getInputStream();   bm = BitmapFactory.decodeStream(inputStream);   } catch (MalformedURLException e) {   // TODO Auto-generated catch block   e.printStackTrace();  } catch (IOException e) {   // TODO Auto-generated catch block   e.printStackTrace();  }      return bm;    } } 

回答1:

Update:

Based on the HTML response, I can tell you that this is not JSON. The response tells me that you have the incorrect URL for your web service.

You need to check your URL.

Extra Info / Previous Answer:

It looks like the simple answer is the right one - your result is not a valid JSON string. See JSON.org website for details on what JSON should look like.

Check out JSON Parser Online - I find its very useful when working with JSON.

It is strange that you are requesting JSON, and it is not returning it properly - perhaps I have missed something.

回答2:

Yes, we get such kind of warning when the given URL is not valid.

Just check the URL once.

回答3:

Remove docType from API. and set content Type Application/json . (as text/html will not read as json . thus you were seeing the error.)

回答4:

I received the same ".." error when working with Google Translate's json URLs. Then, I found this code somewhere and it worked :

        BasicHttpParams basicHttpParams = new BasicHttpParams();         HttpConnectionParams.setConnectionTimeout((HttpParams)basicHttpParams, (int)10000);         HttpConnectionParams.setSoTimeout((HttpParams)basicHttpParams, (int)10000);         HttpConnectionParams.setTcpNoDelay((HttpParams)basicHttpParams, (boolean)true);         DefaultHttpClient defaultHttpClient = new DefaultHttpClient((HttpParams)basicHttpParams);         HttpGet httpGet = new HttpGet(url);         BasicResponseHandler basicResponseHandler = new BasicResponseHandler();          JSONObject json=null;         try {             json = new JSONObject((String)defaultHttpClient.execute((HttpUriRequest)httpGet, (ResponseHandler)basicResponseHandler));         } catch (ClientProtocolException e) {             e.printStackTrace();         } catch (JSONException e) {             e.printStackTrace();         } catch (IOException e) {             e.printStackTrace();         } 

文章来源: org.json.JSONException: Value ype java.lang.String cannot be converted to JSONObject