Question

这里我想使用API密钥显示JSON内容。但我无法获得身份验证。

我在JsonObject中收到错误：

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

在我的Android应用程序中，我只是传递API密钥和URL id以在以下URL中获取JSON响应。我使用JSON数组显示JSON内容。

但如果我：

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;
   }
}

Answer 1

更新

基于HTML响应，我可以告诉你这不是JSON。该回复告诉我您的网络服务的网址不正确。

您需要检查自己的网址。

额外信息/上一个答案：

看起来简单的答案是正确的 - 您的结果不是有效的JSON字符串。有关JSON应该是什么样的详细信息，请参阅JSON.org网站。

查看JSON Parser Online - 我发现它在使用JSON时非常有用。

很奇怪你正在请求JSON，而且它没有正确地返回它 - 也许我错过了一些东西。

Answer 2

是的，当给定的网址无效时，我们会收到此类警告。

只需检查一次网址。

Answer 3

从API中删除docType。并设置内容类型Application / json。（因为text / html不会读作json。因此你看到了错误。）

Answer 4

使用Google翻译的json网址时，我收到了同样的"<!Doctype..."错误。然后，我在某个地方找到了这个代码并且它有效：

        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();
        }

Answer 5

可能对您有帮助

https://teamtreehouse.com/community/solved-exception-cannot-convert-string-type-to-json-object

已解决。

原来，运行时错误还可以追溯到上一个视频。

我在做

JSONObject当前=新的JSONObject（“ currently”）; 代替

JSONObject currently = forecast.getJSONObject("currently");

所以我的猜测是Android认为我正在尝试设置一个全新的JSON对象，而不是尝试从现有对象中检索信息！ :)现在控制台完美显示了时间！

Answer 6

我也遇到了这个问题，我将我的Internet连接更改为另一个网络，并且可以正常工作。

问题是 ISP 不接受http访问。

另一种解决方案，您可以打开 VPN ，然后重试，也许可行...

org.json.JSONException：值

6 个答案:

更新

额外信息/上一个答案：