android - 在android中用jsonarray解析json对象

Question

你好你好我是 android 的新手。我正在从服务器收到响应。但我得到以下回应..

出现以下错误

Value {"Elec":"Cup1"} at 1 of type java.lang.String cannot be converted to JSONObject

响应如下

{"Electronics":["{\"Elec\":\"Cup0\"}","{\"Elec\":\"Cup1\"}","{\"Elec\":\"Cup2\"}","{\"Elec\":\"Cup3\"}","{\"Elec\":\"Cup4\"}","{\"Elec\":\"Cup5\"}","{\"Elec\":\"Cup6\"}","{\"Elec\":\"Cup7\"}","{\"Elec\":\"Cup8\"}","{\"Elec\":\"Cup9\"}"],"Printable":["{\"Print\":\"Mug0\"}","{\"Print\":\"Mug1\"}","{\"Print\":\"Mug2\"}","{\"Print\":\"Mug3\"}","{\"Print\":\"Mug4\"}","{\"Print\":\"Mug5\"}","{\"Print\":\"Mug6\"}","{\"Print\":\"Mug7\"}","{\"Print\":\"Mug8\"}","{\"Print\":\"Mug9\"}"]}

我发送和接收的响应如下

    class SliderImages extends AsyncTask<String, String, String> {

    @Override
    protected String doInBackground(String... urls) {

        return POST1(urls[0]);
    }

    @Override
    public void onPostExecute(String result) {

    }

}

public String POST1(String url) {
    InputStream inputStream;
    String result = "";
    try {
        arraylist = new ArrayList<HashMap<String, String>>();

        // 1. create HttpClient
        HttpClient httpclient = new DefaultHttpClient();

        // 2. make POST request to the given URL
        HttpPost httpPost = new HttpPost(url);

        String json = "";

        // 3. build jsonObject
        JSONObject jsonObject = new JSONObject();


        // 4. convert JSONObject to JSON to String
        json = jsonObject.toString();
        Log.d("JsonStringSend", json);
        // ** Alternative way to convert Person object to JSON string usin Jackson Lib
        // ObjectMapper mapper = new ObjectMapper();
        // json = mapper.writeValueAsString(person);

        // 5. set json to StringEntity
        StringEntity se = new StringEntity(json);

        // 6. set httpPost Entity
        httpPost.setEntity(se);

        // 7. Set some headers to inform server about the type of the content
        httpPost.setHeader("Accept", "application/json");
        httpPost.setHeader("Content-type", "application/json");

        // 8. Execute POST request to the given URL
        HttpResponse httpResponse = httpclient.execute(httpPost);

        // 9. receive response as inputStream
        inputStream = httpResponse.getEntity().getContent();

        // 10. convert inputstream to string
        if (inputStream != null) {
            result = convertInputStreamToString(inputStream);
            Log.d("JSONResponce", result);
            JSONObject jsonObj = new JSONObject(result);
            contacts = jsonObj.getJSONArray("Electronics");
            for (int i = 1; i < contacts.length()-1; i++) {
                JSONObject c = contacts.getJSONObject(i);
                String id = c.getString("Print");
                Log.e("Errrrrrrr",""+id);
            }
        }
    } catch (IOException e) {
        e.printStackTrace();
    } catch (JSONException e) {
        e.printStackTrace();
    }
    // 11. return result
    return result;
}


private String convertInputStreamToString(InputStream inputStream) throws IOException {
    BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
    String line = "";
    String result = "";
    while ((line = bufferedReader.readLine()) != null)
        result += line;
    inputStream.close();
    return result;

}

所以得到以下错误

Value {"Elec":"Cup1"} at 1 of type java.lang.String cannot be converted to JSONObject

请帮帮我

提前致谢

Answer 1

这是在 jsoneditoronline.org 中格式化你的 json 后的结果

请注意，“Electronics”数组中的对象不是 JSONObject，而是 String。

一个简单的解决方法是首先将对象作为 String 获取，然后将其解析为 JSONObject，例如:

    // 10. convert inputstream to string
    if (inputStream != null) {
        result = convertInputStreamToString(inputStream);
        Log.d("JSONResponce", result);
        JSONObject jsonObj = new JSONObject(result);
        contacts = jsonObj.getJSONArray("Electronics");

        // Note contacts.length() NOT contacts.length() - 1
        for (int i = 1; i < contacts.length(); i++) {

            String cString = contacts.getString(i);
            JSONObject c = new JSONObject(cString);

            String id = c.getString("Print");
            Log.e("Errrrrrrr",""+id);
        }
    }

然而，最好的选择是更正从服务器返回的数据。