当前位置: 首页 > 知识库问答 >
问题:

不能将org.json.jsonArray类型的值转换为JSONObject

经佐
2023-03-14

被这个错误卡住了:

public class JSONParser extends AsyncTask<Void, Void, Boolean> {

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);

        ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
        nameValuePairs.add(new BasicNameValuePair("username", LoginActivity.Suser));
        nameValuePairs.add(new BasicNameValuePair("password", LoginActivity.Spass));
        httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

        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;

}
@Override
protected Boolean doInBackground(Void... params) {
    getJSONFromUrl("http://flapplabs.se/development/snapchat/friends.php");
    return null;
}
}

共有1个答案

唐声
2023-03-14

[..]意味着它应该是JSONArray{..}意味着它应该是JSONObject

因此:

try {
        JSONArray jObj = new JSONArray(json);
    } catch (JSONException e) {
        Log.e("JSON Parser", "Error parsing data " + e.toString());
    }
 类似资料: