获取错误：解析数据时出错org.json.JSONException：org.json.JSONArray类型的值无法转换为JSONObject

Question

Stuck and need help. JSON parser cannot convert data so I can use the username and password to login.

jsonParser = new JSONParser();
public JSONObject loginUser(String username, String password) {
    // Building Parameters
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("tag", login_tag));
    params.add(new BasicNameValuePair("username", username));
    params.add(new BasicNameValuePair("password", password));
    JSONObject json = jsonParser.getJSONFromUrl(loginURL, params);
    return json;
}

PHP part of the code:

while($r = mysql_fetch_assoc($result)) {
    $json[] = $r;
}
print json_encode($json);

Answer 1

With the comments of the @njzk2 the solution should be to change your PHP coding:

But then you just get the first:

while($r = mysql_fetch_assoc($result)) {
    $json = $r;
    break;
}
print json_encode($json);

Or change your Java Coding to a JsonArray

jsonParser = new JSONParser();
public JSONArray loginUser(String username, String password) {
    // Building Parameters
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("tag", login_tag));
    params.add(new BasicNameValuePair("username", username));
    params.add(new BasicNameValuePair("password", password));
    // change the following line, to the correct method. Which parser do you use?
    JSONArray json = jsonParser.getJSONArrayFromUrl(loginURL, params);
    return json;
}