Question

我正在写一个JSONArray，以便从我的Android应用程序的数据库中返回多个数据。

我编写了以下代码，但它根本没有返回任何数据，因为此行的错误为java.lang.NullPointerException：jArray = new JSONArray(result);

如何更改代码以便JSONArray有效？对JSONArray的工作方式不太确定。

提前感谢您的帮助！

php文件：

$user=$_POST["username"];

$query = "SELECT longitude,latitude FROM friends INNER JOIN coordinates ON friends.username = coordinates.username WHERE friends.friend_of='$user'";
$sql=mysqli_query($conn, $query);

if (!$sql) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}

$rows = array();
while($r = mysqli_fetch_assoc($sql)) {
    $rows[] = $r;
}
print json_encode($rows);

JSONArray类：

List<NameValuePair> nameValuePairList = new ArrayList<NameValuePair>();
nameValuePairList.add(new BasicNameValuePair("username", username));
try {
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://powerbankk.16mb.com/check_if_friends.php");
    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairList));
    HttpResponse response = httpclient.execute(httppost);
    HttpEntity entity = response.getEntity();
    is = entity.getContent();
} catch (Exception e) {
    Log.e("log_tag", "Error in http connection" + e.toString());
}

//convert response to string
try {
    BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"), 8);
    sb = new StringBuilder();
    sb.append(reader.readLine() + "\n");
    String line = "0";

    while ((line = reader.readLine()) != null) {
        sb.append(line + "\n");
    }

    is.close();
    result = sb.toString();
} catch (Exception e) {
    Log.e("log_tag", "Error converting result " + e.toString());
}

//paring data
Double longitude, latitude;

try {
    jArray = new JSONArray(result);
    JSONObject json_data = null;

    for (int i = 0; i < jArray.length(); i++) {
        json_data = jArray.getJSONObject(i);
        longitude = json_data.getDouble("longitude");
        latitude = json_data.getDouble("latitude");
        System.out.println(longitude);
        System.out.println(latitude);
    }
} catch (JSONException e1) {
    Toast.makeText(getBaseContext(), "Location not found", Toast.LENGTH_LONG).show();
}

Answer 1

尝试在功能IOUtils.ToString()中构建以将流转换为字符串

  InputStream is = new InputStream(is, "iso-8859-1");
     String result= IOUtils.toString( is );
       jArray = new JSONArray(result);

Answer 2

我会尝试将php的最后一行更改为：

print json_encode(Array("Posts" => $rows));

并将Java修改为：

//paring data
        Double longitude, latitude;

        try {
            JSONObject object = new JSONObject(result);
            jArray = object.getJSONArray("Posts"); // get specific array from php.
            JSONObject json_data = null;

            for (int i = 0; i < jArray.length(); i++) {
                json_data = jArray.getJSONObject(i);
                longitude = json_data.getDouble("longitude");
                latitude = json_data.getDouble("latitude");
                System.out.println(longitude);
                System.out.println(latitude);
            }

        } catch (JSONException e1) {
            Toast.makeText(getBaseContext(), "Location not found", Toast.LENGTH_LONG).show();
        }

JSONArray不返回数据

2 个答案: