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我已經成立了一個PHP腳本來創建JSON here，但是當我嘗試顯示JSONArray我有一些像這樣的錯誤在我的Android監..ANDROID - 如何顯示JSONArray到TextView的
值（HTML）（正文）（java.lang.String類型的腳本不能轉換爲JSONArray
有人可以告訴我怎麼解決？
MainActivity.java
package flix.yudi.okhttp1;
import android.app.ProgressDialog;
import android.os.AsyncTask;
import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;
import android.util.Log;
import android.widget.ListAdapter;
import android.widget.ListView;
import android.widget.SimpleAdapter;
import android.widget.Toast;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;
import java.util.ArrayList;
import java.util.HashMap;
public class MainActivity extends AppCompatActivity {
private String TAG = MainActivity.class.getSimpleName();
private ProgressDialog pDialog;
private ListView lv;
// URL to get contacts JSON
private static String url = "http://zxccvvv.cuccfree.com/send_data.php";
ArrayList<HashMap<String, String>> contactList;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
contactList = new ArrayList<>();
lv = (ListView) findViewById(R.id.list);
new GetContacts().execute();
}
/**
* Async task class to get json by making HTTP call
*/
private class GetContacts extends AsyncTask<Void, Void, Void> {
@Override
protected void onPreExecute() {
super.onPreExecute();
// Showing progress dialog
pDialog = new ProgressDialog(MainActivity.this);
pDialog.setMessage("Please wait...");
pDialog.setCancelable(false);
pDialog.show();
}
@Override
protected Void doInBackground(Void... arg0) {
HttpHandler sh = new HttpHandler();
// Making a request to url and getting response
String jsonStr = sh.makeServiceCall(url);
Log.e(TAG, "Response from url: " + jsonStr);
if (jsonStr != null) {
try {
JSONArray jsonObj = new JSONArray(jsonStr);
// Getting JSON Array node
JSONArray pertanyaan = jsonObj.getJSONArray("pertanyaan");
// looping through All Contacts
for (int i = 0; i < pertanyaan.length(); i++) {
JSONObject c = pertanyaan.getJSONObject(i);
String id = c.getString("id");
String ask = c.getString("ask");
// tmp hash map for single contact
HashMap<String, String> pertanyaans = new HashMap<>();
// adding each child node to HashMap key => value
pertanyaans.put("id", id);
pertanyaans.put("ask", ask);
// adding contact to contact list
contactList.add(pertanyaans);
}
} catch (final JSONException e) {
Log.e(TAG, "Json parsing error: " + e.getMessage());
runOnUiThread(new Runnable() {
@Override
public void run() {
Toast.makeText(getApplicationContext(),
"Json parsing error: " + e.getMessage(),
Toast.LENGTH_LONG)
.show();
}
});
}
} else {
Log.e(TAG, "Couldn't get json from server.");
runOnUiThread(new Runnable() {
@Override
public void run() {
Toast.makeText(getApplicationContext(),
"Couldn't get json from server. Check LogCat for possible errors!",
Toast.LENGTH_LONG)
.show();
}
});
}
return null;
}
@Override
protected void onPostExecute(Void result) {
super.onPostExecute(result);
// Dismiss the progress dialog
if (pDialog.isShowing())
pDialog.dismiss();
/**
* Updating parsed JSON data into ListView
* */
ListAdapter adapter = new SimpleAdapter(
MainActivity.this, contactList,
R.layout.list_item, new String[]{"ask"}, new int[]{R.id.ask});
lv.setAdapter(adapter);
}
}
}
HttpHandler.java
package flix.yudi.okhttp1;
import android.util.Log;
import java.io.BufferedInputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.ProtocolException;
import java.net.URL;
public class HttpHandler {
private static final String TAG = HttpHandler.class.getSimpleName();
public HttpHandler() {
}
public String makeServiceCall(String reqUrl) {
String response = null;
try {
URL url = new URL(reqUrl);
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setRequestMethod("GET");
// read the response
InputStream in = new BufferedInputStream(conn.getInputStream());
response = convertStreamToString(in);
} catch (MalformedURLException e) {
Log.e(TAG, "MalformedURLException: " + e.getMessage());
} catch (ProtocolException e) {
Log.e(TAG, "ProtocolException: " + e.getMessage());
} catch (IOException e) {
Log.e(TAG, "IOException: " + e.getMessage());
} catch (Exception e) {
Log.e(TAG, "Exception: " + e.getMessage());
}
return response;
}
private String convertStreamToString(InputStream is) {
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
StringBuilder sb = new StringBuilder();
String line;
try {
while ((line = reader.readLine()) != null) {
sb.append(line).append('\n');
}
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
is.close();
} catch (IOException e) {
e.printStackTrace();
}
}
return sb.toString();
}
}
我從here
任何想法的參考代碼/另一參考方法來解決？
編輯
send_data.php
<?php
include 'dbconfig.php';
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, ask FROM pertanyaan";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row[] = $result->fetch_assoc()) {
$json = json_encode($row);
}
} else {
echo "0 results";
}
mysql_close($dbname);
echo $json;
?>
我應該刪除我> JSONArray pertanyaan = jsonObj.getJSONArray（ 「pertanyaan」）;先生？ – Flix
@Flix是啊你不需要它，並且在'pertanyaan'處使用'jsonObj'就像我的帖子 –
中顯示的那樣仍然會出現同樣的錯誤先生，我應該聲明我的PHP代碼來創建JSON嗎？ – Flix