【发布时间】：2018-08-01 14:31:36
【问题描述】：
实际上在我的项目中，我被阻止了。所以，我第一次询问 Stackoverflow 社区。我是开发新手。
所以，我有一个带有我的数据的 MySql，我不想在我的应用程序中看到用户的项目。
为此，我有这个：
public class SuccessActivity extends AppCompatActivity {
public static final int CONNECTION_TIMEOUT = 10000;
public static final int READ_TIMEOUT = 15000;
private ListView listView;
protected String meubles[] = new String[100];
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_success);
Intent intent = getIntent();
String id = intent.getStringExtra("id");
this.listView = (ListView) findViewById(R.id.liste);
new SuccessActivity.Recup().execute(id);
}
//PRIVATE CLASSE POUR AFFICHER LES MEUBLES
private class Recup extends AsyncTask<String, String, String> {
HttpURLConnection conn;
URL url = null;
@Override
protected String doInBackground(String... params) {
try {
//url d'ou reside mon fichier php
url = new URL("http://opix-dev.fr/mytinyhomme/personne/afficher.meuble.php");
} catch (MalformedURLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
return "exception";
}
try {
// parametrage du HttpURLConnection pour recevoir et envoyer des donner à mysql
conn = (HttpURLConnection) url.openConnection();
conn.setReadTimeout(READ_TIMEOUT);
conn.setConnectTimeout(CONNECTION_TIMEOUT);
conn.setRequestMethod("POST");
// setDoInput and setDoOutput method depict handling of both send and receive
conn.setDoInput(true);
conn.setDoOutput(true);
// Append parameters to URL
Uri.Builder builder = new Uri.Builder()
.appendQueryParameter("id", params[0]);
String query = builder.build().getEncodedQuery();
// Open connection for sending data
OutputStream os = conn.getOutputStream();
BufferedWriter writer = new BufferedWriter(
new OutputStreamWriter(os, "UTF-8"));
writer.write(query);
writer.flush();
writer.close();
os.close();
conn.connect();
} catch (IOException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
return "exception";
}
try {
int response_code = conn.getResponseCode();
// Check if successful connection made
if (response_code == HttpURLConnection.HTTP_OK) {
// Read data sent from server
InputStream input = conn.getInputStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(input));
StringBuilder result = new StringBuilder();
String line;
while ((line = reader.readLine()) != null) {
result.append(line);
}
// Pass data to onPostExecute method
return (result.toString());
} else {
return ("unsuccessful");
}
} catch (IOException e) {
e.printStackTrace();
return "exception";
} finally {
conn.disconnect();
}
}
@Override
protected void onPostExecute(String resulta) {
//this method will be running on UI thread
if (resulta.equalsIgnoreCase("false")) {
} else if (resulta.equalsIgnoreCase("exception") || resulta.equalsIgnoreCase("unsuccessful")) {
} else {
try {
JSONArray nom = new JSONArray(resulta);
System.out.println(nom);
String meubles[] = new String[100];
for (int i = 0; i < nom.length(); i++){
JSONObject jsonobject = nom.getJSONObject(i);
meubles[i]= jsonobject.getString("nom");
System.out.println(jsonobject);
System.out.println(meubles);
item.setText( meubles[i]);
}
System.out.println(meubles);
} catch (JSONException e) {
e.printStackTrace();
}
}
}
}
}
file.php 是正确的，因为JSONArray in System.print 没问题 但是我尝试使用一些 TextView 在最后显示板，但我没有成功。
我如何使用meuble[0] , meuble[1], meuble[2](it's board of String name of items) in a ListView ?
【问题讨论】：
标签： android mysql xml database