所以即时使用createQuery连接我的symfony数据库与Doctrine这样
// JSON ACTION
public function jsonAction()
{
$em = $this->getDoctrine()->getEntityManager();
$query = $em->createQuery('SELECT pa FROM CBMedBundle:Patients pa');
$myArray = $query->getArrayResult();
return new JsonResponse($myArray);
}
在网址:http://localhost/Symfony/web/app_dev.php/json 现在我想在Android(eclipse android)上导出结果(jsonreponse)并在那里显示响应! 我怎么能这样做?
我在谷歌上发现这个代码,关于在Android上显示json代码,但我不知道如何根据我的问题调整它。感谢
public class MainActivity extends Activity {
TextView t;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
t = (TextView) findViewById(R.id.text1);
StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
StrictMode.setThreadPolicy(policy);
t.setText(getServerData(strURL));
}
public static final String strURL = "http://10.0.2.2/connexion.php";
private String getServerData(String returnString) {
InputStream is = null;
String result ="";
ArrayList<NameValuePair> nameValuePaires = new ArrayList<NameValuePair>();
nameValuePaires.add(new BasicNameValuePair("tblville","tblville"));
// Envoie de la commande Http
try{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(strURL);
httppost.setEntity(new UrlEncodedFormEntity(nameValuePaires));
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());
}
//conversion de la réponse en chaine de caractère
try
{
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"UTF-8"));
StringBuilder sb = new StringBuilder();
String line = null;
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());
}
//recuperation des donnees json
try{
JSONArray jArray = new JSONArray(result);
for(int i=0;i<jArray.length();i++)
{
JSONObject json_data = jArray.getJSONObject(i);
Log.i("log_tag","ID_ville : "+json_data.getString("ID_ville")+" , Nom_ville : "+json_data.getString("Nom_ville")
);
returnString += "\n\t" + jArray.getJSONObject(i);
}
}
catch(JSONException e){
Log.e("log_tag","Error Parsing data"+e.toString());
}
return returnString;
}
}
编辑:JSONresponse结果
[{"id":1,"nom":"Kane","prenom":"Samba","DateNaissance":{"date":"1995-03-08 00:00:00","timezone_type":3,"timezone":"Europe\/Paris"},"LieudeNaissance":"Nouakchott","sexe":"Homme","adresse":"Tunis","profession":"Etudiant","NumerodeTel":"+21629760962","Email":"kane_samba4@yahoo.fr","SituationFamilial":"Mari\u00e9"},{"id":2,"nom":"Bebeskho","prenom":"AllStar","DateNaissance":{"date":"1995-02-08 00:00:00","timezone_type":3,"timezone":"Europe\/Paris"},"LieudeNaissance":"Bamako","sexe":"Homme","adresse":"Tunis","profession":"Cuisinier","NumerodeTel":"+2167583486","Email":"bebeskho@gmail.com","SituationFamilial":"C\u00e9libataire"},{"id":3,"nom":"Vecenzo","prenom":"Gaimno","DateNaissance":{"date":"1995-01-01 00:00:00","timezone_type":3,"timezone":"Europe\/Paris"},"LieudeNaissance":"Tunis","sexe":"Homme","adresse":"Ariana","profession":"Etudiant","NumerodeTel":"+21654758","Email":"vecenzogo","SituationFamilial":"Mari\u00e9"},{"id":4,"nom":"Diara","prenom":"Bakary","DateNaissance":{"date":"1995-06-10 00:00:00","timezone_type":3,"timezone":"Europe\/Paris"},"LieudeNaissance":"Dakar","sexe":"Homme","adresse":"Dakar Sacr\u00e9 Coeur","profession":"Etudiant","NumerodeTel":"+2214753526","Email":"diaBakis@gmail.com","SituationFamilial":"C\u00e9libataire"}]
现在,我只想在我的Android模拟器应用程序(eclipse)上显示此结果
答案 0 :(得分:1)
要根据您的需要调整代码 - 您需要更改URL以指向您的服务器。将test rax, 2049
更改为public static final String strURL = "http://10.0.2.2/connexion.php";
(我假设HTTP-GET请求返回您粘贴的JSON数组)。然后,您可以修改JSON处理代码以提取JSON文件中包含的详细信息 - 如下所示:
http://<your-server-ip-address>/Symfony/web/app_dev.php/json
但我真的建议您使用像Gson这样的东西将传入的JSON有效负载解析为&#34; Person&#34;例如,对象。作为一个例子,假设你已经创建了一个Person类,你可以这样做:
for(int i=0;i<jArray.length();i++)
{
JSONObject personJSONData= jArray.getJSONObject(i);
String id = personJSONData.getString("id");
String nom = personJSONData.getString("nom");
//do the same for the other attributes
//note that according to your JSON file, DateNaissance is a JSONObject - so you have to do something like
JSONObject dateNaissanceObject = personJSONData.getJSONObject("DateNaissance");
}
。 Here就是一个很好的例子 - here是另一个使用Gson的好例子 - 那里也是一个很好的教程here。