我在製作Android應用程序,並試圖從遠程數據庫中檢索數據。從遠程數據庫獲取數據Android
我的問題是如何檢查查詢結果是否包含數據並且不是空的?
我正在使用JSON,但我對它很陌生。這裏是我的代碼:
public class MainActivity extends Activity {
private TextView txt;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
LinearLayout rootLayout = new LinearLayout(getApplicationContext());
txt = new TextView(getApplicationContext());
rootLayout.addView(txt);
setContentView(rootLayout);
txt.setText("Connexion...");
txt.setText(getServerData(strURL));
}
public static final String strURL = "http://.../marecherche/adresse.php";
private String getServerData(String returnString) {
InputStream is = null;
String result = "";
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("adresse","adr casa"));
try{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(strURL);
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
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());
}
try{
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),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());
}
try{
JSONArray jArray = new JSONArray(result);
for(int i=0;i<jArray.length();i++){
JSONObject json_data = jArray.getJSONObject(i);
Log.i("log_tag","raison sociale: "+json_data.getString("raison_social")+
", Adresse: "+json_data.getString("adresse")
);
returnString += "\n\t" + jArray.getJSONObject(i);
}
}catch(JSONException e){
Log.e("log_tag", "Error parsing data " + e.toString());
}
return returnString;
}
}
和PHP文件:
<?php
mysql_connect("localhost", "root", "passwd");
mysql_select_db("dbname");
$mots = explode(' ', $_REQUEST['adresse']);
$like = array();
foreach ($mots AS $mot) {
$like[] = ' "%' . mysql_real_escape_string($mot) . '%" ';
}
$condition = implode(' OR adresse LIKE ', $like);
$sql = mysql_query("SELECT * FROM entreprise WHERE adresse like " . $condition);
while ($row = mysql_fetch_assoc($sql))
$output[] = $row;
print(json_encode($output));
mysql_close();
?>
謝謝你的回答,itried那但它給了我錯誤: – 2013-04-10 09:43:20
謝謝你的答案,它的工作,但爲測試:如果(is_array($輸出))總是正確的,我將其更改爲!empty($ output)非常感謝 – 2013-04-10 10:11:04