このコードを試していましたが、null になります。場所の名前を教えてくれるはずです。私のコードの問題は何ですか?場所IDを指定して場所の詳細を取得したい。しかし、私もデバッグしましたが、常にnullを返していました
コード
String result = "";
InputStream is = null;
// the year data to send
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("place_id", "2"));
// http post
try {
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(
"http://example.com/getAllPeopleBornAfter.php");
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());
}
// convert response to string
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());
}
// parse json data
try {
JSONArray jArray = new JSONArray(result);
for (int i = 0; i < jArray.length(); i++) {
JSONObject json_data = jArray.getJSONObject(i);
TextView z = (TextView)findViewById(R.id.textView1);
z.setText(json_data.getString("place_id"));
}
} catch (JSONException e) {
Log.e("log_tag", "Error parsing data " + e.toString());
}
}
});
PHP
<?php
include "db_config.php";
$q=mysql_query("SELECT 'name' FROM places WHERE place_id='".$_REQUEST['place_id']."'");
while($e=mysql_fetch_assoc($q))
$output[]=$e;
print(json_encode($output));
mysql_close();
?>