【发布时间】:2016-11-30 23:23:21
【问题描述】:
我正在尝试从我的数据库中检索数据,我编写了这段代码
<?php
require "conn.php";
$mysql_qry = "select * from domande";
$res = mysqli_query($conn,$mysql_qry);
$result = array();
while($row = mysqli_fetch_array($res)){
array_push($result,
array('id_domanda'=>$row[0],
'domanda'=>$row[1],
'username'=>$row[2]
));
}
echo json_encode(array("result"=>$result));
mysqli_close($conn);
?>
它应该返回一个数组。我已经能够以这种方式在警报对话框中显示数组:
class BackgroundWorkerDomande extends AsyncTask<String,Void,String> {
Context context;
AlertDialog alertDialog;
BackgroundWorkerDomande (Context ctx){
context = ctx;
}
@Override
protected String doInBackground(String... params) {
String type = params[0];
if (type=="getquestion"){
try {
URL url = new URL(params[3]);
HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
httpURLConnection.setRequestMethod("POST");
httpURLConnection.setDoOutput(true);
httpURLConnection.setDoInput(true);
InputStream inputStream = httpURLConnection.getInputStream();
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream, "iso-8859-1"));
String result = "";
String line = "";
while ((line = bufferedReader.readLine()) != null) {
result += line;
}
bufferedReader.close();
inputStream.close();
httpURLConnection.disconnect();
return result;
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
return null;
}
@Override
protected void onPreExecute() {
alertDialog = new AlertDialog.Builder(context).create();
alertDialog.setTitle("Status");
}
@Override
protected void onPostExecute(String result) {
alertDialog.setMessage(result);
alertDialog.show();
}
@Override
protected void onProgressUpdate(Void... values) {
super.onProgressUpdate(values);
}
}
但是如果我想将结果插入到 MainActivity 中声明的新数组中,我该怎么办?我只是按照互联网上的一些教程来编写这段代码,所以我并不完全知道它是如何工作的。 (对不起我的英语不好)
【问题讨论】: