[英]JSON data cannot be retrieved in android from php
我想获取已登录用户的详细信息。我在警报对话框中得到null或无值(变量-名称)
我对所有事物都是陌生的,所以很抱歉,如果这个问题不值得,我已经尝试找到解决方案,但是还没有运气,请提供帮助。
if (type.equals("linkuser")) {
try {
URL url = new URL(Link_url);
HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
httpURLConnection.setRequestMethod("POST");
httpURLConnection.setDoOutput(true);
httpURLConnection.setDoInput(true);
OutputStream outputStream = httpURLConnection.getOutputStream();
BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(outputStream, "UTF-8"));
String post_data = URLEncoder.encode("user", "UTF-8") + "=" + URLEncoder.encode(user, "UTF-8");
bufferedWriter.write(post_data);
bufferedWriter.flush();
bufferedWriter.close();
outputStream.close();
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();
}
Your php query is wrong
Initialize array first Note that $result is not your array
$menus= array();
while($row = mysqli_fetch_array($result )){
array_push($menus,
array('name'=>$row[0],
'ign'=>$row[1],
'email'=>$row[2]
));
}
mysqli_close($con);
echo json_encode(array("menu"=>$menus));
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.