Question

错误我的PHP代码面临的问题： -

    Notice: Undefined index: latitude in C:\xampp1\htdocs\tutorialspoint\pass.php on line 9
    Notice: Undefined index: longitude in C:\xampp1\htdocs\tutorialspoint\pass.php on line 10
     success

以下是将值传递给.php文件的java代码：

public void writePost(double latitude,double longitude){


    String urlSuffix = "&latitude="+latitude+"&longitude="+longitude;
    class RegisterUser extends AsyncTask<String, Void, String> {




        @Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);

        }

        @Override
        protected String doInBackground(String... params) {
            String s = params[0];
            BufferedReader bufferedReader = null;
            try {
                URL url = new URL(GETWRITE_URL+s); //getwriteurl= url of ur php uptop .php
                HttpURLConnection con = (HttpURLConnection) url.openConnection();
                con.setDoInput(true);
                bufferedReader = new BufferedReader(new InputStreamReader(con.getInputStream()));

                String result;

                result = bufferedReader.readLine();
                return result;
            }catch(Exception e){
                return null;
            }
        }
    }

    RegisterUser ru = new RegisterUser();
    ru.execute(urlSuffix);
}

以下是我的php代码。我接受来自我的Android应用程序的值。

 <?php
 $con=mysqli_connect("localhost","******","******","route69");

 if (mysqli_connect_errno($con))
 {
 echo "Failed to connect to MySQL: " . mysqli_connect_error();
 }

 $latitude = $_GET['latitude']; // accepting the current latitude from android app
 $longitude = $_GET['longitude'];//accepting the current longitude from android app
 $check = "INSERT INTO taxi1(Latitude, Longitude) values('$latitude','$longitude')";
 $result = mysqli_query($con,$check);

 if(isset($result)){
 echo "success";
 } else {
 echo "failed";
 }
 mysqli_close($con);
 ?>

请建议我如何解决问题？。

Answer 1

基本上，您要使用的$_GET变量未设置。为了确保应用程序传递正确的变量，请使用var_dump($_GET);查看它传递的变量。

有关错误的详细信息，请参阅：php notice undefined variable and notice undefined index

未定义的索引：在php代码中（xampp）

1 个答案: