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錯誤,我有我的PHP代碼面向: -未定義指數:在PHP代碼(XAMPP)
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
以下是傳遞價值PHP文件java代碼:
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代碼。我從我的android應用程序接受值。
<?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($con);
?>
請建議我如何解決這個問題?
那麼,什麼是你的問題嗎?你得到的通知的原因在StackOverflow中僅在這裏解釋了53492649263484個答案:'$ _GET'數組根本沒有這樣的密鑰,這意味着它們沒有被髮送。 – arkascha
另一方面:您想了解「sql注入漏洞」以及如何使用「準備好的語句」和「參數綁定」來防止它。 – arkascha
你可以在這裏發佈解決方案 – Aravind