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我一直在試圖建立一個寄存器中,使用PHP和MySQL與Android Studio登錄Android應用程序。看來這些值正在輸入到數據庫中,但它沒有給出迴應。表的名字是哈利和PHP文件是:響應未收到對PHP文件
<?php
define('DBUSER', 'id650955_gokulm100');
define('DBPASS', 'gokulm100');
define('DBHOST', 'localhost');
define('DBNAME', 'id650955_harry');
$conn = new mysqli(DBHOST, DBUSER, DBPASS, DBNAME);
if (!$conn) {
die('error connecting to database');
}
echo 'you have created case';
if(isset($_POST["username"]) && isset($_POST["email"]) && isset($_POST["phone"]) && isset($_POST["password"]))
{
$username = $_POST["username"];
$email = $_POST["email"];
$phone = $_POST["phone"];
$password = $_POST["password"];
$statement = mysqli_prepare($conn, "INSERT INTO harry (username, email, phone, password) VALUES (?, ?, ?, ?)");
mysqli_stmt_bind_param($statement, "ssis", $username, $email, $phone, $password);
mysqli_stmt_execute($statement);
$response = array();
$response["success"] = 1;
echo json_encode($response);
}
else{
echo "not set";
}
?>
你'回聲「您創建情況」;輸出的'會的一部分。這會破壞你的json字符串。 – Philip