我从包含我的数据库的网络主机收到这些警告。我正在尝试在Android工作室中开发一个Android应用程序,以便将数据从注册用户活动发送到数据库。我想我有一个PHP脚本错误..
以下是我注册用户的PHP代码:
<?php
$con = mysqli_connect("localhost", "user", "pass", "db");
if (isset($_POST["name"], $_POST["email"], $_POST["username"], $_POST["password"]))
{
$name = $_POST["name"];
$email = $_POST["email"];
$username = $_POST["username"];
$password = $_POST["password"];
}
$statement = mysqli_prepare($con, "INSERT INTO user (name, username, email, password) VALUES (?, ?, ?, ?)");
mysqli_stmt_bind_param($statement, "siss", $name, $username, $email, $password);
mysqli_stmt_execute($statement);
$response = array();
$response["success"] = true;
echo json_encode($response);
?>
答案 0 :(得分:0)
您已检查错误: -
<?php
//comment these two lines when code started working fine
error_reporting(E_ALL);
ini_set('display_errors',1);
$con = mysqli_connect("localhost", "id2833909_split421", "pass123", "id2833909_splitw");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
if (isset($_POST["name"], $_POST["email"], $_POST["username"], $_POST["password"])) {
$name = $_POST["name"];
$email = $_POST["email"];
$username = $_POST["username"];
$password = $_POST["password"];
$statement = mysqli_prepare($con, "INSERT INTO `user` (`name`, `username`, `email`, `password`) VALUES (?, ?, ?, ?)");
mysqli_stmt_bind_param($statement, "ssss", $name, $username, $email, $password); // i need to be s
$response = array();
if(mysqli_stmt_execute($statement)){
$response["message"] = "success";
}else{
$response["message"] = "error";
}
echo json_encode($response);
}
?>