运行php代码后,屏幕上会显示hello1,但不会显示hello2。我想我的连接代码有问题。
我找不到我的代码有什么问题。不幸的是,我的代码似乎正确,即使经过多次检查。我该怎么修?
顺便说一句,我正在MacBookAir上运行mamp。
<?php
echo "hello1";
$connect = mysqli_connect("localhost:8888", "Capstone", "", "capstone");
$mysqli->set_charset('utf8');
echo "hello2";
if (!$connect) {
printf("Connection failed: %s\n", $mysqli->connect_error);
die();
echo "hello3";
}
session_start();
if (isset($_POST["Sign Up"]))
{
if (empty($_POST["Email"]) || empty($_POST["Password"]))
{
echo '<script> alert ("Both Feldsa are required)</script">';
}
else
{
$_SESSION['email'] = $_POST['Email'];
$_SESSION['password'] = $_POST['Password'];
$_SESSION['Repeatpassword'] = $_POST['Repeatpassword'];
$_SESSION['name'] = $_POST['name'];
$_SESSION['weight'] = $_POST['weight'];
$_SESSION['feet'] = $_POST['feet'];
$_SESSION['inches'] = $_POST['inches'];
$_SESSION['age'] = $_POST['age'];
$_SESSION['goal'] = $_POST['Goal'];
// Escape all $_POST variables to protect against SQL injection
$email = $mysqli->escape_string($_POST['email']);
$password = $mysqli->escape_string(password_hash($_POST['password'], PASSWORD_BCRYPT));
$RepPassword = $mysqli->escape_string(password_hash($_POST['Repeatpassword'], PASSWORD_BCRYPT));
$name = $mysqli->escape_string($_POST['name']);
$Weight = $mysqli->escape_string($_POST['weight']);
$feet = $mysqli->escape_string($_POST['feet']);
$inches = $mysqli->escape_string($_POST['inches']);
$age = $mysqli->escape_string($_POST['age']);
$goal = $mysqli->escape_string($_POST['goal']);
$hash = $mysqli->escape_string(md5(rand(0, 1000)));
// Check if user with that email already exists
// We know user email exists if the rows returned are more than 0
$result = $mysqli->query("SELECT * FROM User WHERE Email_Address='$email'") or die($mysqli->error);
if ($result->num_rows > 0) {
$_SESSION['message'] = 'User with this email already exists!';
}
else { // Email doesn't already exist in a database, proceed...
// active is 0 by DEFAULT (no need to include it here)
$sql = "INSERT INTO User (Email_Address, Password, Full Name, Weight, Feet, Inches, Age, Goal, hash) "
. "VALUES ('$email', 'password', 'name', 'Weight', 'feet', 'inches', 'age', 'goal', 'hash')";
}
if (! $mysqli->query($sql)
{
$_SESSION['message'] = 'Registration successfully';
echo $_SESSION['message'];
header("location: loginaccount.html");
}
}
else {
$_SESSION['message'] = 'Registration failed!';
echo $_SESSION['message'];
}
}
if (isset($_POST["Login"]))
{
$email = $mysqli->escape_string($_POST['Email']);
$result = $mysqli->query("SELECT * FROM User WHERE Email_Address='$email'");
if ($result->num_rows == 0) { //
{
$_SESSION['message'] = "User with that email doesn't exist!";
echo $_SESSION['message'];
}
else {
$user = $result->fetch_assoc();
if (password_verify($_POST['password'], $user['Password'])) {
$_SESSION['email'] = $user['Email_Address'];
$_SESSION['name'] = $user['Full Name'];
$_SESSION['weight'] = $user['Weight '];
$_SESSION['feet'] = $user['Feet '];
$_SESSION['inches'] = $user['Inches '];
$_SESSION['age'] = $user['Age '];
$_SESSION['goal'] = $user['Goal '];
$_SESSION['logged_in'] = true;
$_SESSION['active'] = $user['Active'];
header("location: loginaccount.html");
}
}
mysqli_close($connect);
session_destroy();
?>
2条答案
按热度按时间gtlvzcf81#
在脚本开始时:
在这里的第3行,您尝试使用$mysqli。这个变量不存在。您还没有声明它,所以在这一点上,当您尝试引用一个对象的方法时,您将得到一个php运行时错误,这个对象实际上是一个不存在的变量。
实际上比这更糟,因为你把程序mysqli和面向对象的mysqli混在一起了。您真正需要的是这个,但明显的问题是mysqli连接变量的名称是
$connect
!nlejzf6q2#
你也可以使用
try/catch
查找有关错误的详细信息p、 s.-英寸
$mysqli->set_charset("utf-8");
$mysqli
未定义,请使用$connect
在这里