我正在尝试创建一个简单的登录功能，但它不起作用，而且我对 mysqli 比较陌生，所以请多多包涵。我只想检查电子邮件地址和密码是否正确，如果正确则让用户登录。在此先感谢。
这是我检查凭据的登录代码:
更新代码 - 我添加了所有报告，但现在出现内部服务器错误
<?php
require_once 'connect.php';
mysqli_report(MYSQLI_REPORT_ALL);
session_start();
if (!isset($_SESSION['email'])) {
$e = trim($_REQUEST['email']);
$email = $mysqli->real_escape_string($e);
$p = trim($_REQUEST['password']);
$password = $mysqli->real_escape_string($p);
/*
if ($result = $mysqli->query("SELECT email, password, user_id" .
" FROM users" .
" WHERE email = '$email' AND password = '$password'")) {
printf("Select returned %d rows.\n", $result->num_rows);
echo 'Total results: ' . $result->num_rows;
}
*/
if ($stmt = $mysqli->prepare("SELECT email, password, user_id FROM users WHERE email=? AND password=?")) {
/* bind parameters for markers */
$stmt->bind_param("ss", $email, $password);
/* execute query */
$stmt->execute();
/* bind result variables */
$stmt->bind_result($email, $password);
/* fetch value */
$stmt->fetch();
if ($stmt->num_rows==1) {
$row = $stmt->fetch_assoc(MYSQLI_NUM);
$user_id = $row['user_id'];
$_SESSION['user_id'] = $user_id;
$_SESSION['email'] = $email;
header("Location: home.php");
/* close statement */
$stmt->close();
} else {
printf("Error message: %s\n", $mysqli->error);
}
/*
if ($result->num_rows==1) {
$row = $result->fetch_assoc(MYSQLI_NUM);
$user_id = $row['user_id'];
if ($query_group = $mysqli->query("SELECT *" .
" FROM user_groups" .
" WHERE user_id = '".$user_id."'")) {
//No more setcookie
$_SESSION['user_id'] = $user_id;
$_SESSION['email'] = $email;
} else {
echo 'Did not work';
}
}
*/
/* free result set */
// $result->close();
}
}
?>
这里是连接到数据库的 connect.php:
<?php
$mysqli = new mysqli("data", "username", "password", "db");
if($mysqli->connect_errno > 0){
die('Unable to connect to database [' . $mysqli->connect_error . ']');
}
?>
最佳答案
您是否尝试删除代码末尾的括号之一。我猜你必须删除其中一个。
关于php 登录表单无法正常工作，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/22309688/