这大约在 5 分钟前起作用，但突然停止了。这是一个简单的登录表单，请参阅下面的代码
<?php
// Initialize the session
session_start();
// Check if the user is already logged in, if yes then redirect him to welcome page
if(isset($_SESSION["loggedin"]) && $_SESSION["loggedin"] === true){
header("location: welcome.php");
exit;
}
// Include config file
require_once "db/config.php";
// Define variables and initialize with empty values
$username = $password = "";
$username_err = $password_err = "";
// Processing form data when form is submitted
if($_SERVER["REQUEST_METHOD"] == "POST"){
// Check if username is empty
if(empty(trim($_POST["username"]))){
$username_err = "Please enter username.";
} else{
$username = trim($_POST["username"]);
}
// Check if password is empty
if(empty(trim($_POST["password"]))){
$password_err = "Please enter your password.";
} else{
$password = trim($_POST["password"]);
}
// Validate credentials
if(empty($username_err) && empty($password_err)){
// Prepare a select statement
$sql = "SELECT id, username, password FROM users WHERE username = ?";
if($stmt = mysqli_prepare($link, $sql)){
// Bind variables to the prepared statement as parameters
mysqli_stmt_bind_param($stmt, "s", $param_username);
// Set parameters
$param_username = $username;
// Attempt to execute the prepared statement
if(mysqli_stmt_execute($stmt)){
// Store result
mysqli_stmt_store_result($stmt);
// Check if username exists, if yes then verify password
if(mysqli_stmt_num_rows($stmt) == 1){
// Bind result variables
mysqli_stmt_bind_result($stmt, $id, $username, $hashed_password);
if(mysqli_stmt_fetch($stmt)){
if(password_verify($password, $hashed_password)){
// Password is correct, so start a new session
session_start();
// Store data in session variables
$_SESSION["loggedin"] = true;
$_SESSION["id"] = $id;
$_SESSION["username"] = $username;
// Redirect user to welcome page
header("location: welcome.php");
} else{
// Display an error message if password is not valid
$password_err = "The password you entered was not valid.";
}
}
} else{
// Display an error message if username doesn't exist
$username_err = "No account found with that username.";
}
} else{
echo "Oops! Something went wrong. Please try again later.";
}
}
// Close statement
mysqli_stmt_close($stmt);
}
// Close connection
mysqli_close($link);
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Login</title>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.css">
<style type="text/css">
body{ font: 14px sans-serif; }
.wrapper{ width: 350px; padding: 20px; }
</style>
</head>
<body>
<div class="wrapper">
<h2>Login</h2>
<p>Please fill in your credentials to login.</p>
<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">
<div class="form-group <?php echo (!empty($username_err)) ? 'has-error' : ''; ?>">
<label>Username</label>
<input type="text" name="username" class="form-control" value="<?php echo $username; ?>">
<span class="help-block"><?php echo $username_err; ?></span>
</div>
<div class="form-group <?php echo (!empty($password_err)) ? 'has-error' : ''; ?>">
<label>Password</label>
<input type="password" name="password" class="form-control">
<span class="help-block"><?php echo $password_err; ?></span>
</div>
<div class="form-group">
<input type="submit" class="btn btn-primary" value="Login">
</div>
<p>Don't have an account? <a href="register.php">Sign up now</a>.</p>
</form>
</div>
</body>
</html>
我得到的错误是 Warning: mysqli_stmt_close() expects parameter 1 to be mysqli_stmt, boolean given in /opt/lampp/htdocs/magic/client/login.php on line 83
这大约在 5 分钟前起作用，突然停止了，不知道为什么。有任何想法吗？
当它工作时，它只是将用户重定向到welcome.php，现在它不是，并给我那个错误。代码取自https://www.tutorialrepublic.com/php-tutorial/php-mysql-login-system.php
看看你问的是什么：
if($stmt = mysqli_prepare($link, $sql)){
// [...]
}
mysqli_stmt_close($stmt);
你总是试图关闭你的语句，即使它没有成功创建。您需要移动该关闭尝试：
if($stmt = mysqli_prepare($link, $sql)){
// [...]
mysqli_stmt_close($stmt);
}
此外，您首先收到此错误这一事实意味着您的 SQL 查询未能准备好，因此其中存在某种错误。您目前没有检查它，如果您想弄清楚为什么您的代码不起作用，这可能会让您感到困难。我建议为此添加一个检查：
if($stmt = mysqli_prepare($link, $sql)){
// [...]
mysqli_stmt_close($stmt);
} else {
echo "Something's wrong with the query: " . mysqli_error($link);
}
请注意，在生产代码中，您不想mysqli_error()直接输出 ，因为它可能会显示有关您的数据库的信息，而这些信息不是任何人的业务而是您的业务，因此请仅在调试时执行此操作。