所以我正在创建一个 PHP 函数,它应该将数据插入到 mySQL 数据库中。这个函数之外没有错误,错误信息说它在函数内部的第一行,“$sql =”INSERT INTO ...”
createUser($conn, $name, $email, $username, $pwd) {
$sql = "INSERT INTO users (usersName, usersEmail, usersUid, usersPwd) VALUES (?, ?, ?, ?);";
$stmt = mysqli_stmt_init($conn);
if (!mysql_stmt_prepare($stmt, $sql)) {
header("location: ../signup.php?error=stmtfailed");
exit();
}
$hashedPwd = password_hash($pwd, PASSWORD_DEFAULT);
mysqli_stmt_bind_param($stmt, "ssss", $name, $email, $username, $hashedPwd);
mysqli_stmt_execute($stmt);
mysqli_stmt_close($stmt);
header("location: ../signup.php?error=none");
exit();
}
我正在使用 VSCode 和 XAMPP,如果有帮助的话......希望有人能找到一些东西,感谢搜索!