if(isset($_POST["submit"]))
{
// Details for inserting into the database
$id = htmlentities($_POST["id"]);
$firstname = htmlspecialchars($_POST["firstname"]);
$lastname = htmlspecialchars($_POST["lastname"]);
$username = htmlspecialchars($_POST["username"]);
$password = htmlspecialchars($_POST["password"]);
// Dealing with inserting
$query = "INSERT INTO `myDatabaseForAll`.`users` (`id`, `firstname`, `lastname`, `username`, `password`)
VALUES (NULL, $firstname, $lastname,$username,$password)";
$result = mysqli_query($connection,$query);
if(!$result) {
die('There were some errors '.mysqli_error($connection));
} else {
redirect_to('index.php');
}
}
$query = "INSERT INTO `myDatabaseForAll`.`users` (`id`, `firstname`, `lastname`, `username`, `password`)
VALUES (NULL, $firstname, $lastname,$username,$password)";
在sql查询中，需要在文本假框周围使用单引号
将上述查询更改为
$query = "INSERT INTO `myDatabaseForAll`.`users` (`id`, `firstname`, `lastname`, `username`, `password`)
VALUES (NULL, '$firstname', '$lastname','$username','$password')";
发生此错误的原因可能是表列名与查询中给定的名称不匹配。如果不是这样，请尝试下面的代码。
$query = "INSERT INTO `myDatabaseForAll`.`users` (id, firstname, lastname, username, password)
VALUES (NULL, '".$firstname."', '".$lastname."','".$username."','".$password."')";
在您的查询中，您将一些字符串部分与php变量连接起来。因此，应该使用连接运算符以上述方式进行连接。