使用 PHP 的单个 MySQL 插入正在输入多个记录

Question

PHP 的 OOP 新手，有点迷茫，在此先感谢。

我遇到的问题是，当我创建新用户时，它会为同一用户插入 5 条记录（每个用户都有新的唯一主 ID）。发现很多人试图学习插入多条记录的问题，但认为这是在此处发布的第一个关于仅尝试插入一条记录时将多条记录插入 MySQL 数据库的问题！

如果您需要查看任何其他代码，请告诉我——谢谢！

用户类

class User{

protected static $table_name="users";
public $id;
public $username;
public $password;
public $first_name;
public $last_name;
public $email;

public static function find_all(){
    return self::find_by_sql("SELECT * FROM users");    
}

public static function find_by_id($id=0){
    global $database;
    $result_array = self::find_by_sql("SELECT * FROM users WHERE id={$id} LIMIT 1");
    return !empty($result_array) ? array_shift($result_array) : false;
}

public static function find_by_sql($sql=""){
    global $database;
    $result_set = $database->query($sql);
    $object_array = array();
    while ($row = $database->fetch_array($result_set)){
        $object_array[] = self::instantiate($row);  
    }
    return $object_array;
}

public static function authenticate($username="", $password=""){
    global $database;
    $username = $database->escape_value($username);
    $password = $database->escape_value($password);
    $sql  = "SELECT * FROM users ";
    $sql .= "WHERE username = '{$username}' ";
    $sql .= "AND password = '{$password}' ";
    $sql .= "LIMIT 1";
    $result_array = self::find_by_sql($sql);
    return !empty($result_array) ? array_shift($result_array) : false;
}

public function full_name(){
    if(isset($this->first_name) && isset($this->last_name)){
        return $this->first_name . " " . $this->last_name;  
    } else {
        return "";  
    }
}

private static function instantiate($record){
    $object = new self;
    foreach($record as $attribute=>$value){
        if($object->has_attribute($attribute)){
            $object->$attribute = $value;   
        }
    }
    return $object;
}

private function has_attribute($attribute){
    $object_vars = get_object_vars($this);
    return array_key_exists($attribute, $object_vars);
}

public function create(){
    global $database;
    $sql  = "INSERT INTO users (";
    $sql .= "username, password, first_name, last_name, email";
    $sql .= ") VALUES ('";
    $sql .= $database->escape_value($this->username) ."', '";
    $sql .= $database->escape_value($this->password) ."', '";
    $sql .= $database->escape_value($this->first_name) ."', '";
    $sql .= $database->escape_value($this->last_name) ."', '";
    $sql .= $database->escape_value($this->email) ."')";
    if($database->query($sql)){
        $this->id = $database->insert_id();
        return true;
    } else {
        return false;
    }
}
}

数据库类

class MySQLDatabase{

private $connection;

function __construct(){
    $this->open_connection();   
}

public function open_connection(){
    $this->connection = mysqli_connect(DB_SERVER, DB_USER, DB_PASS, DB_NAME);
    if(mysqli_connect_errno()){
        die("Database Connection Failed: " .
             mysqli_connect_error() .
             " (" . mysqli_connect_errno() . ")"
        );
    }
}

public function close_connection(){
    if(isset($this->connection)){
        mysqli_close($this->connection);
        unset($this->connection);
    }
}

public function query($sql){
    $result = mysqli_query($this->connection, $sql);
    $this->confirm_query($result);
    return $result;
}

private function confirm_query($result){
    if(!$result) {
        die("Database query failed.");  
    }
}

public function escape_value($string){      
    $escaped_string = mysqli_real_escape_string($this->connection, $string);
    return $escaped_string;
}

public function fetch_array($result_set){
    return mysqli_fetch_array($result_set); 
}

public function num_rows($result_set){
    return mysqli_num_rows($result_set);    
}

public function insert_id(){
    return mysqli_insert_id($this->connection);
}

public function affected_rows(){
    return mysqli_affected_rows($this->connection); 
}

}

实例化

$user = new User();
$user->username = "test";
$user->password = "testing";
$user->first_name = "Bob";
$user->last_name = "Smith";
$user->email = "bsmith@gmail.com";
$user->create();

Answer 1

Blokeish 是正确的，问题在于框架中在定义之前发生的包含/要求。