Question

我有一个类命名用户，其代码如下

require_once(LIB_PATH.DS.'database.php');
class User extends DatabaseObject {
    protected static $table_name="users";
    public $id;
    public $username;
    public $password;
    public $first_name;
    public $last_name;
 }

我从一个DatabaseObject类中扩展了这个User类，我刚刚在下面创建了一个Create方法，我为数据库CRUD提供了常用的方法是我的DatabaseObject类

require_once(LIB_PATH.DS.'database.php');

class DatabaseObject {

   public static function attributes() { 
        $class_name = get_called_class();
        $object = new $class_name;
        return get_object_vars($object);
    }

   public static function sanitized_attributes() {
      global $database;
      $clean_attributes = array();
      foreach(static::attributes() as $key => $value){
         $clean_attributes[$key] = $database->escape_value($value);
      }
   return $clean_attributes;
  }

    public static function create() {
        global $database;
        $attributes = static::sanitized_attributes();
      $sql = "INSERT INTO ".static::$table_name." (";
      $sql .= join(", ", array_keys($attributes));
      $sql .= ") VALUES ('";
      $sql .= join("', '", array_values($attributes));
      $sql .= "')";
      if($database->query($sql)) {
        return $sql;
      } else {
     return false;
      }
   }
}

当我从test.php运行测试时，检查Create是否有效使用此代码

 $user = new User();
 $user->username = "johnsmith";
 $user->password = "abcd12345";
 $user->first_name = "John";
 $user->last_name = "Smith";
 echo User::create();

它只返回带有空值的查询

   INSERT INTO users 
  (id, username, password, first_name, last_name) VALUES ('', '', '', '', '')

当我检查phpmyadmin时插入一行但是空值是什么，这是我正在制作的错误请帮助

此致

Tapos

Answer 1

您的create()函数是static，因此它与您正在处理的实例没有关联。

然后，您的sanitized_attributes()会创建类[$object = new $class_name;]的无关实例，并对其属性进行迭代，根据定义，这些属性为空。

顺便说一下，用户准备了凌乱的字符串连接操作语句：PDO正等着你！

Answer 2

将您的方法更改为public删除static关键字，并使用下面的代码将指针引用传递给create方法

    $user = new User();
 $user->username = "johnsmith";
 $user->password = "abcd12345";
 $user->first_name = "John";
 $user->last_name = "Smith";
 echo $user->create($user);

当您将对象传递给thye create方法时，其余代码将如下所示

    public function attributes(&$obj) { 
      return get_object_vars($obj);
     }

public function sanitized_attributes(&$obj) {
 global $database;
  $clean_attributes = array();

foreach($obj->attributes($obj) as $key => $value){
    $clean_attributes[$key] = $database->escape_value($value);
  }
  return $clean_attributes;
}


public function create(&$obj) {
    global $database;

    $attributes = $obj->sanitized_attributes($obj);
   $sql = "INSERT INTO ".static::$table_name." (";
    $sql .= join(", ", array_keys($attributes));
  $sql .= ") VALUES ('";
    $sql .= join("', '", array_values($attributes));
    $sql .= "')";
  if($database->query($sql)) {
   } else {
    return false;
  }$obj->sanitized_attributes($obj);
}

希望这有效！

类抽象不起作用

2 个答案: