Question

我在OOP Php中有这个代码

include ('connection.php');


class NestedSet
{
/*Properties*/

/**
 * Mysqli object
 * @var object
 */
protected $db;

/**
 * Name of the database table
 * @var string
 */
public $table = 'tree';

/**
 * Primary key of the database table
 * @var string
 */
public $pk = 'id';

/**
 * Namefield in the database table
 * @var unknown_type
 */
public $name = 'name';

/*Methods*/

/**
 * Stores a Mysqli object for further use
 * @param object $mysqli Mysqli object
 * @return boolean true
 */
public function __construct() {
    $this->db = mysqliConn::init();
    return true;
}

 protected static $instance = NULL;

// public static function get_instance()
// {
//     //if ( NULL === self::$instance )
//      //   self::$instance = new self;

//    // return self::$instance;
// }

/**
 * Creates the root node
 * @param string $name Name of the new node
 * @return boolean true
 */


public function createRootNode($name) {
    $this->db->query("LOCK TABLES " . $this->table . " WRITE");
    $sql = "SELECT rgt FROM " . $this->table . " ORDER BY rgt DESC LIMIT 1";
    $result = $this->db->query($sql);
    if ($this->db->affected_rows == 0) {
        $lft = 1;
        $rgt = 2;
    } else {
        $obj = $result->fetch_object();
        $lft = $obj->rgt + 1;
        $rgt = $lft + 1;
    }
    $sql = "INSERT INTO " . $this->table . " (" . $this->name . ", lft, rgt) VALUES ('" . $name . "', " . $lft . ", " . $rgt . ");";
    $this->db->query($sql);
    $this->db->query("UNLOCK TABLES");
    return true;
}




}

?>

我在另一个名为index.php

的文件中为类NestedSet创建一个新对象

    <?php


    include("nested_set.php");




    $nested = new NestedSet(); //Create a NestedSet object
    $nested->createRootNode('root');

    ?>

我可以在db上写，但$ rgt和$ lft保持2和1; 并显示此错误：

"Notice: Undefined property: mysqliConn::$affected_rows in C:\wamp\www\hr-test\nested_set.php on line 67"

关于我做错什么的任何想法？谢谢!!

CODE FOR connection.php

   <?php

   define('SERVER', 'localhost');
   define('USERNAME', 'root');
   define('PASSWORD', '');
   define('DATABASE', 'hr_test2');

 class mysqliConn
 {
 private static $instance;
 private $connection;

 private function __construct()
 {
    $this->connection = new mysqli(SERVER,USERNAME,PASSWORD,DATABASE);
 }

 public static function init()
 {
    if(is_null(self::$instance))
    {
        self::$instance = new mysqliConn();
    }

    return self::$instance;
 }


 public function __call($name, $args)
 {
    if(method_exists($this->connection, $name))
    {
         return call_user_func_array(array($this->connection, $name),    $args);
    } else {
         trigger_error('Unknown Method ' . $name . '()', E_USER_WARNING);
         return false;
    }
}

} ？＆GT;

Answer 1

由于mysqli->query()会返回一个mysqli_result对象，其中包含有关查询结果的信息，您需要使用$result而不是$this->db->

mysqli_result对象也不包含affected_rows属性，您应该使用确实存在的num_rows属性，而是使用$result对象。

您也可以简化您创建的查询字符串的连接，尽管您应该使用预准备语句。

public function createRootNode($name) {
    $this->db->query("LOCK TABLES " . $this->table . " WRITE");
    $sql = "SELECT rgt FROM " . $this->table . " ORDER BY rgt DESC LIMIT 1";
    $result = $this->db->query($sql);

    // if ($this->db->affected_rows == 0) {
    if ($result->num_rows == 0) {
        $lft = 1;
        $rgt = 2;
    } else {
        $obj = $result->fetch_object();
        $lft = $obj->rgt + 1;
        $rgt = $lft + 1;
    }

    $sql = "INSERT INTO {$this->table} ( {$this->name}, lft, rgt) 
            VALUES ('$name', $lft , $rgt)";

    $this->db->query($sql);
    $this->db->query("UNLOCK TABLES");
    return true;
}

Php错误。注意：未定义的属性：mysqliConn :: $ affected_rows

1 个答案: