Question

我读了很多类似的话题，但我无法理解我的错误。

问题：

Uncaught exception 'PDOException' with message 'SQLSTATE[HY093]: Invalid parameter number: parameter was not defined' in I:\home\lm.ru\www\classes\db.class.php:39 Stack trace: #0 I:\home\lm.ru\www\classes\db.class.php(39): PDOStatement->execute(Array) #1 I:\home\lm.ru\www\classes\page.class.php(34):

代码：

class Page
{
    public $id = "";
    public $link = ""; //+
    public $links_count = 0;//+
    public $places_count = 0;
    public $external_links = 0; //+
    public $context = 0;
    public $has_text = 0;
    public $text = "";
    public $ya_index = 0;
    public $ya_index_data = 0;
    public $google_pr = 0;
    public $google_pr_date = 0;
    public $level = 0; // 0 - main, 1 - 2 level... +
    public $title = ""; // +
    public $is_active = 1;
    public $deleted = 0;
    public $site_id = 0; //+

    function Add($data)
    {
        foreach($data as $key => $value)
        {
            $this->$key = $value;
        }

        print_r((array)$this);

        PDO_wrap::Change("INSERT INTO pages (id, link, links_count, places_count, external_links, context, has_text, text, ya_index, ya_index_date, google_pr, google_pr_date, level, title, is_active, deleted, site_id) VALUES(:id, :link, :links_count, :places_count, :external_links, :context, :has_text, :text, :ya_index, :ya_index_date, :google_pr, :google_pr_date, :level, :title, :is_active, :deleted, :site_id)", (array)$this);

        $this->id = PDO_wrap::LastInsertId();
    }

}

PDO_wrap方法：

public static function Change($query, $params = array())
    {
        $sql = self::$pdo->prepare($query);
        $sql->execute($params);
        return $sql->rowCount();
    }

Answer 1

类道具和值列表之间存在不匹配

$ya_index_data;

vs

:ya_index_date

Answer 2

我遇到了同样的问题。 PDO要求数组中的条目数与SQL语句中的元素数相匹配。我试图通过给PDO $ _POST数组做同样的事情。只有在将元素复制到另一个数组后，PDO才会接受该数组。

带有消息'SQLSTATE [HY093]的未捕获异常'PDOException'：参数号无效：参数未定义'

2 个答案: