Question

有一个无法正常工作的删除方法......任何人都可以找出它返回0的原因吗？该项目存在于表格中。

public function deleteItem($item_id)
{
    $userItemIDS = array();

    $userItemIDS = $this->helperClass->userItemIDS();

    if( !in_array($item_id, $userItemIDS) )
    {
        return false;
    }

    $q = $this->db->mysqli->prepare("DELETE i, s FROM items i
        LEFT JOIN save_list s
            ON i.id = s.item_id
        WHERE i.id = ? AND s.item_id = ?
        AND
            NOT EXISTS (SELECT id FROM pending_wins WHERE item_id = ?)
        AND
            NOT EXISTS (SELECT id FROM bids WHERE item_id = ?)");
    if($q)
    {
        $q->bind_param("iiii", $item_id, $item_id, $item_id, $item_id);
        $a_r = $q->affected_rows;
        $q->execute();
        $q->close();

        return $a_r;
    }
}

Answer 1

评论非常有帮助，我非常感谢他们。但是我的问题出在查询本身。取出AND s.item_id = ?会删除来自items表和save_list（如果有）的条目。

 $q = $this->db->mysqli->prepare("DELETE i, s FROM items i
    LEFT JOIN save_list s
        ON i.id = s.item_id
    WHERE i.id = ?
    AND
        NOT EXISTS (SELECT id FROM pending_wins WHERE item_id = ?)
    AND
        NOT EXISTS (SELECT id FROM bids WHERE item_id = ?)");

mysqli - 删除方法不删除

1 个答案: