php - mysqli - 如何在 0 个受影响的行上返回 false

Question

即使项目未被删除，deleteItem 方法也会返回 true。在这种特殊情况下，如何使用 $mysqli->affected_rows 来检查数据库是否没有更改并返回 false？

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

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

    $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( $q != false && in_array($item_id, $userItemIDS) )
    {
        $q->bind_param("iiii", $item_id, $item_id, $item_id, $item_id);
        $q->execute();
        $q->close();
        return true;
    }
        return false;
}

Answer 1

事情本来就是这样的。

public function deleteItem($item_id)
{
    $userItemIDS = $this->helperClass->userItemIDS();
    if( !in_array($item_id, $userItemIDS) )
    {
        return FALSE;
    }
    $sql = "...";
    $q = $this->db->mysqli->prepare($sql);
    $q->bind_param("iiii", $item_id, $item_id, $item_id, $item_id);
    $q->execute();
    return (bool)$this->db->mysqli->affected_rows;
}