得到一个不工作的删除方法…有人知道为什么它返回0
吗?
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;
}
}
这些评论很有帮助,我非常感激。然而,我的问题是在查询本身。取出AND s.item_id = ?
将删除
items
表和save_list
表中获取。
$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 = ?)");