php - sql左联接不起作用

我想从表inv_site_item中的'item_id' in inv_sie_item = 'item_code'中的inv_items中获取转移的物品总数，我也从打包表中获取了打包内容，在此查询中，只有inv_site_item存在问题，打包表才能正常工作。

错误是：Unknown column 'inv_site_item.site_id' in 'field list'

$where .= " AND inv_items.item_code = $item_code";

$query = "SELECT inv_items.*,packing.name_en `packing_name`,"
    . " COUNT(inv_site_item.site_id) `transfer_out`, COUNT(inv_site_item.location_site_id) `transfer_in`  FROM inv_items"
            . " left join "
            . "inv_packing as packing on packing.id=inv_items.packing"
            . " left join "
            . "inv_site_item as transfer on transfer.item_id=inv_items.item_code"
    . " WHERE item_code !='' " . $where . "";

最佳答案

您必须改用表别名transfer，所以：

从

inv_site_item.site_id

至

transfer.site_id

也将inv_site_item.location_site_id设置为transfer.location_site_id

关于php - sql左联接不起作用，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/35622635/