Question

我有以下SQL查询

SELECT *
FROM inventory_audits
WHERE id in (
SELECT distinct inventory_audits.id
FROM inventory_audits
LEFT JOIN pick_lists ON (pick_lists.number = substr(inventory_audits.reference,0,8))
LEFT JOIN pick_list_items ON (pick_list_items.pick_list_id = pick_lists.id)
LEFT JOIN load_plan_items ON (load_plan_items.load_plan_id = pick_list_items.load_plan_id)
LEFT JOIN sales_order_items ON (load_plan_items.sales_order_item_id = sales_order_items.id)
LEFT JOIN invoices ON (invoices.id = sales_order_items.invoice_id)
WHERE inventory_adjustment_reason_description = 'Picked' 
AND (invoices.number = '230367' OR invoices.number = '230368')
)

并且，我需要将其转换为推进结果为

$cr = new Criteria();
$cr->addJoin(PickListPeer::ID, PickListItemPeer::PICK_LIST_ID);
$cr->addJoin(PickListItemPeer::LOAD_PLAN_ID, LoadPlanItemPeer::LOAD_PLAN_ID);
$cr->addJoin(LoadPlanItemPeer::SALES_ORDER_ITEM_ID, SalesOrderItemPeer::ID);
$cr->addJoin(SalesOrderItemPeer::INVOICE_ID, InvoicePeer::ID);
$cr->add(InventoryAuditPeer::INVENTORY_ADJUSTMENT_REASON_DESCRIPTION, 'Picked');
$cr->add(InvoicePeer::NUMBER, $invoiceNumbers, Criteria::IN);
$cr->setDistinct(true);

但我有点困惑，我怎么能在推进中转换这个联接...

LEFT JOIN pick_lists ON（pick_lists.number = SUBSTR（inventory_audits.reference，0,8））

，我试过这个但没有工作..

$ cr-＆gt; add（PickListPeer :: NUMBER，substr（InventoryAuditPeer :: REFERENCE， 0,8））;

我正在使用推进1.2.1 任何引用都赞赏:) tyvm ..

Answer 1

Nvm，我通过使用这条线来实现它。

$cr->addJoin('substr(inventory_audits.REFERENCE, 0, 8)', PickListPeer::NUMBER, Criteria::LEFT_JOIN);

如何在propel中添加子串的连接

1 个答案: