Question

我进行此查询，并返回了88个ID

SELECT DISTINCT so.id
FROM stock_picking sp
                       INNER JOIN stock_move sm ON sp.id = sm.picking_id
                       INNER JOIN procurement_order po ON sm.procurement_id = po.id
                       INNER JOIN sale_order_line sol ON po.sale_line_id = sol.id 
                       INNER JOIN sale_order so ON sol.order_id = so.id

                       INNER JOIN sale_order_invoice_rel so_inv_rel on so.id = so_inv_rel.order_id
                       INNER JOIN account_invoice inv on so_inv_rel.invoice_id = inv.id

    WHERE 

            so.invoice_status = 'to invoice'
            and sp.state = 'done'       
            and inv.state != 'draft'

现在我想更新这些ID，并像这样尝试了一下，但是出现错误，如何正确执行？

我没有选择，而是尝试了这一行

update so set invoice_status = 'invoiced' but get an error

ERROR:  relation "so" does not exist
LINE 1: update so set invoice_status = 'invoiced'

Answer 1

[so]是您为sale_order表设置的别名，并且代码只能在设置了别名的查询中识别它。话虽如此，如果您运行更新查询，则希望更新sale_order表中的所有行。要仅更新SELECT查询返回的ID的表，请使用以下命令：

update sale_order set invoice_status = 'invoiced' 
where id in 
(
SELECT DISTINCT so.id
FROM stock_picking sp
                   INNER JOIN stock_move sm ON sp.id = sm.picking_id
                   INNER JOIN procurement_order po ON sm.procurement_id = po.id
                   INNER JOIN sale_order_line sol ON po.sale_line_id = sol.id 
                   INNER JOIN sale_order so ON sol.order_id = so.id

                   INNER JOIN sale_order_invoice_rel so_inv_rel on so.id = 
                   so_inv_rel.order_id
                   INNER JOIN account_invoice inv on so_inv_rel.invoice_id = inv.id

WHERE 

        so.invoice_status = 'to invoice'
        and sp.state = 'done'       
        and inv.state != 'draft'
)

更新选择的ID

1 个答案: