- SQL code
SELECT SJ_CangChu.pinming as 品名, SJ_CangChu.xinghao as 型号, SJ_CangChu.tiaoma as 条码,sum(SJ_Order.num) as 数量FROM dbo.SJ_Order INNER JOIN dbo.SJ_CangChu ON SJ_Order.outer_sku_id = SJ_CangChu.tiaoma where SJ_Order.status='WAIT_SELLER_SEND_GOODS' group by SJ_CangChu.pinming, SJ_CangChu.xinghao, SJ_CangChu.tiaoma union allSELECT SJ_CangChu.pinming as 品名, SJ_CangChu.xinghao as 型号, SJ_CangChu.tiaoma as 条码,sum(SJ_Order.num) as 数量FROM dbo.SJ_Order INNER JOIN dbo.SJ_CangChu ON SJ_Order.outer_iid = SJ_CangChu.tiaoma where SJ_Order.status='WAIT_SELLER_SEND_GOODS' group by SJ_CangChu.pinming, SJ_CangChu.xinghao, SJ_CangChu.tiaoma
- SQL code
SELECT SJ_CangChu.pinming as 品名, SJ_CangChu.xinghao as 型号, SJ_CangChu.tiaoma as 条码,sum(SJ_Order.num) as 数量FROM dbo.SJ_Order INNER JOIN dbo.SJ_CangChu ON SJ_Order.outer_sku_id = SJ_CangChu.tiaoma or SJ_Order.outer_iid = SJ_CangChu.tiaoma where SJ_Order.status='WAIT_SELLER_SEND_GOODS' group by SJ_CangChu.pinming, SJ_CangChu.xinghao, SJ_CangChu.tiaoma
SJ_Order的新旧2个字段对应（outer_sku_id ，SJ_Order.outer_iid ）分别对应SJ_CangChu.tiaoma
这2个SQL有什么区别
------解决方案--------------------
结果几乎一样,只不过一个是or条件,一个是用的union all.
如果把union all改为union 则结果一模一样了.
不过貌似用or比较好.