支付与订单时间差

[sqlsuper1]

如图所示，求代码

[shoppo0505]

select *, datediff(hour, 订单时间，支付时间) from A union all select *, datediff(hour, 订单时间，支付时间) from B

[二月十六]

SELECT *,DATEDIFF(DAY,订单时间,支付时间) AS 时间差 FROM A WHERE DATEDIFF(DAY,订单时间,支付时间)>3
UNION ALL
SELECT *,DATEDIFF(DAY,订单时间,支付时间) AS 时间差 FROM B WHERE DATEDIFF(DAY,订单时间,支付时间)>3

[听雨停了]

SELECT *,DATEDIFF(DAY,订单时间,支付时间) AS 时间差 FROM A WHERE DATEDIFF(DAY,订单时间,支付时间)>3 UNION SELECT *,DATEDIFF(DAY,订单时间,支付时间) AS 时间差 FROM B WHERE DATEDIFF(DAY,订单时间,支付时间)>3 去除了重复的结果

[sqlsuper1]

我认为应该是union比union all 更适合一些

[0与1之间]

union 是去掉是重复的 union all 是所有的