#方法一：使用 not in ,容易理解,效率低 ~执行时间为：1.395秒~ SELECT COUNT(1) FROM ecs_goods WHERE ecs_goods.goods_id NOT IN (SELECT ecs_member_price.goods_id FROM ecs_member_price); #方法二：使用 left join...on... , "B.ID isnull" 表示左连接之后在B.ID 字段为 null的记录 ~执行时间：0.739秒~ SELECT COUNT(1) FROM ecs_goods LEFT JOIN ecs_member_price ON ecs_goods.goods_id=ecs_member_price.goods_id WHERE ecs_member_price.goods_id IS NULL; #方法三：逻辑相对复杂,但是速度最快 ~执行时间: 0.570秒~ SELECT COUNT(1) FROM ecs_goods c WHERE (SELECT COUNT(1) AS num FROM ecs_member_price WHERE ecs_member_price.goods_id=ecs_goods.goods_id) = 0;
