mysql 多表查询 比较两个字段最大、最小值，并显示对应字段

Question

有两个表，表A和表B,结构相同，但是具体字段不同，在表A.date = B.date条件下，查询出以下结果：MAX(A.ticker_buy-B.ticker_sell) 和MIN(A.ticker_buy-B.ticker_sell) ，即同一时间下两个表不同字段的差值的最大值和最小值，并显示对应最大值、最小值对应的date字段，我尝试用sql语句写了下，但是结果不对（用excel大致比较过）。我的语句如下：

select max(okcomfuturetickerquarter.ticker_buy-okcomfuturetickernextweek.ticker_sell) as "最大差价",min(okcomfuturetickerquarter.ticker_buy-okcomfuturetickernextweek.ticker_sell) as "最小差价",okcomfuturetickerquarter.date as "时间" from okcomfuturetickerquarter,okcomfuturetickernextweek where okcomfuturetickerquarter.date=okcomfuturetickernextweek.date and okcomfuturetickerquarter.ticker_buy is not null and okcomfuturetickernextweek.ticker_sell is not null ，

请各位大神帮助，写出正确查询语句。

Answer 1

First let me complain about the very long table name...

SELECT a.date as "时间", max(a.ticker_buy-b.ticker_sell) AS "最大差价",min(a.ticker_buy-b.ticker_sell) AS "最小差价" FROM a,b 
WHERE a.date = b.date 
AND a.ticker_buy IS NOT NULL
AND b.ticker_sell IS NOT NULL
GROUP BY a.date;

Answer 2

The parameter of max should be the column name. First calculate the difference between ticker_buy and ticker_sell in each row, and then use order by to sort. Take the first one
select (a.ticker_buy-b.ticker_sell) as ticker from a,b where a.date = b.date GROUP BY a.date order by ticker;