MYSQL查询返回两天之间价格涨幅最高的值

Question

I have two tables:

PRICE
ticker date close
AAPL 3/20/2019 $100
AAPL 3/21/2019 $101.5
AAPL 3/22/2019 $106.5
GOOG 3/20/2019 $100
GOOG 3/21/2019 $130
GOOG 3/22/2019 $110
MSFT 3/20/2019 $184.5
MSFT 3/21/2019 $188.5
MSFT 3/22/2019 $210
IBM 3/20/2019 $72
IBM 3/21/2019 $70
IBM 3/22/2019 $10

STOCK
ticker exchange
AAPL NASDAQ
GOOG NASDAQ
MSFT NASDAQ
IBM NYSE

and I want to return the ticker that is in 'NYSE' with the highest increase in closing price from 2019-03-20 to 2019 03-21. I have:

SELECT DISTINCT T.ticker 
FROM PRICE T 
WHERE T.ticker IN SELECT D1.ticker, MAX(D1.close-D2.close) 
                  FROM PRICE D1, PRICE D2, STOCK S 
                  WHERE S.exchange='NYSE' AND D1.date = '2019-03-21' AND D2.date = '2019-03-20' AND D1.ticker = D2.ticker GROUP BY D1.ticker) 
GROUP BY T.ticker;

But it returns 'ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'SELECT D1.ticker, MAX(D1.close-D2.close) FROM PRICE D1, PRICE D2, STOCK S WHERE ' at line 1'

IBM should return as output, as it is the only ticker in NYSE.

I'm completely new to SQL and any suggestions would be appreciated :)

Answer 1

Use a self-join:

select p1.ticket
from price p1 join
     price p2
     on p1.ticker = p2.ticker and
        p1.date = '2019-03-20' and
        p2.date = '2019-03-21'
order by p2.price - p1.price desc
limit 1;

You only need the join to stock if you need the exchange (which your query suggests you want, but the explanation doesn't mention):

select p1.ticket
from price p1 join
     price p2
     on p1.ticker = p2.ticker and
        p1.date = '2019-03-20' and
        p2.date = '2019-03-21' join
     stock s
     on p1.ticker = s.ticker
where s.exchange = 'NYSE'
order by p2.price - p1.price desc
limit 1;