今年から昨年の売上を差し引いてmysqlを使った各クライアントの売上（より多くのエントリーを1年合計）

Question

すでにタイトルに書かれているように、私はまず、昨年のすべての売上を合計と今年を入力し、今年の去年を各クライアントの(last year - this year)から引きます。

編集：

clientid | ....date..... | quant | product. | client | place | price | 
......1... |2016-01-10  | ...1.... | product1 | anna | Italy... | .100. | 
......1... |2016-04-12  | ...2.... | product2 | anna | Italy... | .300. | 
......1... |2017-01-10  | ...1.... | product1 | anna | Italy... | .200. | 
......1... |2017-04-12  | ...2.... | product1 | anna | Italy... | .400. | 

だから、最後の年にアンナの結果は400である必要があり、今年600 diffが、私はすでに試した200

する必要がありますクエリはいくつかありますが、私にとっては複雑すぎます。

SELECT 
    `date`, `quant`, `clientid`, `product`, `client`, `place`, lastyear.price, thisyear.price , (lastyear.price - thisyear.price) AS diff 
FROM 
    (SELECT 
     `date`, `quant`, `product`, `client`, `place`, SUM(price) 
    FROM 
     `verkauf` 
    WHERE 
     `date` BETWEEN DATE_FORMAT(CURDATE() , '%Y-%m-01') - INTERVAL 24 MONTH AND DATE_FORMAT(CURDATE() , '%Y-%m-31') - INTERVAL 12 MONTH) 

     )as lastyear, 

    (SELECT 
     `date`, `quant`, `product`, `client`, `place`, SUM(price) 
    FROM 
     `verkauf` 
    WHERE 
     `date` BETWEEN DATE_FORMAT(CURDATE() , '%Y-%m-01') - INTERVAL 12 MONTH AND DATE_FORMAT(CURDATE() , '%Y-%m-31') 

     )as thisyear 
WHERE 
    `clientid` = :id 
ORDER BY 
    diff ASC 
LIMIT 
    :start, :limit 

Answer 1

お試しください。 1つのサブクエリでは、今年の合計と別のサブクエリで前年の合計をとり、その両方を使用して別のサブクエリを取得しています。

Select m1.clientid, m1.cyear, m0.pyear, m1.sum csum, m0.sum psum, ( m1.sum - m0.sum ) diff from ( select clientid, cyear, sum from ( select clientid, year(date) cyear, sum(price) sum from buget group by clientid, cyear ) T where cyear = year(curdate()) ) m1, ( ( select clientid, pyear, sum from ( select clientid, year(date) pyear, sum(price) sum from buget group by clientid, pyear ) T where pyear = ( year(curdate()) ) - 1) m0 ) where m1.clientid = m0.clientid

Answer 2

更新回答

select ABS(t3.price2 - t2.price1),t1.client from buget t1 inner join (select sum(price) as price1 from buget where year(date) = year(now()) group by client) t2 on year(date) = year(now()) inner join (select sum(price) as price2 from buget where year(date) = year(now()) - 1 group by client) t3 group by client; 

いくつかのデータと