id和日期的MySQL SUM行值[重复]答案

【问题标题】：MySQL SUM row value by id and date [duplicate]id和日期的MySQL SUM行值[重复]
【发布时间】：2018-05-30 08:40:17
【问题描述】：

我需要根据 VARCHAR 列值和年份范围对 MySQL 表的列值求和。也许这很令人困惑，所以我将举一个基本的例子：

列：id（对此无用）、fruit_type、date_of_sale、sale_type、price

基本查询：

SELECT * FROM fruits

给我：

  1  APPLES   2018-05-30  GOOD   50.50 
  2  APPLES   2018-05-30  BAD   -20.50 
  3  APPLES   2018-05-30  GOOD   40.00 
  4  APPLES   2018-05-30  BAD   -10.50 
  5  APPLES   2017-05-29  GOOD   39.50 
  6  APPLES   2017-05-29  BAD   -10.00 
  7  APPLES   2017-05-29  GOOD   30.00 
  8  APPLES   2017-05-29  BAD   -20.00 
  9  BANANAS  2018-05-29  GOOD   20.00 
 10  BANANAS  2018-05-29  BAD    20.00 
 11  BANANAS  2018-05-29  GOOD   10.00 
 12  BANANAS  2018-05-29  BAD    -5.00 
 13  BANANAS  2017-05-29  GOOD   50.00 
 14  BANANAS  2017-05-29  BAD    -3.00 
 15  BANANAS  2017-05-29  GOOD   10.00 
 16  BANANAS  2017-05-29  BAD    -3.00 
 17  ORANGES  2018-05-28  GOOD    5.00 
 18  ORANGES  2018-05-28  BAD    -1.00 
 19  ORANGES  2017-05-28  GOOD   10.00 
 20  ORANGES  2017-05-28  BAD    -1.00

我需要完成什么：

  Fruit      Sales 2017    Bad Sales 2017    Total 2017    Sales 2018    Bad Sales 2018    Total 2018       
  APPLES     69.50         -30.00            39.50         90.50         -30.50            60.00            
  BANANAS    60.00         -6.00             54.00         30.00         -10.00            20.00            
  ORANGES    10.00         -1.00             9.00          5.00          -1.00             5.00

我一直在尝试的查询是：

SELECT fruit_type AS Fruit,\
(SELECT SUM(price) FROM fruits WHERE sale_type='GOOD' AND date_of_sale BETWEEN '2017-01-01' AND '2017-12-31') AS Sales_2017,\
(SELECT SUM(price) FROM fruits WHERE sale_type='BAD' AND date_of_sale BETWEEN '2017-01-01' AND '2017-12-31') AS Bad_Sales_2017,\
(SELECT SUM(Sales_2017-Bad_Sales_2017)),\
(SELECT SUM(price) FROM fruits WHERE sale_type='GOOD' AND date_of_sale BETWEEN '2018-01-01' AND '2018-12-31') AS Sales_2018,\
(SELECT SUM(price) FROM fruits WHERE sale_type='BAD' AND date_of_sale BETWEEN '2018-01-01' AND '2018-12-31') AS Bad_Sales_2018,\
(SELECT SUM(Sales_2018-Bad_Sales_2018)),\
FROM fruits
GROUP BY fruit_type;

问题是查询返回的值是整列的总和，其值为 GOOD 或 BAD，而不是按水果类型匹配的值 GOOD。

需要一些技巧来解决这个问题。

【问题讨论】：

跳过子查询。改为使用 case 表达式进行条件聚合。
我认为您可以使用附加总和，例如 select sum(case when sale_type='GOOD' AND date_of_sale BETWEEN '2017-01-01' AND '2017-12-31' then price else 0 ) ... 等等
在子查询中添加fruit_type条件。假设水果别名是 f1，在子查询中水果别名是 f2，然后添加 f1.fruit_type = f2.fruit_type。
认真考虑处理应用代码中数据显示的问题

标签： mysql sql

【解决方案1】：

使用sum + case when condition 总结您的销售情况

试试看：

select 
  fruit_type AS Fruit,
  sum(
    case when date_of_sale between '2017-01-01 00:00:00' and '2017-12-31 00:00:00' then
      case when sale_type = 'GOOD' then price else 0 end
    else 0 end
  ) "Sales 2017",
  sum(
    case when date_of_sale between '2017-01-01 00:00:00' and '2017-12-31 00:00:00' then
      case when sale_type = 'BAD' then price else 0 end
    else 0 end
  ) "Bad Sales 2017", 
  sum(
    case when date_of_sale between '2017-01-01 00:00:00' and '2017-12-31 00:00:00' then
      price
    else 0 end
  ) "Total 2017",  
  sum(
    case when date_of_sale between '2018-01-01 00:00:00' and '2018-12-31 00:00:00' then
      case when sale_type = 'GOOD' then price else 0 end
    else 0 end
  ) "Sales 2018",
  sum(
    case when date_of_sale between '2018-01-01 00:00:00' and '2018-12-31 00:00:00' then
      case when sale_type = 'BAD' then price else 0 end
    else 0 end
  ) "Bad Sales 2018", 
  sum(
    case when date_of_sale between '2018-01-01 00:00:00' and '2018-12-31 00:00:00' then
      price
    else 0 end
  ) "Total 2018"
FROM fruits
GROUP BY fruit_type;

SQL Fiddle Demo Link

【讨论】：

Thx 像魔术一样工作

【解决方案2】：

SELECT T.FRUIT, 
  SUM(IF(T.YEAR='2017' AND T.sale_type='GOOD',T.price_per_year,0)) `Sales 2017`,
  SUM(IF(T.YEAR='2017' AND T.sale_type='BAD',T.price_per_year,0)) `Bad Sales 2017`,
  SUM(IF(T.YEAR='2017',T.price_per_year,0)) `Total 2017`,
  SUM(IF(T.YEAR='2018' AND T.sale_type='GOOD',T.price_per_year,0)) `Sales 2018`,
  SUM(IF(T.YEAR='2018' AND T.sale_type='BAD',T.price_per_year,0)) `Bad Sales 2018`,
  SUM(IF(T.YEAR='2018',T.price_per_year,0)) `Total 2018`
FROM (SELECT
  FRUIT_TYPE FRUIT,
  YEAR(date_of_sale) YEAR,
  sale_type,
  SUM(price) price_per_year
FROM fruits 
GROUP BY 
FRUIT_TYPE,
YEAR(date_of_sale),
sale_type) T
GROUP BY FRUIT
ORDER BY T.FRUIT;

查看工作中的DEMO on SQL Fiddle。

【讨论】：

我更正了@Strawberry :) 感谢您的链接

【解决方案3】：

试试这个，只需为你想要的每一行添加更多的子选择。

SELECT all2017.fruit_type as Fruit
,Sales_2017
,Bad_Sales_2017
from (SELECT fruit_type
,SUM(ABS(price)) Sales_2017
FROM fruits
WHERE year(date) = 2017
GROUP BY fruit_type) as all2017
join (SELECT fruit_type
,SUM(ABS(price)) Bad_Sales_2017
FROM fruits
WHERE year(date) = 2017
AND sales_type = 'BAD'
GROUP BY fruit_type) as bad2017
on all2017.fruit_type = bad2017.fruit_type

【讨论】：