使用Group by时的Sum（）总计不正确

Question

我有一个没有正确求和的MYSQL查询，我在查询中看不到错误。

SELECT s.source, vanMoney.userid, vanMoney.Username, vanMoney.vandate, vanMoney.FirstBuy, vanMoney.LastBuy, vanMoney.SourceID, vanMoney.TodayDate,
Sum(inv_temp.cost* inv_temp.quantity) as TotalCost,
Sum(inv_temp.quantity) as TotalQTY
FROM source AS s 
Left JOIN inv_temp ON inv_temp.source_id = s.source_id 
Left JOIN vanMoney ON inv_temp.source_id = vanMoney.SourceID 
WHERE s.buyer_type_id = 6 and quantity>0 
group by vanMoney.vandate, vanMoney.userid

TotalCost和TotalQTY列是整个Source的汇总，不仅仅是source和vandate

我已编辑添加以下信息

 the output i should get is 
for userid 186 on 3/9/2018 is TotalCost=7 and TotalQTY=2
for userid 186 on 3/10/2018 is TotalCost=77 and TotalQTY=2
for userid 186 on 3/11/2018 is TotalCost=4 and TotalQTY=2
for userid 179 on 3/26/2018 is TotalCost=30 and TotalQTY=2

Answer 1

您不会显示费用或数量列，但它看起来像这样：

SELECT s.source, vanMoney.userid, vanMoney.Username, vanMoney.vandate, 
vanMoney.FirstBuy, vanMoney.LastBuy, anMoney.SourceID, vanMoney.TodayDate,
sum(TABLE.cost) as TotalCost
sum(TABLE.QTY) as TotalQTY 
FROM source AS s 
Left JOIN inv_temp ON inv_temp.source_id = s.source_id 
Left JOIN vanMoney ON inv_temp.source_id = vanMoney.SourceID 
WHERE s.buyer_type_id = 6 
  and vandate='2018-03-10'
group by s.source, vanMoney.userid, vanMoney.Username, vanMoney.vandate,
vanMoney.FirstBuy, vanMoney.LastBuy, vanMoney.SourceID, vanMoney.TodayDate

如果您只想按vandate和userid进行分组，则需要排除其他列。否则你会有重复的总数。

Answer 2

当您使用GROUP BY子句时，您需要声明所有字段而不进行聚合 功能

在子查询上使用SUM GROUP BY source_id聚合totle，source_id可以JOIN，然后SELECT s.source, vanmoney.userid, vanmoney.username, vanmoney.vandate, vanmoney.firstbuy, vanmoney.lastbuy, vanmoney.sourceid, vanmoney.todaydate, inv_temp.TotalCost, inv_temp.TotalQTY FROM source AS s LEFT JOIN ( SELECT source_id, Sum(cost * quantity) AS TotalCost, Sum(quantity) AS TotalQTY FROM inv_temp WHERE quantity > 0 GROUP BY source_id ) inv_temp ON inv_temp.source_id = s.source_id LEFT JOIN vanmoney ON inv_temp.source_id = vanmoney.sourceid WHERE s.buyer_type_id = 6 。

你可以试试这个。

inv_temp

修改

SELECT source_id, date_created, Sum(cost * quantity) AS TotalCost, Sum(quantity) AS TotalQTY FROM inv_temp WHERE quantity > 0 GROUP BY source_id, date_created 期望结果查询可能像

SELECT s.source, 
       vanmoney.userid, 
       vanmoney.username, 
       date_format(str_to_date(vanmoney.vandate, '%m/%d/%Y'),'%Y%m'), 
       vanmoney.firstbuy, 
       vanmoney.lastbuy, 
       vanmoney.sourceid, 
       vanmoney.todaydate, 
       Sum(inv_temp.cost * inv_temp.quantity) AS TotalCost, 
       Sum(inv_temp.quantity)                 AS TotalQTY 
FROM   source AS s 
       LEFT JOIN inv_temp 
              ON inv_temp.source_id = s.source_id 
       LEFT JOIN vanmoney 
              ON inv_temp.source_id = vanmoney.sourceid 
WHERE  s.buyer_type_id = 6 
       AND quantity > 0 
GROUP  BY  s.source, 
           vanmoney.userid, 
           vanmoney.username, 
           date_format(str_to_date(vanmoney.vandate, '%m/%d/%Y'),'%Y%m'), 
           vanmoney.firstbuy, 
           vanmoney.lastbuy, 
           vanmoney.sourceid, 
           vanmoney.todaydate

  

https://www.db-fiddle.com/f/9yVD8cdV9iGnUQGiJ9vmFk/0

从你的编辑答案我想你想要这样使用。

a = [['w1' , 'w3'] , ['w2' , 'w4']]

df=pd.DataFrame({'a': a })
print(df)

mlb = MultiLabelBinarizer()
print(np.array(mlb.fit_transform(df['a'].as_matrix())))

  

https://www.db-fiddle.com/f/9yVD8cdV9iGnUQGiJ9vmFk/2

Answer 3

因此，为了使查询不对整行进行求和，而只是对应于日期的行，我执行了where = statement

SELECT
vanMoney.userid,
vanMoney.Username,
vanMoney.vandate,
Max(vanMoney.FirstBuy) AS firstbuy,
Max(vanMoney.LastBuy) AS lastbuy,
vanMoney.SourceID,
inv_temp.date_created,
Sum(inv_temp.cost* inv_temp.quantity) AS TotalCost,
Sum(inv_temp.quantity) AS TotalQTY,
source.source
FROM
inv_temp
LEFT JOIN vanMoney ON inv_temp.purchased_by = vanMoney.userid
Left JOIN source ON inv_temp.source_id = source.source_id
where date_created=vandate
GROUP BY  
vanMoney.vandate,
      vanMoney.userid