mysql sum（）使用多个联接返回双值

我使用多种货币，如美元、人民币、卢比。 它工作正常，但今天我注意到sum（）的双倍值，比如b.GrandTotal应该是11500，但它的返回值是23000

select 
a.ClientID,
f.Currency,
a.OrganizationName,
COALESCE(sum(b.GrandTotal),0) as SaleGrandTotal,
COALESCE(sum(g.AmountReceived),0) as AmountReceived,
COALESCE(sum(b.GrandTotal - g.AmountReceived),0) as SaleBalanceRemaining, 
COALESCE(sum(d.GrandTotal), 0) as PurchaseGrandTotal, 
COALESCE(sum(e.AmountPaid), 0) as AmountPaid,
COALESCE(sum(d.GrandTotal - e.AmountPaid),0) as PurchaseBalanceRemaining,
COALESCE(sum(b.GrandTotal - g.AmountReceived),0) - COALESCE(sum(d.GrandTotal - e.AmountPaid),0) as Total 
from na_clients as a
join na_currency as f 

left join na_transaction as b
on a.ClientID = b.ClientID and b.CurrencyID = f.CurrencyID and b.IsActive = 1

left join na_recoverylogs as g
on b.TID = g.TID

left join na_purchase as d 
on a.ClientID = d.ClientID and d.CurrencyID = f.CurrencyID  and d.IsActive = 1

left join na_purchaselogs as e
on e.PID = d.PID

group by a.OrganizationName,f.Currency
order by a.OrganizationName

因此，结果应该是：

Table Client:
clientid,name,organizationName
1,client1,OrgName
2,client2,OrgName

Table Currency:
currencyid,cname
1,Dollar
2,Rupees

Table Transaction:
tid,clientid,currencyid,grandTotal,amountReceived,balanceremaining
1,1,1,11000,0,11000
2,1,1,500,0,500

Table recoveryLogs: // Another Error Here
id,tid,amountreceived
1,1,0
2,2,0
3,2,2000     // Again sum() multiply value - because of PID 2 is repeating


Table Purchase:
pid,clientid,currencyid,grandTotal,amountPaid,balanceRemaining
1,1,1,25000,0,25000
1,2,2,2,3000,1000,2000
Now I am using sum(b.grandTotal) instead of 11500 it return 23000

Table PurchaseLogs: // Another Error Here
id,pid,amountpaid
1,1,0
2,2,1000
3,1,1000            // Again sum() multiply value - because of PID 1 is repeating

但我得到的结果是：

Client: Client1
SaleGrandTotal: 11500
AmountReceived: 0
SaleBalanceRemaining: 11500
PurchaseGrandTotal: 25000
AmountPaid: 0
PurchaseBalanceRemaining: 25000
Total Amount: -13500

---

如果我从查询中删除purchase子句（d和e）或transaction（b和g）子句，它可以单独工作。

数据加倍的原因是您的

ClientID在transaction和purchase表中出现的次数不同，因此不是1对1匹配
ClientID=1
和
CurrencyID=1
在交易中出现两次，在购买中只出现一次。当您加入表时，将产生一组1 x 2=2
ClientID
记录，其中一些字段重复数据。因此，重复输入的总和将加倍。如图所示：

    Client: Client1
    SaleGrandTotal: 23000
    AmountReceived: 0
    SaleBalanceRemaining: 23000
    PurchaseGrandTotal: 50000
    AmountPaid: 0
    PurchaseBalanceRemaining: 50000
    Total Amount: -27000

考虑使用派生表分离两个表之间的聚合。然后，加入四个底层聚合（事务、购买、恢复日志、购买日志）进行最终查询。如果在
ClientID
和
CurrencyID
、
TID
和
PID
上进行聚合，则联接将以1对1进行匹配

      Transaction Data    | Purchase Data
row1: 1,1,1,11000,0,11000 | 1,1,1,25000,0,25000 
row2: 2,1,1,500,0,500     | 1,1,1,25000,0,25000

您不使用客户机id分组的原因是什么？我只是每隔一个RDM在查看页面@WilliamBurnhamIn中使用ClientID作为参考，此聚合SQL查询将失败，因为您包含一个非聚合列，
ClientID
，但不在
GROUP BY
子句中指定它。MySQL运行在多个模式下，其中一个是
仅完整分组方式（仅在MySQL 5.7.5中是默认设置）。只需不使用MySQL“扩展”来分组方式即可。始终在GROUPBY子句中包含所有非聚合列，这样可以避免错误。当您在“non determinate”（即您的猜测和我的一样）中从group by子句中排除列时，MySQL会做什么？实际上，它会为group by子句中未指定的非聚合列任意选择一些值。您真的在使用Microsoft的SQL Server吗？结果：-返回错误的结果，货币美元没有条目，它返回多行，如：巴基斯坦卢比，巴基斯坦卢比，美元，美元。上面的查询返回双结果，而不是2货币，其返回4-2用于transAgg，2用于purchAgg@Parfait请参阅将
CurrencyID
添加到派生表的
JOIN
中的更新。非常感谢@Parfait的回答和解释：）再次使用sum（）返回双精度值。我刚刚在purchaseLogs和recoveryLogs中添加了条目。我已经更新了问题。请看一看@ParfaitYou将不得不分离日志（删除它们当前的左连接），并将它们单独聚合到另外两个派生表中，这一次连接到相应的transAgg和purchAgg派生表中。
SELECT
  transAgg.ClientID, transAgg.Currency, transAgg.OrganizationName,
  transAgg.SaleGrandTotal, recovLogAgg.SumOfAmtReceived, 
  (transAgg.SaleGrandTotal - recovLogAgg.SumOfAmtReceived) as SaleBalanceRemaining, 
  purchAgg.PurchaseGrandTotal, purchLogAgg.SumOfAmtPaid, 
  (purchAgg.PurchaseGrandTotal - purchLogAgg.SumOfAmtPaid) as PurchaseBalanceRemaining,
  ((transAgg.SaleGrandTotal - recovLogAgg.SumOfAmtReceived) - 
   (purchAgg.PurchaseGrandTotal - purchLogAgg.SumOfAmtPaid)) As [Total]

FROM
  (SELECT
      a.ClientID, f.CurrencyID, f.Currency, a.OrganizationName,
      COALESCE(sum(b.GrandTotal),0) as SaleGrandTotal
  FROM na_clients as a
  INNER JOIN na_currency as f     
  LEFT JOIN na_transaction as b
       ON a.ClientID = b.ClientID 
       AND b.CurrencyID = f.CurrencyID 
       AND b.IsActive = 1              
  GROUP BY a.ClientID, a.OrganizationName, f.CurrencyID, f.Currency
  ORDER BY a.OrganizationName) As transAgg    

INNER JOIN

  (SELECT
      a.ClientID, f.CurrencyID, f.Currency, a.OrganizationName,
      COALESCE(sum(d.GrandTotal), 0) as PurchaseGrandTotal
  FROM na_clients as a
  INNER JOIN na_currency as f     
  LEFT JOIN na_purchase as d 
       ON a.ClientID = d.ClientID 
       AND d.CurrencyID = f.CurrencyID  
       AND d.IsActive = 1          
  GROUP BY a.ClientID, a.OrganizationName, f.CurrencyID, f.Currency
  ORDER BY a.OrganizationName) As purchAgg  

ON transAgg.ClientID = purchAgg.ClientID 
AND transAgg.CurrencyID = purchAgg.CurrencyID

INNER JOIN

 (SELECT
      g.TID, COALESCE(sum(g.AmountReceived),0) As SumOfAmtReceived
  FROM na_recoverylogs as g          
  GROUP BY g.TID) As recovlogAgg  

ON transAgg.TID = recovlogAgg.TID

INNER JOIN

 (SELECT
      e.PID, COALESCE(sum(e.AmountPaid),0) As SumOfAmtPaid
  FROM na_purchaselogs as e 
  GROUP BY e.PID) As purchlogAgg

ON purchAgg.PID = purchlogAgg.PID