Question

我有两张桌子：

付款：

id  amount  type    code
1   1200    0       111
2   100     1       111
3   200     0       111
4   50      0       112
5   500     2       112
6   300     3       113

票据：

id  details     code
-----------------------
1   bill-1      111 
2   bill-2      112
3   bill-3      113
4   bill-4      114

我想在amounts表格中对payments求和，然后将其加入bills，如下所示

结果：

bills.code  type0Sum    type1Sum    type2Sum    type3Sum
-------------------------------------------------------------------------
111         1400        100         0           0
112         50          0           500         0
113         0           0           0           300
114         0           0           0           0

对不起，如果这是一个新手问题

[编辑]
我使用了类似的查询如下：

SELECT *
FROM bills, 
     (SELECT SUM(amount) AS type0Sum, code
             FROM payments
             WHERE type = 0
             GROUP BY code) 
      AS sub1,
      (SELECT SUM(amount) AS type1Sum, code
              FROM payments
              WHERE type = 1
              GROUP BY ref_code) 
      AS sub2
      WHERE bills.code = sub1.code
      AND bills.code = sub2.code

但我只得到那些 type 的行：

bills.code  type0Sum    type1Sum    type2Sum    type3Sum
    -------------------------------------------------------
    111         1400        100

Answer 1

我已经修改了最终查询以进行正确的连接，而不是你正在做的旧连接（阅读笛卡尔连接）。给你一个机会，看看它是否有效;

SELECT b.code
    ,sub1.type0Sum
    ,sub2.type1Sum
FROM bills b
LEFT JOIN (
    SELECT SUM(amount) AS type0Sum
        ,code
    FROM payments
    WHERE type = 0
    GROUP BY code
    ) AS sub1 ON b.code = sub1.code
LEFT JOIN (
    SELECT SUM(amount) AS type1Sum
        ,code
    FROM payments
    WHERE type = 1
    GROUP BY ref_code
    ) AS sub2 ON b.code = sub2.code

还有其他方法可以提高效率，但我会继续查询，以帮助您学习。

SQLite：需要将一个表的结果与另一个结果相结合

1 个答案: