Question

+----------------+
| Tables_in_test |
+----------------+
| contribution   |
| expectedamount |
| registration   |
+----------------+

registration table
+----+--------+-------------+
| id | reg_no | fullname    |
+----+--------+-------------+
|  1 | TTI001 | JOHN JAMES  |
|  2 | TTI002 | DAVID CERES |
|  3 | TTI003 | JOYCE LEE   |
|  4 | TTI004 | JOEL MARTIN |
+----+--------+-------------+

espectedamount
+----+--------+---------+---------+---------+
| id | reg_no | number1 | number2 | number3 |
+----+--------+---------+---------+---------+
|  1 | TTI001 |     500 |     500 |     500 |
|  2 | TTI002 |     500 |     500 |     500 |
|  3 | TTI003 |     500 |     500 |     500 |
|  4 | TTI004 |     500 |     500 |     500 |
|  5 | TTI001 |     400 |     400 |     400 |
|  6 | TTI001 |    1000 |    1000 |    1000 |
|  7 | TTI002 |    1000 |    1000 |    1000 |
|  8 | TTI003 |    1000 |    1000 |    1000 |
|  9 | TTI004 |    1000 |    1000 |    1000 |
+----+--------+---------+---------+---------+

contribution table
+----+--------+---------+---------+---------+
| id | reg_no | number1 | number2 | number3 |
+----+--------+---------+---------+---------+
|  1 | TTI001 |     200 |     400 |     600 |
|  2 | TTI002 |     100 |      50 |     250 |
|  3 | TTI001 |     100 |     400 |     400 |
|  4 | TTI002 |     300 |     400 |     600 |
|  5 | TTI003 |     300 |     100 |      50 |
|  6 | TTI004 |      50 |      60 |      40 |
|  7 | TTI004 |     500 |     300 |     400 |
+----+--------+---------+---------+---------+

我创建了以下查询以联接表注册，期望值和贡献，其中我想从总贡献表中减去总和表金额表，但结果不正确

select registration.reg_no
      ,registration.fullname
      ,sum(expectedamount.number1-contribution.number1) as contribution1
      ,sum(expectedamount.number2-contribution.number2) as contribution2
      ,sum(expectedamount.number3-contribution.number3) as contribution3 
FROM registration
INNER JOIN expectedamount ON registration.reg_no = expectedamount.reg_no 
INNER JOIN contribution ON expectedamount.reg_no = contribution.reg_no
GROUP BY reg_no;

+--------+-------------+---------------+---------------+---------------+
| reg_no | fullname    | contribution1 | contribution2 | contribution3 |
+--------+-------------+---------------+---------------+---------------+
| TTI001 | JOHN JAMES  |           700 |           200 |             0 |
| TTI002 | DAVID CERES |           600 |           550 |           150 |
| TTI003 | JOYCE LEE   |           200 |           400 |           450 |
| TTI004 | JOEL MARTIN |           450 |           640 |           560 |
+--------+-------------+---------------+---------------+---------------+

预期结果

+-------+---------------+---------------+------------------+--------------+
|reg_no | fullname      | contribution1 | contribution2    | contribution3|
+-------+---------------+---------------+------------------+--------------+
|TTI001 | JOHN JAMES    | 1600          | 1100             |   900        |
|TTI002 | DAVID CERES   | 1000          | 950              |   550        |
|TTI003 | JOYCE LEE     | 1200          | 1400             |   1450       |
|TTI004 | JOEL MARTIN   |  950          | 1140             |   1060       |
+-------+---------------+---------------+------------------+--------------+

伙计们帮忙。

Answer 1

我认为您需要-

select registration.reg_no
  ,registration.fullname
  ,sum(expected.num1-contrib.num1) as contribution1
  ,sum(expected.num2-contrib.num2) as contribution2
  ,sum(expected.num3-contrib.num3) as contribution3 
FROM registration
INNER JOIN (SELECT reg_no, SUM(number1) num1, SUM(number2) num2, SUM(number3) num3
            FROM expectedamount
            GROUP BY reg_no) expected ON registration.reg_no = expected.reg_no 
INNER JOIN (SELECT reg_no, SUM(number1) num1, SUM(number2) num2, SUM(number3) num3
            FROM contribution
            GROUP BY reg_no) contrib ON expected.reg_no = contrib.reg_no
GROUP BY reg_no;

MYSQL数据库求和函数

1 个答案: