我正在使用以下查询从我的数据库的不同表中获取报告,检查以下...
SELECT s.id, s.name, c.name AS course_name,
s.open_bal AS open_balance, sum(i.amount) AS gross_fee,
sum(i.discount) AS discount, sum(i.amount) - sum(i.discount) AS net_payable,
SUM(r.reg_fee+r.tut_fee+r.other_fee) AS net_recieved,
(sum(i.amount) - sum(i.discount)) - SUM(r.reg_fee+r.tut_fee+r.other_fee) AS balance_due
FROM students s
INNER JOIN courses c on c.id = s.course_id
LEFT JOIN invoices i on i.student_id = s.id
LEFT JOIN recipts r on r.student_id = s.id;
发票
| id | student_id | amount | discount | dnt |
+----+------------+----------+----------+-------------+
| 2 | 22 | 35000 | 0 | 2011/01/01 |
+----+------------+----------+----------+-------------+
未从 gross_fee 和 net_payable 获得正确的值。
谢谢。
最佳答案
鉴于选择中的 SUM
,我想 GROUP BY s.id
应该可以解决问题。无论如何,GROUP BY
似乎丢失了 :)
SELECT s.id, s.name, c.name AS course_name,
s.open_bal AS open_balance,
SUM(r.reg_fee+r.tut_fee+r.other_fee) AS net_recieved,
(sum(i.amount) - sum(i.discount)) - SUM(r.reg_fee+r.tut_fee+r.other_fee) AS balance_due
FROM students s
INNER JOIN courses c on c.id = s.course_id
LEFT JOIN invoices i on i.student_id = s.id
LEFT JOIN recipts r on r.student_id = s.id
GROUP BY s.id;
编辑
单独的查询允许检索所有发票的 gross_fee 和 net_payable
SELECT sum(amount) AS gross_fee,
sum(discount) AS discount,
sum(amount) - sum(discount) AS net_payable,
FROM invoices;
关于mysql 内连接,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/5648597/