未分组所有选择项的余额到期SQL

我已经提供了我的表的摘要，并且在SQL上有了一个良好的开端，但是我一直在想如何限制返回的项的数量。我应该能够选择一个或多个条款，并从这些条款中取回到期的余额

一个学生应该有一个记录，他们可以在几个条款上有几个预订，但付款不是针对预订，而是针对学生。这就是让我感到厌烦的部分

表结构、日期和我的SQL的开始日期如下。有人能帮我吗？该结果不应显示Sue Smith从第3条款中支付的500美元

我正在使用PostgreSQL，但我认为这是一个非常基本的问题，不需要任何特定于Postgres的东西

当前结果集： 模式： 查询：

---

以下是我的查询更新版本：

SELECT      
    s.student_sis_id AS "Student ID",        
    s.student_last_name AS Last,
    s.student_first_name AS First,
    SUM(r.reservation_fee_amount) AS "Total Fees",
    COUNT(r.reservation_id) AS "Reservation Count",
    COALESCE(
        SUM(
            p.payment_check_amount
            + p.payment_cash_amount
            + p.payment_credit_card_amount
        ), 0.00
    ) AS "Amount Paid",
    SUM(r.reservation_fee_amount) - (
        COALESCE(
            SUM(
                p.payment_check_amount
                + p.payment_cash_amount
                + p.payment_credit_card_amount
            ), 0.00
        )  
    ) AS "Amount Due" 
FROM 
    student s
    INNER JOIN reservation r ON s.student_id = r.student_id  
    LEFT JOIN payment p ON p.student_id = r.student_id AND p.term_id = r.term_id
WHERE r.term_id IN (1,2)
GROUP BY
    s.student_id, 
    s.student_sis_id,        
    s.student_last_name,
    s.student_first_name
ORDER BY 
    s.student_sis_id
;

注意事项：

• 我在主（外部）查询中包含了

  付款

  ，以避免子查询

• 联接类型为

  LEFT[OUTER]联接

  ，因此缺少任何

  付款

  行不会阻止其他数据出现在结果集中

• 连接条件包括

  term\u id

  （我认为这基本上就是您丢失的地方）

• 最后，我使用了短表别名来提高可读性

---

我希望这就是您想要的。

找到了2付款输入问题的解决方案（我在最初的问题中没有认识到）。答案如下：

set search_path to temp, public;
SELECT      
s.student_sis_id AS "Student ID",        
s.student_last_name AS "Last Name",
s.student_first_name AS "First Name",
SUM(r.reservation_fee_amount) AS "Total Fees",
COALESCE(p.paid, 0.00) AS "Amount Paid",
COALESCE(SUM(r.reservation_fee_amount) - p.paid, 0.00) AS "Amount Due"    
FROM 
    student s       
    INNER JOIN reservation r ON s.student_id = r.student_id
    left outer join 
    (
      select student_id, term_id,
        SUM(
            p.payment_check_amount
            + p.payment_cash_amount
            + p.payment_credit_card_amount
        ) AS "paid" 
      from payment p
      group by student_id, term_id
    ) as p 
    ON p.student_id = r.student_id AND p.term_id = r.term_id
        WHERE r.reservation_completed  AND  r.term_id IN (1,2)                      
GROUP BY
    s.student_sis_id,        
    s.student_last_name,
    s.student_first_name,
    p.paid
ORDER BY 
    s.student_sis_id

---

谢谢Dezzo和davek

就是这样，谢谢。谢谢你给我问题的标题，我没想到。我试着问一个高质量的问题——它似乎吸引了高质量的答案，就像你的一样。@user973828——你的问题确实质量很好。一个小的补充：你可以通过使用.Hi，附加问题来避免粘贴太多的代码（比如

INSERT

s）。。如果我在From子句中添加以下内容：s.student\u id上的From student s内部加入保留r=r.student\u id上的离开加入付款p.student\u id=r.student\u id和p.term\u id=r.term\u id，其中r.reservation\u完成和r.reservation\u删除的日期为空，而不是r.reservation\u等待列表和r.term\u id（$list）我收到的预订费是假的。我不明白为什么？因为学生为一个预订支付了2笔费用，所以费用增加了一倍。我需要有付款的子查询来处理这种情况吗？

SELECT      
    student.student_sis_id AS "Student ID",        
    student.student_last_name AS Last,
    student.student_first_name AS First,
    SUM(reservation.reservation_fee_amount) AS "Total Fees",
    (
        SELECT COUNT(reservation.reservation_id)
        FROM reservation
        WHERE student.student_id = reservation.student_id
    ) AS "Reservation Count",
    (
        SELECT 
            COALESCE(SUM(
                payment.payment_check_amount
                + payment.payment_cash_amount
                + payment.payment_credit_card_amount
            ), 0.00)  
        FROM payment 
        WHERE payment.student_id = student.student_id
    ) AS "Amount Paid",
    SUM(reservation.reservation_fee_amount) - (
        SELECT 
            COALESCE(SUM(
                payment.payment_check_amount
                + payment.payment_cash_amount
                + payment.payment_credit_card_amount
            ), 0.00)  
        FROM payment WHERE payment.student_id = student.student_id
    ) AS "Amount Due" 
FROM 
    student
    INNER JOIN reservation ON student.student_id = reservation.student_id  
WHERE reservation.term_id IN (1,2)
GROUP BY
    student.student_id, 
    student.student_sis_id,        
    student.student_last_name,
    student.student_first_name
ORDER BY 
    student.student_sis_id
;

SELECT      
    s.student_sis_id AS "Student ID",        
    s.student_last_name AS Last,
    s.student_first_name AS First,
    SUM(r.reservation_fee_amount) AS "Total Fees",
    COUNT(r.reservation_id) AS "Reservation Count",
    COALESCE(
        SUM(
            p.payment_check_amount
            + p.payment_cash_amount
            + p.payment_credit_card_amount
        ), 0.00
    ) AS "Amount Paid",
    SUM(r.reservation_fee_amount) - (
        COALESCE(
            SUM(
                p.payment_check_amount
                + p.payment_cash_amount
                + p.payment_credit_card_amount
            ), 0.00
        )  
    ) AS "Amount Due" 
FROM 
    student s
    INNER JOIN reservation r ON s.student_id = r.student_id  
    LEFT JOIN payment p ON p.student_id = r.student_id AND p.term_id = r.term_id
WHERE r.term_id IN (1,2)
GROUP BY
    s.student_id, 
    s.student_sis_id,        
    s.student_last_name,
    s.student_first_name
ORDER BY 
    s.student_sis_id
;

set search_path to temp, public;
SELECT      
s.student_sis_id AS "Student ID",        
s.student_last_name AS "Last Name",
s.student_first_name AS "First Name",
SUM(r.reservation_fee_amount) AS "Total Fees",
COALESCE(p.paid, 0.00) AS "Amount Paid",
COALESCE(SUM(r.reservation_fee_amount) - p.paid, 0.00) AS "Amount Due"    
FROM 
    student s       
    INNER JOIN reservation r ON s.student_id = r.student_id
    left outer join 
    (
      select student_id, term_id,
        SUM(
            p.payment_check_amount
            + p.payment_cash_amount
            + p.payment_credit_card_amount
        ) AS "paid" 
      from payment p
      group by student_id, term_id
    ) as p 
    ON p.student_id = r.student_id AND p.term_id = r.term_id
        WHERE r.reservation_completed  AND  r.term_id IN (1,2)                      
GROUP BY
    s.student_sis_id,        
    s.student_last_name,
    s.student_first_name,
    p.paid
ORDER BY 
    s.student_sis_id