當第二個查詢中沒有記錄時，如何減去兩個表的列值

Question

在我的學校申請書中，我想知道學生的應有數量，以便在我的

fee_class_mapping表我將費用分配給學生的班級，如下所示

+----------------------+------------+----------+------------+
| fee_class_mapping_id | fee_cat_id | class_id | fee_amount |
+----------------------+------------+----------+------------+
|                    1 |          2 |        1 | 10000      |
|                    2 |          2 |        2 | 15000      |
|                    3 |          3 |        1 | 2000       |
+----------------------+------------+----------+------------+

在student_present_class_details表中，我會有學生班級信息

+----------------------+----------+
| student_admission_id | class_id |
+----------------------+----------+
|                    1 | 1        |
|                    2 | 2        |
|                    3 | 1        |
|                    4 | 1        |
+----------------------+----------+

通過映射student_present_class_details class_id和fee_class_mapping class_id我將獲得分配給學生多少費用

還有另一個表格，其中包含fee_category詳細信息

+------------+--------------+-----------+
| fee_cat_id | fee_cat_name | parent_id |
+------------+--------------+-----------+
|          1 | School Fee   |         0 |
|          2 | School Fee   |         1 |
|          3 | books        |         1 |
|          4 | Dhobi        |         0 |
+------------+--------------+-----------+

我將通過以下查詢獲得特定學生總費用的fee_category明智金額

select f.fee_cat_name, sum(fcm.fee_amount) from  fee_class_mapping fcm ,student_present_class_details pd ,fee_category f 

where pd.class_id=fcm.class_id and f.fee_cat_id=fcm.fee_cat_id and pd.student_admission_id=4 group by fcm.fee_cat_id;

淘汰

+--------------+---------------------+
| fee_cat_name | sum(fcm.fee_amount) |
+--------------+---------------------+
| School Fee   |               10000 |
| books        |                2000 |
+--------------+---------------------+

我還有另一個稱為Fee_transactions的表，其中我將按學生費用類別明細處理交易明細，表結構為

+---------------------+------------------+------+-----+---------+----------------+
| Field               | Type             | Null | Key | Default | Extra          |
+---------------------+------------------+------+-----+---------+----------------+
| transaction_id      | int(11)          | NO   | PRI | NULL    | auto_increment |
| student_id          | int(11)          | NO   | MUL | NULL    |                |
| fee_cat_id          | int(11)          | NO   | MUL | NULL    |                |
| amount_paid         | varchar(255)     | YES  |     | NULL    |                |
| paid_on             | date             | YES  |     | NULL    |                |
| trans_receipt_no    | varchar(255)     | YES  |     | NULL    |                |
| is_concession_given | enum('Yes','No') | YES  |     | NULL    |                |
| payment_mode        | varchar(45)      | YES  |     | NULL    |                |
| refrence_number     | varchar(45)      | YES  |     | NULL    |                |
| tdm_id              | int(11)          | YES  |     | NULL    |                |
+---------------------+------------------+------+-----+---------+----------------+

從上表中，我將按以下方式獲得已支付費用類別的總和：

select sum(ft.amount_paid) from fee_transactions ft where ft.student_id=4 group by ft.fee_cat_id;

淘汰

+---------------------+
| sum(ft.amount_paid) |
+---------------------+
|                3000 | //school fee
+---------------------+

我要減去總金額-已付金額

所以我寫了以下查詢

select f.fee_cat_name, sum(fcm.fee_amount)-(select sum(ft.amount_paid) from fee_transactions ft where ft.student_id=4 and ft.fee_cat_id=fcm.fee_cat_id group by ft.fee_cat_id) from  fee_class_mapping fcm ,student_present_class_details pd ,fee_category f

where pd.class_id=fcm.class_id and f.fee_cat_id=fcm.fee_cat_id and pd.student_admission_id=4 group by fcm.fee_cat_id;

輸出

+--------------+------------+
| fee_cat_name | paidAmount |
+--------------+------------+
| School Fee   |       7000 |
| books        |       NULL |
+--------------+------------+

在fee_transaction表中，我沒有書費條目，有學費條目

總學費= 10000-支付的金額3000 = 7000是正確的

對於圖書費用，我沒有在fee_transaction表中記錄，因此其未返回值

我想要的是

 +--------------+------------+
    | fee_cat_name | paidAmount |
    +--------------+------------+
    | School Fee   |       7000 |
    | books        |       NULL | //2000 instead of NULL
    +--------------+------------+

我將如何實現它。

請幫幫我。 謝謝！！。

Answer 1

在子查詢的結果中，使用coalesce()將NULL替換為0。

SELECT f.fee_cat_name,
       sum(fcm.fee_amount) - coalesce((SELECT sum(ft.amount_paid)
                                              FROM fee_transactions ft
                                              WHERE ft.student_id = 4
                                                    AND ft.fee_cat_id = fcm.fee_cat_id
                                              GROUP BY ft.fee_cat_id), 0)
       FROM fee_class_mapping fcm
            INNER JOIN student_present_class_details pd
                       ON pd.class_id = fcm.class_id
            INNER JOIN fee_category f
                       ON f.fee_cat_id = fcm.fee_cat_id
       WHERE pd.student_admission_id = 4
       GROUP BY fcm.fee_cat_id;

我還重寫了它以使用顯式的JOIN語法，建議使用該語法，因為它更易於閱讀和理解。