Question

以下是我的bill_detail表，其中我存储了用户购买物品的账单

create table bill_details(
bill_no bigint(20) unsigned NOT NULL AUTO_INCREMENT,
customer_id bigint(20) unsigned NOT NULL,
sub_total decimal(15,2),
vat decimal(15,2),
total decimal(15,2),
payment_type varchar(30) NOT NULL,
status varchar(20),
add_date timestamp DEFAULT 0,
CONSTRAINT fk_cust1 FOREIGN KEY (customer_id)
REFERENCES customer_details(id),
cid bigint(20) unsigned NOT NULL,
CONSTRAINT cli_fk15 FOREIGN KEY (cid) REFERENCES client (id) ON DELETE CASCADE ON UPDATE CASCADE,
primary key(bill_no))ENGINE=InnoDB DEFAULT CHARSET=utf8;

这里有多个用户同时使用我的应用程序。我的问题是用户没有获得序列式的账单号。

我该如何解决这个问题？

Answer 1

好吧，请考虑以下内容......

DROP TABLE IF EXISTS my_table;

CREATE TABLE my_table(id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,user_id INT NOT NULL, bill_no INT NULL);

INSERT INTO my_table (user_id) VALUES (1),(1),(2),(1),(2),(3),(1),(2),(3),(1),(4);

SELECT * FROM my_table;
+----+---------+---------+
| id | user_id | bill_no |
+----+---------+---------+
|  1 |       1 |    NULL |
|  2 |       1 |    NULL |
|  3 |       2 |    NULL |
|  4 |       1 |    NULL |
|  5 |       2 |    NULL |
|  6 |       3 |    NULL |
|  7 |       1 |    NULL |
|  8 |       2 |    NULL |
|  9 |       3 |    NULL |
| 10 |       1 |    NULL |
| 11 |       4 |    NULL |
+----+---------+---------+

SELECT x.*
     , COUNT(*) rank 
  FROM my_table x  
  JOIN my_table y 
    ON y.user_id = x.user_id 
   AND y.id <= x.id 
 GROUP 
    BY x.user_id
     , x.id;
+----+---------+---------+------+
| id | user_id | bill_no | rank |
+----+---------+---------+------+
|  1 |       1 |    NULL |    1 |
|  2 |       1 |    NULL |    2 |
|  4 |       1 |    NULL |    3 |
|  7 |       1 |    NULL |    4 |
| 10 |       1 |    NULL |    5 |
|  3 |       2 |    NULL |    1 |
|  5 |       2 |    NULL |    2 |
|  8 |       2 |    NULL |    3 |
|  6 |       3 |    NULL |    1 |
|  9 |       3 |    NULL |    2 |
| 11 |       4 |    NULL |    1 |
+----+---------+---------+------+

UPDATE my_table a 
  JOIN 
     ( SELECT x.*,COUNT(*) rank 
         FROM my_table x  
         JOIN my_table y  
           ON y.user_id = x.user_id 
          AND y.id <= x.id 
        GROUP 
           BY x.user_id
            , x.id
     ) b 
    ON b.id = a.id 
   SET a.bill_no = b.rank 
 WHERE a.bill_no IS NULL; -- <-- You may want to discard or rethink this line

SELECT * FROM my_table;
+----+---------+---------+
| id | user_id | bill_no |
+----+---------+---------+
|  1 |       1 |       1 |
|  2 |       1 |       2 |
|  3 |       2 |       1 |
|  4 |       1 |       3 |
|  5 |       2 |       2 |
|  6 |       3 |       1 |
|  7 |       1 |       4 |
|  8 |       2 |       3 |
|  9 |       3 |       2 |
| 10 |       1 |       5 |
| 11 |       4 |       1 |
+----+---------+---------+

编辑：每个用户最新的bill_no ...

 SELECT x.* 
   FROM my_table x 
   JOIN 
      ( SELECT user_id,MAX(bill_no) max_bill_no FROM my_table GROUP BY user_id ) y 
     ON y.user_id = x.user_id 
    AND y.max_bill_no = x.bill_no;
+----+---------+---------+
| id | user_id | bill_no |
+----+---------+---------+
|  8 |       2 |       3 |
|  9 |       3 |       2 |
| 10 |       1 |       5 |
| 11 |       4 |       1 |
+----+---------+---------+

为每个用户顺序生成账单号码

1 个答案: