Question

我尝试使用SQL进行计算。

我的数据库下面有前三列，我需要计算第四列：

Month |Name |Monthly Premium Paid |Number of consecutive months paid
Jan 2010 | Bob | Y | 1
Feb 2010 | Bob | Y | 2
Mar 2010 | Bob | N | 0
Apr 2010 | Bob | Y | 1
May 2010 | Bob | Y | 2
Jul 2010 | Bob | N | 0

Answer 1

您可以尝试

select month, name, premiun_paid,
    @row_number:=CASE
        WHEN premiun_paid = 'Y' THEN @row_number + 1
        ELSE 0 
        END consecutive_month
from table,(select @row_number:=0)r

参考：http://www.mysqltutorial.org/mysql-row_number/

模式（MySQL v5.7）

create table fiddle(month text,name text,
                    premium text,number int);

insert into fiddle values('Jan','Bob','Y',0),
                         ('Feb','Bob','Y',0),
                         ('Mar','Bob','N',0),
                         ('Apr','Bob','Y',0),
                         ('May','Bob','Y',0),
                         ('June','Bob','N',0),
                         ('Jul','Bob','Y',0),
                         ('Aug','Bob','Y',0),
                         ('Sep','Bob','Y',0);

查询＃1

select month, name, premium,
    @row_number:=CASE
        WHEN premium='Y' THEN @row_number + 1
        ELSE 0 
        END as consecutive_months 
from fiddle,(select @row_number:=0)r;

| month | name | premium | consecutive_months |
| ----- | ---- | ------- | ------------------ |
| Jan   | Bob  | Y       | 1                  |
| Feb   | Bob  | Y       | 2                  |
| Mar   | Bob  | N       | 0                  |
| Apr   | Bob  | Y       | 1                  |
| May   | Bob  | Y       | 2                  |
| June  | Bob  | N       | 0                  |
| Jul   | Bob  | Y       | 1                  |
| Aug   | Bob  | Y       | 2                  |
| Sep   | Bob  | Y       | 3                  |

在SQL中连续计数0

1 个答案: