Question

这是我的数据集：

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我需要的是产生以下结果的查询：

id  user_id amount  added   
1   1   10.00   2018-09-11  
2   2   10.00   2018-09-12
3   3   10.00   2018-09-13
4   1   8.00    2018-09-14
5   2   6.00    2018-09-15

我想获取用户ID的状态，无论它是否发生。如果不是，则显示id user_id amount added status 1 1 10.00 2018-09-11 new 2 2 10.00 2018-09-12 new 3 3 10.00 2018-09-13 new 4 1 8.00 2018-09-14 recurring 5 2 6.00 2018-09-15 recurring，否则显示new。

Answer 1

您想知道记录的日期是否是用户的最短日期：

select id, user_id, amount, added,
  case when (user_id, added) in (select user_id, min(added) from mytable group by user_id)
    then 'new' else 'recurring' as status
from mytable
order by id;

当然也可以通过加入子查询或使用相关的EXISTS子句来实现。从MySQL 8.0开始，我将使用MIN() OVER()。

Answer 2

尝试此查询：

Demo

Answer 3

尝试以下方法：http://sqlfiddle.com/#!9/6ef4c6/11

select *, case when num=1 then 'New' when num=2 then 'Recurring' end as status from 
(SELECT id,
    @row_number:=CASE
        WHEN @customer_no = userid THEN @row_number + 1
        ELSE 1
    END AS num,
    @customer_no:=userid as CustomerNumber

FROM
    test, (SELECT @customer_no:=0,@row_number:=0) as t
ORDER BY userid)a

Answer 4

您可以简单地JOIN使用分组数据：

SELECT
    userdata.*,
    CASE WHEN userdata.added = groupsq.first_date THEN 'new' ELSE 'recurring' END AS status
FROM userdata
INNER JOIN (
    SELECT user_id, MIN(added) AS first_date
    FROM userdata
    GROUP BY user_id
) AS groupsq ON userdata.user_id = groupsq.user_id

Answer 5

SELECT id,user_id , amount , added    , 'new' AS status FROM new_table GROUP BY user_id
UNION ALL 
SELECT id, user_id , amount , added    , 'recursive' AS status FROM new_table 
WHERE id  NOT IN (SELECt * FROM ( SELECT idnew_table FROM   new_table GROUP BY user_id ) AS t )

MySQL获取当前计数直到结果

5 个答案: