尝试此查询：

select @user_id_lag := 0;
select id, user_id, amount, added, status from (
    select case when @user_id_lag = user_id then 'recurring' else 'new' end status,
           @user_id_lag := user_id user_id,
           id, amount, added
    from tbl
    order by user_id, added
) a order by added

演示

请尝试以下操作：http://sqlfiddle.com/#!2011年9月6日

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

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，也可以通过连接子查询或者使用相关的 EXISTS 条款。从mysql 8.0开始，我将使用 MIN() OVER() .

你可以简单地 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

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 )