Question

如何获得三个记录，即他的记录，以前的等级记录和等级记录我在下面的小提琴里有SP：

 DELIMITER $$

 CREATE DEFINER=`ntc`@`%` PROCEDURE `stckrank`()
BEGIN
SELECT AgentID, 
    current_day_amount, 
    month_amount, 
    year_amount,
    @Position:=@Position + 1 AS `Rank`
FROM
(
SELECT just_agent.AgentID, 
        total_current_date.Totoalamountperday AS current_day_amount, 
        total_current_month.Totoalamountpermonth AS month_amount, 
        total_year.Totoalamountperyear AS year_amount
     FROM
(
    SELECT DISTINCT AgentID
    FROM collection
) just_agent
LEFT OUTER JOIN
(
    select AgentID, SUM(AmountRecevied) as Totoalamountperday
    from collection
    where  day(Date_Time) = day(CURRENT_DATE())
    group by AgentID
) total_current_date
ON just_agent.AgentID = total_current_date.AgentID
LEFT OUTER JOIN
(
    select AgentID, sum(AmountRecevied) as Totoalamountpermonth
    from collection
    where  date_time between DATE_FORMAT(NOW(), '%Y-%m-01') and LAST_DAY(now() - interval 0 month) 
    group by AgentID
) total_current_month
ON just_agent.AgentID = total_current_month.AgentID
LEFT OUTER JOIN
(
    select AgentID, sum(AmountRecevied) as Totoalamountperyear
    from collection
    where  year(Date_Time) = YEAR(CURRENT_DATE()) 
    group by AgentID
) total_year
ON just_agent.AgentID = total_year.AgentID
ORDER BY total_year.Totoalamountperyear DESC
) Sub1
 CROSS JOIN (SELECT @Position:=0) Sub2;
END

输出：

agentiid dayamount monthamount yearamount rank
 2        3000       4000          7000   1
 1        2000      3000           5000   2
 3         200       300            500   3
 4         100       100            200    4

现在，如果我选择代理商ID 2，我需要获得以下记录并且没有上述记录，所以null或0应该显示

选择代理商ID -2 需要展示或获得他的排名记录和记录之后

agentiid dayamount monthamount yearamount rank
    2        3000       4000          7000   1
    1        2000      3000           5000   2

Answer 1

您可以按等级条件将表格连接到自身：

SELECT t2.*
FROM yourTable t1
JOIN yourTable t2
ON ABS(t1.rank - t2.rank) <= 1.5 --is this what you want ?
WHERE t1.agentiid = 2

This SQL fiddle提供了理想的结果。

使用此连接，原始表中的每一行都与自身（始终）以及具有上一级和下一级（如果存在）的行相关联。

如何根据agentID获取三条记录并在mysql中排名

1 个答案: