一、排名

/*普通排名:从1开始,顺序往下排*/
SELECT cs.*,@r :=@r + 1 AS rank
FROM cs,(SELECT @r := 0) r
ORDER BY score;

 

/*并列排名:相同的值是相同的排名*/
SELECT cs.* ,
CASE 
WHEN @p=score THEN @r
WHEN @p:=score THEN @r:=@r+1
END rank
FROM cs,(SELECT @r:=0,@p:=NULL)r
ORDER BY score;

 

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/*并列排名:相同的值名次相同,与上例中的并列排名不同*/
SELECT city,score,rank
FROM
(
SELECT cs.*,
@c:=IF(@p=score,@c,@r) AS rank,
@p:=score,
@r:=@r+1
FROM cs ,(SELECT @p:=NULL,@r:=1,@c:=0)r
ORDER BY score
)c

 

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 二、分组后组内排名

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/*分组普通排名:顺序排名*/
SELECT city,score,rank
FROM
(
SELECT cs.*,IF(@p=city,@r:=@r+1,@r:=1) AS rank,
    @p:=city
FROM cs,(SELECT @p:=NULL,@r:=0)r
ORDER BY city,score
)s;
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/* 分组后并列排名:组内相同数值排名相同*/
SELECT city,score,rank
FROM
(
SELECT *,
IF(@p=city,
    CASE 
       WHEN @s=score THEN @r
       WHEN @s:=score THEN @r:=@r+1
    END,
   @r:=1 ) AS rank,
@p:=city,
@s:=score
FROM cs,(SELECT @p:=NULL,@s:=NULL,@r:=0)r
ORDER BY city,score 
)s;

 

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 三、分组后取各组的前两名

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/*取每组分数高的前两个,法一*/
SELECT city,score,rank
FROM
(
SELECT *,
IF(@p=city,
    CASE 
        WHEN @s=score THEN @r
        WHEN @s:=score THEN @r:=@r+1
    END,
  @r:=1 ) AS rank,
@p:=city,
@s:=score
FROM cs,(SELECT @p:=NULL,@s:=NULL,@r:=0)r
ORDER BY city,score DESC 
)s
WHERE rank <3;

 

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/*分组后取前两个,法二*/
SELECT * FROM cs c
WHERE (
    SELECT count(*) FROM cs
    WHERE city=c.city AND score>c.score )<2
  ORDER BY city,score DESC

 

 

 

如何用mysql 查询出数据出数据排名,然后加编号;

 

表字段说明:查询e_Task_Result表 ,score代表分数,task_result_id 代表id

SELECT
(@rowNO := @rowNo+1) AS rowno,
task_result_id ,score FROM (
(SELECT task_result_id ,score FROM e_task_result ORDER BY score DESC) a,(SELECT @rowNo :=0) b )

 

如何将查询出来的排名更新到表中
表字段说明:查询e_Task_Result表 ,score代表分数,task_result_id 代表id,rank代表排名

UPDATE e_task_result d
LEFT JOIN (SELECT
(@rowNO := @rowNo+1) AS rowno,
task_result_id
FROM (SELECT
task_result_id
FROM e_task_result etr
ORDER BY task_result_status DESC,etr.score DESC,end_date) a,
(SELECT
@rowNO :=0) b) c
ON c.task_Result_id = d.task_result_id
SET d.rank = rowno

 

再如:

UPDATE 
(SELECT @a := @a + 1 AS rank, user_code, task_type 
FROM ( 
select * from user_score_quality 
where  task_type =1  
ORDER BY user_score_quality.last_finish_task_num desc, user_score_quality.update_time asc
) h, (SELECT @a := 0 ) t 
) t1, user_score_quality t2 

SET t2.last_rank = t1.rank 
WHERE t2.user_code = t1.user_code  AND  t2.task_type=t1.task_type

 

 

 

参考:

https://blog.csdn.net/w329636271/article/details/51751282 

https://www.cnblogs.com/niniya/p/9046449.html

https://www.jianshu.com/p/bb1b72a1623e

http://blog.sina.com.cn/s/blog_4c197d420101e408.html

posted on 2019-03-27 11:39  yuluoxingkong  阅读(1891)  评论(0编辑  收藏  举报