如何根据php中的两个值查找排名？

Question

我在 php 中有一个与锻炼相关的 Web 应用程序，其中包含锻炼表，该表根据活动类型（如跑步、游泳、骑自行车等）存储用户的日常锻炼数据。我需要根据高排名系统生成排行榜。

Default workout table

SELECT 
  SUM(avg_speed) AS average_speed, 
  SUM(max_speed) AS maximum_speed, 
  date_created, 
  user_id 
FROM workouts 
WHERE date_created LIKE '%2019-12%' 
GROUP BY user_id

通过这个查询，我得到如下输出：

output after running select query

我需要的输出应该基于average_speed

Answer 1

尝试像这样玩查询。

select user_id,avg,max from users nu
join
(
    select sum(avg_speed) as avg,sum(max_speed) as max,user_id FROM workouts group by user_id
)t2 on t2.user_id=nu.id 
 where avg<30 and max <50
 and date_created LIKE '%2019-12%'
group by t2.user_id order by avg desc;

在您的 php 代码中循环此结果以添加您的排名逻辑，如下例所示。 （以下逻辑和代码未经测试）。

$rank=0;
$same_rank=0;
foreach($workouts as $key=>$item)
{
 $avg=$item->avg;
 if($avg!=$prev_item)
 { 
    $add_rank=$same_rank!=0?$same_rank:1;
    $rank=$rank+$add_rank;
    $same_rank=0;
 }
 else
 {
  $same_rank++; //increments if same value repeated
 }

 $workouts[$key]['rank']=$rank;
 $prev_item=$avg;


 }
echo $workouts;

这可能对你有帮助

Answer 2

你可以试试这个 SQL 语句：

SELECT s1.avg_speed, s1.max_speed, s1.user_id,
       @row_num:=IF(s1.avg_speed = @last_avg_speed, @row_num, @origin_row_num+1) rank_no,
       @last_avg_speed:=s1.avg_speed last_avg_speed,
       @origin_row_num:=@origin_row_num+1 origin_row_num
FROM (
    SELECT SUM(avg_speed) avg_speed, SUM(max_speed) max_speed, user_id FROM workouts
        WHERE date_created LIKE '2019-12%'
        GROUP BY user_id
) s1, (SELECT @origin_row_num:=0, @row_num:=0, @last_avg_speed:=0) R
WHERE s1.avg_speed < 30 AND s1.max_speed < 50
ORDER BY s1.avg_speed DESC

这是我的测试代码：

create table workouts
(
    id int auto_increment
        primary key,
    user_id int not null,
    date_created date null,
    activity_type varchar(10) null,
    avg_speed float null,
    max_speed float null
);
DELETE FROM workouts WHERE user_id IN (1, 2, 3)
INSERT INTO workouts (id, user_id, date_created, activity_type, avg_speed, max_speed)
VALUES (NULL, 1, '2019-12-01', 'Walk', 1, 10),
       (NULL, 1, '2019-12-02', 'Walk', 1, 10),
       (NULL, 1, '2019-12-03', 'Walk', 1, 10),
       (NULL, 1, '2019-12-04', 'Walk', 1, 10),

       (NULL, 2, '2019-12-01', 'Walk', 1, 10),
       (NULL, 2, '2019-12-02', 'Walk', 5, 10),
       (NULL, 2, '2019-12-03', 'Walk', 1, 10),
       (NULL, 2, '2019-12-04', 'Walk', 1, 10),

       (NULL, 3, '2019-12-01', 'Walk', 1, 10),
       (NULL, 3, '2019-12-02', 'Walk', 5, 10),
       (NULL, 3, '2019-12-03', 'Walk', 1, 10),
       (NULL, 3, '2019-12-04', 'Walk', 1, 10);

SELECT s1.avg_speed, s1.max_speed, s1.user_id,
       @row_num:=IF(s1.avg_speed = @last_avg_speed, @row_num, @origin_row_num+1) rank_no,
       @last_avg_speed:=s1.avg_speed last_avg_speed,
       @origin_row_num:=@origin_row_num+1 origin_row_num
FROM (
    SELECT SUM(avg_speed) avg_speed, SUM(max_speed) max_speed, user_id FROM workouts
        WHERE date_created LIKE '2019-12%'
        GROUP BY user_id
) s1, (SELECT @origin_row_num:=0, @row_num:=0, @last_avg_speed:=0) R
WHERE s1.avg_speed < 30 AND s1.max_speed < 50
ORDER BY s1.avg_speed DESC

输出：

|--------------------------------------------------|
|avg_speed|max_speed|user_id|rank_no|last_avg_speed|
|--------------------------------------------------|
|8        |40       |2      |1      |8             |
|--------------------------------------------------|
|8        |40       |3      |1      |8             |
|--------------------------------------------------|
|4        |40       |1      |3      |4             |
|--------------------------------------------------|

在最后一列last_avg_speed总是等于avg_speed，你可以忽略它。

顺便说一句，如果你使用的是 MySQL 8.0.0 或更高版本，你可以使用ROW_NUMBER() 和RANK() 函数来解决它，非常非常简单。 here 是文档。