Mysql 如何在交叉表查询中聚合不重复的数据

我在MySQL中有这样一个示例表：（实际表中有更多的种族和学校）

首先，我将汇总每个种族的数量并按降序排列。查询如下：（我知道如何处理此部分）

最后，我想得到一个如下的交叉表查询：（根据每个种族的人数提取前3名。此外，最终的查询是按每个种族的总人数排序的）

我尝试了下面列出的代码。我无法获取“总和”列。在“麻省理工学院”和“福特汉姆大学”的领导下，每个种族的人数是相同的，并且有重复的。（参考） 我想这是因为我想按每个种族的总人数排序，然后按总人数排序。然而，不同学校中每个种族的人数顺序不同

SELECT race,
   COUNT(IF(school="MIT",race,0)) as MIT,
   COUNT(IF(school="Fordham",race,0)) as Fordham
FROM table_name
GROUP by race

它应该是

SUM（）

，而不是

COUNT（）

---

希望这有帮助。

您可以使用动态sql查询来实现这一点

查询

set @query = null;
select
group_concat(distinct
    concat(
      'count(case when school = ''',
      school, ''' then school end) as ',school
    )
  ) into @query
from your_table_name ;

set @query = concat('select race, ', @query, ',count(*) as `SUM` from your_table_name 
                  group by race order by count(*) desc limit 3
');

prepare stmt from @query;
execute stmt;
deallocate prepare stmt;

输出

+----------+-----+---------+-----+-----+
|     race | MIT | Fordham | NYU | SUM |
+----------+-----+---------+-----+-----+
|    Asian |   4 |       4 |   3 |  11 |
|    White |   3 |       2 |   1 |   6 |
| Hispanic |   1 |       2 |   1 |   4 |
+----------+-----+---------+-----+-----+

首先：

select * from 
(
    select count(*) as count,race from table1 group by race
) as data
order by data.count desc;

最后

select data.* from (
SELECT race,
      SUM(IF(school='MIT',1,0)) as MIT,
      SUM(IF(school='Fordham',1,0)) as Fordham,
      SUM(IF(school='NYU',1,0)) AS NYU,
      COUNT(*) AS total_Sum
FROM table1
GROUP by race ) as data
order by data.total_Sum desc;

---

谢谢这很有帮助。就在附近。但是总数（*）将对包括美国印第安人在内的所有种族进行汇总。但是，如果我只想对前三名的人数或我想展示的学校数量进行汇总？

set @query = null;
select
group_concat(distinct
    concat(
      'count(case when school = ''',
      school, ''' then school end) as ',school
    )
  ) into @query
from your_table_name ;

set @query = concat('select race, ', @query, ',count(*) as `SUM` from your_table_name 
                  group by race order by count(*) desc limit 3
');

prepare stmt from @query;
execute stmt;
deallocate prepare stmt;

+----------+-----+---------+-----+-----+
|     race | MIT | Fordham | NYU | SUM |
+----------+-----+---------+-----+-----+
|    Asian |   4 |       4 |   3 |  11 |
|    White |   3 |       2 |   1 |   6 |
| Hispanic |   1 |       2 |   1 |   4 |
+----------+-----+---------+-----+-----+

select * from 
(
    select count(*) as count,race from table1 group by race
) as data
order by data.count desc;

select data.* from (
SELECT race,
      SUM(IF(school='MIT',1,0)) as MIT,
      SUM(IF(school='Fordham',1,0)) as Fordham,
      SUM(IF(school='NYU',1,0)) AS NYU,
      COUNT(*) AS total_Sum
FROM table1
GROUP by race ) as data
order by data.total_Sum desc;