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问Mysql合并计数多列
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Stack Overflow用户
提问于 2020-03-03 23:46:57
回答 1查看 39关注 0票数 2

我有一个表(Tb_data),它是这样的

代码语言:javascript
复制
+---------+---------------------+---------------------+---------------------+---------------------+ 
| Disease | Additional_Disease1 | Additional_Disease2 | Additional_Disease3 | Additional_Disease4 |
+---------+---------------------+---------------------+---------------------+---------------------+ 
| A01     | A03                 | A03                 |                     |                     |
| A03     | A02                 |                     |                     |                     |
| A03     | A05                 |                     |                     |                     |
| A03     | A05                 |                     |                     |                     |
| A02     | A05                 | A01                 | A03                 |                     | 
+---------+---------------------+---------------------+---------------------+---------------------+ 

我的问题是如何让它变成这样

代码语言:javascript
复制
+---------+-------+ 
| Disease | Total |
+---------+-------+
| A03     | 6     |
| A05     | 3     |
| A01     | 2     |
| A02     | 2     |
+---------+-------+

哦,这是我失败的尝试

代码语言:javascript
复制
select Disease, 
count(Disease + Additional_Disease1 + Additional_Disease2 + Additional_Disease3 + Additional Disease_4) as Total
from tb_data
group by Disease
order by Disease desc

我也试过this,但它不起作用,它显示“未知的列‘疾病’在‘字段列表’,因为我真的不明白哪里错了

EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2020-03-03 23:49:23

您可以使用union all取消透视数据集,然后进行聚合:

代码语言:javascript
复制
select disease, count(*) total
from (
    select disease from mytable
    union all select additional_disease1 from mytable
    union all select additional_disease2 from mytable
    union all select additional_disease3 from mytable
    union all select additional_disease4 from mytable
) t
group by disease
order by total desc, disease
票数 2
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/60510875

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