Question

我有一个MySQL表

id | string
------------------------------
 1 | one, two, three
 2 | two, three, five
 3 | one, three, four
 4 | three, five
 5 | one, two, three
 6 | two, three, five
 7 | one, three, four
 8 | one, three, five
 9 | two, three, four, five
10 | one, two, three, five

我希望获取three列中包含子字符串string的元素的比率。我可以轻松地执行两个查询并手动分割：

SELECT COUNT(*) FROM `table` WHERE `string` LIKE '%three%'

并将其除以

SELECT COUNT(*) FROM `table`

但我怎么能只用一个查询来执行此操作呢？

Answer 1

您可以将SUM与IF一起使用，在括号内给出1或0： -

SELECT SUM(IF(`string` LIKE '%three%', 1, 0))/COUNT(*)  
FROM `table`

Answer 2

选择if条件，如果条件为真，则返回1其他明智的0。这里count(*)返回总数或记录。

SELECT (SUM(IF(`string` LIKE '%three%', 1, 0)) * 100)/COUNT(*)  
FROM `table`

Answer 3

试试这个。

select sum(case when string like '%three%' then 1 else 0 end)/count(*) from table.

Answer 4

select sum(find_in_set('three', replace(`string`, ' ', '')) > 0) * 100 / count(*) as percentage
from your_table

但你应该考虑改变你的桌面设计。永远不会永远不会在一列中存储多个值！

如何在一个查询中执行两次COUNT（*）次提取？

4 个答案: