我需要根据百分比限制记录,但 MYSQL 不允许这样做。我需要 (count(
User Id)/max(Total_Users_bynow)
的 10% User Id
我的代码如下:
select * from flavia.TableforThe_top_10percent_of_the_user where `User Id` in (select distinct(`User Id`) from flavia.TableforThe_top_10percent_of_the_user group by `User Id` having count(distinct(`User Id`)) <= round((count(`User Id`)/max(Total_Users_bynow))*0.1)*count(`User Id`));
请帮忙。
最佳答案
考虑将您的问题分解。您可以使用用户变量来获取您需要的内容。 Quoting from this question's answers :
You don't have to solve every problem in a single query.
所以...让我们完成这件事。我不会提供完整的查询,但会提供一些示例:
-- Step 1. Get the total of the rows of your dataset
set @nrows = (select count(*) from (select ...) as a);
-- --------------------------------------^^^^^^^^^^
-- The full original query (or, if possible a simple version of it) goes here
-- Step 2. Calculate how many rows you want to retreive
-- You may use "round()", "ceiling()" or "floor()", whichever fits your needs
set @limrows = round(@nrows * 0.1);
-- Step 3. Run your query:
select ...
limit @limrows;
罢工><罢工>罢工>
<小时/>查了一下,发现this post这说明我的上述方法行不通。然而,还有一个替代方案:
-- Step 1. Get the total of the rows of your dataset
set @nrows = (select count(*) from (select ...) as a);
-- --------------------------------------^^^^^^^^^^
-- The full original query (or, if possible a simple version of it) goes here
-- Step 2. Calculate how many rows you want to retreive
-- You may use "round()", "ceiling()" or "floor()", whichever fits your needs
set @limrows = round(@nrows * 0.1);
-- Step 3. (UPDATED) Run your query.
-- You'll need to add a "rownumber" column to make this work.
select *
from (select @rownum := @rownum+1 as rownumber
, ... -- The rest of your columns
from (select @rownum := 0) as init
, ... -- The rest of your FROM definition
order by ... -- Be sure to order your data
) as a
where rownumber <= @limrows
希望这会有所帮助(我认为这次它会正常工作)
关于MySQL 限制解决方法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32362562/