Mysql 计算组计数的百分比（*）

如何获取给定格式的百分比值

Select * from Namelist;
Name      Age
Sathish   25
Sathish   65
Sathish   55
Sathish   45
Sathish   35
Jana      55
Jana      25
Jana      10
Bala      55
Bala      26

请共享sql查询？

此查询（未测试）应能正常工作：

Name   Count   Percentege
Sathish  5     50%
Jana     3     30%
Bala     2     20%

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替换列名并尝试以下操作：

SELECT Name,
COUNT(*) AS Count,
(COUNT(*) / _total ) * 100 AS Percentege
FROM Namelist,
(SELECT COUNT(*) AS _total
  FROM Namelist) AS myTotal
GROUP BY Name;

输出：

SELECT  iName, 
    COUNT(iName) AS `Count`, 
    concat(FORMAT(((COUNT(iName) * 100) / NewPeople.iCount),2),'%') AS `Percentage`
FROM   people, (SELECT COUNT(iName) AS iCount FROM people) NewPeople 
GROUP BY iName;

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这是一些其他答案的一个稍微性感的版本-注意使用了

sum（100）

，以避免更长（更普通）的

count（*）*100

：）

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平均值不产生百分比-它产生平均值！萨提斯-你愿意接受其中一个答案吗？已经好几个星期了。我们为您投入了工作-您至少可以接受一个或评论，这些都不适用于您不要在SQL上这样做（需要很长时间），请使用您的编程语言。

Name   Count   Percentage
Sathish  5     50.00%
Jana     3     30.00%
Bala     2     20.00%

select
name,
count(name) as `count`,
count(name)/(select count(*) from namelist)*100 as pct
from namelist
group by name

select name, count(*) as count, sum(100) / total as percentage
from namelist
cross join (select count(*) as total from namelist) x
group by 1