mysql如何分组统计并求出百分比
吾爱主题
阅读:173
2022-11-24 15:55:00
评论:0
mysql分组统计并求出百分比
1、mysql 分组统计并列出百分比
SELECT point_id, pname_cn, play_num, round( play_num / total * 100, 2 ) as `ratio` FROM ( SELECT * FROM ( SELECT ANY_VALUE ( `point_id` ) AS point_id, ANY_VALUE ( `pname_cn` ) AS pname_cn, sum( `play` ) AS play_num FROM `dt_collect_antique` WHERE`add_time` BETWEEN "2020-07-02" AND "2020-07-05" GROUP BY `point_id` ) t1 INNER JOIN ( SELECT sum( `play` ) AS total FROM `dt_collect_antique` WHERE`add_time` BETWEEN "2020-07-02" AND "2020-07-05" ) t2 ON 1 = 1 ) t ORDER BY `play_num` DESC LIMIT 0, 10;
--查出符合条件并且分组, 统计出每组数量
SELECT ANY_VALUE ( `point_id` ) AS point_id, ANY_VALUE ( `pname_cn` ) AS pname_cn, sum(`like`) as like_num FROM `dt_collect_antique` WHERE`add_time` BETWEEN "2020-07-02" AND "2020-07-05" GROUP BY `point_id` ) t1
--查出符合条件,总数量
(SELECT sum( `play` ) AS total FROM `dt_collect_antique` WHERE`add_time` BETWEEN "2020-07-02" AND "2020-07-05" ) t2
2、按年龄段分组,并求个年龄段占比
SELECT age_group, age_total, round( age_total / total * 100, 2 ) as `ratio` FROM ( SELECT * FROM ( SELECT SUM(total) AS age_total, CASE WHEN age >= 0 AND age < 18 THEN "18岁以下" WHEN age >= 18 AND age <= 25 THEN "18岁到25岁" WHEN age >= 26 AND age <= 35 THEN "26岁到35岁" WHEN age >= 36 AND age <= 45 THEN "36岁到45岁" WHEN age >= 46 AND age <= 60 THEN "46岁到60岁" ELSE "60岁以上" END AS age_group FROM dt_collect_age WHERE `add_time` BETWEEN ".$time[0]." AND ".$time[1]." GROUP BY age_group ) t1 INNER JOIN ( SELECT SUM( `total` ) AS total FROM `dt_collect_age` WHERE `add_time` BETWEEN ".$time[0]." AND ".$time[1]." ) t2 ON 1 = 1 ) t LIMIT 0, 6;
mysql求百分比的几种方法
函数介绍
1、ROUND(X,D)和ROUND(X)
round函数用于数据的四舍五入,x指要处理的数,d是指保留几位小数
round(x) ,其实就是 round(x,0)
d可以是负数,代表指定小数点左边的d位整数位为0,同时小数位均为0
2、FORMAT(X,D)
提供数据内容格式化功能,可以格式化数据为整数或者浮点数,能四舍五入
D为负数时,按0处理
3、LEFT(str,len)
left函数是一个字符串函数,它返回具有指定长度的字符串的左边部分,str为要处理的字符串,len为长度
left函数为字符串截取,不能四舍五入
4、CONCAT(str1,str2,...)
concat函数用于将多个字符串连接成一个字符串
利用round,format,left与concat求百分比
注意:使用left按位截取百分比时,位数要根据需要合理设置,否则容易出现意外BUG
以上为个人经验,希望能给大家一个参考,也希望大家多多支持服务器之家。
原文地址:https://blog.csdn.net/qq_38776443/article/details/107167702
声明
1.本站遵循行业规范,任何转载的稿件都会明确标注作者和来源;2.本站的原创文章,请转载时务必注明文章作者和来源,不尊重原创的行为我们将追究责任;3.作者投稿可能会经我们编辑修改或补充。