用sql语句实现年龄分段统计

SELECT CASE WHEN (age >= 10 AND age <= 20) THEN '10-20' WHEN (age >= 21 AND age <= 30) THEN '21-30' ELSE '30-' END 'eag_layer', count(*) emps FROM address_book GROUP BY CASE WHEN (age >= 10 AND age <= 20) THEN '10-20' WHEN (age >= 21 AND age <= 30) THEN '21-30' ELSE '30-' END ORDER BY 1;

 

SELECT '10-20' 年龄段, COUNT(*) 人数 FROM [Table] WHERE [年龄] BETWEEN 10 AND 20 UNION ALL SELECT '21-30' 年龄段, COUNT(*) 人数 FROM [Table] WHERE [年龄] BETWEEN 21 AND 30 UNION ALL SELECT '31' 年龄段, COUNT(*) 人数 FROM [Table] WHERE [年龄] > 30 

 

select case when [年龄] BETWEEN 10 AND 20 then '10-20' when [年龄] BETWEEN 20 AND 30 then '20-30' when [年龄] > 30 then '30以上' end as '年龄段', count(*) as '人数' FROM [Table] 

 

先将年龄除10取整

select floor(年龄/10) as age from 表

再根据年龄整数分组统计

select age ,count(age) from ( select floor(年龄/10) as age from 表 ) group by age

 

这样基本效果就出来了，达到楼主的要求就要加如函数计算了

sql语法

select convert(varchar，age*10)+'--'+convert(varchar，(age+1)*10) ,count(age) from ( select floor(年龄/10) as age from 表 ) group by age

 

oracle语法

select age*10 || '--'|| (age+1)*10 ,count(age) from ( select floor(年龄/10) as age from 表 ) group by age