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
- 1
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