题解 | #高级操作符练习(2）#

三种写法：

SELECT IF(age<25 OR age IS NULL,'25岁以下','25岁及以上')  as age_cut,COUNT(device_id) as Number
FROM user_profile
GROUP BY age_cut

#方法2：联合查询
SELECT '25岁以下' as age_cut,COUNT(device_id) as Number
FROM user_profile
WHERE age < 25 OR age IS NULL
UNION
SELECT '25岁及以上' age_cut,COUNT(device_id) Number
FROM user_profile
WHERE age >= 25

#方法3：CASE函数
SELECT CASE WHEN age >= 25 THEN '25岁及以上'
            WHEN age < 25 OR age IS NULL THEN '25岁以下'    #此句可直接改为：else '25岁以下'
            END as age_cut,COUNT(*) as number
FROM user_profile
GROUP BY age_cut