MySQL按年龄区间统计男性患者各类疾病计数的实现方法
SQL调整方案
你可以通过条件聚合的方式实现三个指标的分别统计,以下是修改后的语句:
基础版(仅返回有数据的年龄区间)
SELECT t.ranges, SUM(CASE WHEN dm = 1 THEN 1 ELSE 0 END) AS dm_count, SUM(CASE WHEN htn = 1 THEN 1 ELSE 0 END) AS htn_count, SUM(CASE WHEN ckd = 1 THEN 1 ELSE 0 END) AS ckd_count FROM ( SELECT age, dm, htn, ckd, CASE WHEN age >= 0 AND age < 20 THEN 'under 20' WHEN age >= 20 AND age < 30 THEN '20-29' WHEN age >= 30 AND age < 40 THEN '30-39' WHEN age >= 40 AND age < 50 THEN '40-49' WHEN age >= 50 AND age < 60 THEN '50-59' WHEN age >= 60 AND age < 70 THEN '60-69' WHEN age >= 70 AND age < 80 THEN '70-79' ELSE '80+' END AS ranges FROM patient_sample WHERE LOWER(sex) = 'm' ) AS t GROUP BY t.ranges
全量区间版(返回所有8个年龄区间,无数据的指标计数显示为0,完全匹配示例输出)
如果需要和你给出的示例结果一致,展示所有年龄段区间,哪怕没有符合条件的患者也显示0计数,可以用以下语句:
SELECT r.ranges, COALESCE(SUM(CASE WHEN p.dm = 1 THEN 1 ELSE 0 END), 0) AS dm_count, COALESCE(SUM(CASE WHEN p.htn = 1 THEN 1 ELSE 0 END), 0) AS htn_count, COALESCE(SUM(CASE WHEN p.ckd = 1 THEN 1 ELSE 0 END), 0) AS ckd_count FROM ( -- 手动构造所有需要展示的年龄区间 SELECT 'under 20' AS ranges UNION ALL SELECT '20-29' UNION ALL SELECT '30-39' UNION ALL SELECT '40-49' UNION ALL SELECT '50-59' UNION ALL SELECT '60-69' UNION ALL SELECT '70-79' UNION ALL SELECT '80+' ) AS r LEFT JOIN ( SELECT age, dm, htn, ckd, CASE WHEN age >= 0 AND age < 20 THEN 'under 20' WHEN age >= 20 AND age < 30 THEN '20-29' WHEN age >= 30 AND age < 40 THEN '30-39' WHEN age >= 40 AND age < 50 THEN '40-49' WHEN age >= 50 AND age < 60 THEN '50-59' WHEN age >= 60 AND age < 70 THEN '60-69' WHEN age >= 70 AND age < 80 THEN '70-79' ELSE '80+' END AS ranges FROM patient_sample WHERE LOWER(sex) = 'm' ) AS p ON r.ranges = p.ranges GROUP BY r.ranges
主要修改点说明
- 内层查询新增
dm/htn/ckd三个字段的查询,供外层聚合计算使用 - 外层将原来的
count(*)替换为三个条件求和语句,仅统计对应指标值为1的行数 - 修正了原SQL中age字段用字符串比较的错误,去掉了数值的单引号,避免非预期的比较结果
- 新增
LOWER(sex)逻辑,兼容sex字段的大小写差异 - 全量区间版通过构造全量区间临时表左关联业务数据,配合
COALESCE函数把空值转为0,完全匹配你需要的输出格式。
内容的提问来源于stack exchange,提问作者user16319635
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