Question

我使用的查询（在链接右侧）生成了此结果（左侧）请参阅http://www.sqlfiddle.com/#!2/f34f1/1

我是 JasperReports 和 MySql 的新手。我想

尚未提及嫌疑人/确认客户
在5天内被推荐的客户数
超过5天被推荐的客户数量。

我能够计算在5天内被推荐的客户

SELECT COUNT(*) from 
(select p.patient_id,

(CASE WHEN st.smear_result <> 'NEGATIVE' OR st.gxp_result='MTB+'
THEN IF(DATEDIFF(r.date_referred,MIN(st.date_smear_tested)) IS NULL,'N/A',(DATEDIFF(r.date_referred,MIN(st.date_smear_tested))))

ELSE
(CASE WHEN st.smear_result='NEGATIVE' OR st.gxp_result='MTB-'
THEN IF(DATEDIFF(r.date_referred,MAX(st.date_smear_tested)) IS NULL,'N/A',(DATEDIFF(r.date_referred,MAX(st.date_smear_tested))))

ELSE 'N/A' end )END) as days_taken,

IF(r.date_referred IS NULL,'N/A',r.date_referred) date_referred
from patient as p 
right outer join sputum_test as st on p.patient_id=st.patient_id 
right outer join referral as r on r.patient_id=st.patient_id

where p.suspected_by is not null and (p.patient_status='SUSPECT' or p.patient_status='CONFIRMED')

group by p.patient_id

having days_taken <=5) AS SUBQUERY;

超过5天被推荐的客户数量。

SELECT COUNT(*) from 
(select p.patient_id,

(CASE WHEN st.smear_result <> 'NEGATIVE' OR st.gxp_result='MTB+'
THEN IF(DATEDIFF(r.date_referred,MIN(st.date_smear_tested)) IS NULL,'N/A',(DATEDIFF(r.date_referred,MIN(st.date_smear_tested))))

ELSE
(CASE WHEN st.smear_result='NEGATIVE' OR st.gxp_result='MTB-'
THEN IF(DATEDIFF(r.date_referred,MAX(st.date_smear_tested)) IS NULL,'N/A',(DATEDIFF(r.date_referred,MAX(st.date_smear_tested))))

ELSE 'N/A' end )END) as days_taken,

IF(r.date_referred IS NULL,'N/A',r.date_referred) date_referred
from patient as p 
right outer join sputum_test as st on p.patient_id=st.patient_id 
right outer join referral as r on r.patient_id=st.patient_id

where p.suspected_by is not null and (p.patient_status='SUSPECT' or p.patient_status='CONFIRMED')
group by p.patient_id
having days_taken > 5) AS SUBQUERY;

但我如何获得计数尚未提交嫌疑人/已确认的客户？

我的计划是以某种方式将结果作为2列：
第1列：显示3个条件，第2列：显示它们旁边的行总和。

我将在 iReport 设计器中传递解决方案查询，将3个条件的饼图作为标签，并显示每个切片的百分比。

Answer 1

这样的事情应该有效：

SELECT SUM(days_taken <= 5) AS within_5_days,
       SUM(days_taken > 5) AS more_than_5,
       SUM(days_taken IS NULL) as not_yet_referred
FROM (...) AS subquery

显然，子查询应为未引用的客户端生成NULL，而不是原始子查询中的N/A。

mysql如何获取查询的结果行的计数，条件是datediff返回的列

1 个答案: