Question

我有一个服务表，我需要列出时间，并且同一客户要求提供3种类型的服务。（ServiceA，serviceB，serviceC）

在低级示例中，我已经有两种服务类型，但我尝试添加第三种服务，但我不能。

很抱歉，其他任何非常重要的事情，我忘了提及我是如何得到每个服务在该列表上完成的日期的？

SELECT * 
FROM service 
WHERE status = 'CONCLUDED' AND (
   CASE WHEN service = 'servicoC' THEN 
   (CASE WHEN data LIKE '2017-05%' THEN 1 ELSE 0 END) ELSE data END) AND servico IN ' ServiceB ') 
GROUP BY id_client HAVING MIN (service) ='servicoA'AND MAX (service) ='servicoB'

表：

Id;
Id client
service;
date;
Status;

结果Ex：

Id_client，date_serviceA，date_serviceB，date_serviceC

Answer 1

您可以使用GROUP_BY和HAVING来执行此操作，例如：

SELECT id_client, COUNT(*)
FROM service
WHERE service IN ('ServiceA', 'ServiceB', 'ServiceC')
GROUP BY id_client
HAVING COUNT(DISTINCT service)) >= 3;

如果您需要按服务计算，则可以将其添加到GROUP BY和SELECT，例如：

SELECT id_client, service, COUNT(*)
FROM service
WHERE service IN ('ServiceA', 'ServiceB', 'ServiceC')
GROUP BY id_client, service
HAVING COUNT(DISTINCT service)) >= 3;

<强>更新

如果您想要每个服务的最长日期，那么您可以使用内部SELECT查询来执行此操作，例如：

SELECT s.id_client, s.service, COUNT(s.*),
(SELECT MAX(date) FROM service where id_client = s.id_client and service = 'ServiceA') AS "date_ServiceA",
(SELECT MAX(date) FROM service where id_client = s.id_client and service = 'ServiceB') AS "date_ServiceB",
(SELECT MAX(date) FROM service where id_client = s.id_client and service = 'ServiceC') AS "date_ServiceC"
FROM service s
WHERE s.service IN ('ServiceA', 'ServiceB', 'ServiceC')
GROUP BY s.id_client, s.service
HAVING COUNT(DISTINCT s.service)) >= 3;

使用Mysql比较数据

1 个答案: