php - 连接多个表并根据第一个表 ID 获取状态

Question

代理数据库表

agent_id  agent_name company_name
--------  ----------  -----------
1         AAA           XXX
2         BBB           YYY
3         CCC           ZZZ
4         DDD           XYZ

驱动程序数据库表

agent_id  driver_id   driver_name last_viewed
--------  ----------  ----------- -----------
2         1           EEE           1
2         2           FFF           0
2         3           GGG           0
1         4           HHH           0
3         5           III           1
3         6           JJJ           1

我想要这样的输出

Agent Details     Driver details

1, AAA,             1 Drivers (0 active | 1 idle)
Company name

2, BBB,             3 Drivers (1 active | 2 idle)
Company name

3, CCC,             2 Drivers (2 active | 0 idle)
Company name

我已经尝试过以下查询

$sql="SELECT a.*,d.*, COUNT(d.driver_id) AS drivers_count FROM ta_agent a JOIN ta_drivers d USING(agent_id) GROUP BY a.agent_id";

我想根据 last_viewed 列显示驱动程序的事件和空闲状态。例如，agent_id 2 有三个驱动程序 (1,2,3)，这 3 个驱动程序的 Last_viewed 列中有 1,0,0。所以，我想显示这样的输出:1 个事件和 2 个空闲...

Answer 1

试试这个，。

select a.agent_id, a.agent_name, a.company_name, ifnull(cnt_all,0) total_drivers,ifnull(cnt_active,0) active_drivers, ifnull(cnt_idle,0) idle_drivers
from agent a left join (select agent_id, count(*) cnt_all
                   from driver
                   group by agent_id) cnt on a.agent_id=cnt.agent_id

left join (select agent_id, count(*) cnt_idle
                   from driver
                   where last_viewed=0
                   group by agent_id) idle on a.agent_id=idle.agent_id

left join (select agent_id, count(*) cnt_active
                   from driver
                   where last_viewed=1
                   group by agent_id) active on a.agent_id=active.agent_id

这里是SQLFiddle