Question

我有3个包含相同字段的表（ticket1，ticket2，ticket3）：

ticket1:       ticket2:        ticket2:
========       ========        ========
ticket_id      ticket_id       ticket_id
status         status          status

我的查询就是这样：

("SELECT (SELECT COUNT( * ) FROM `ticket1` WHERE `status` =9) AS done,
                        (SELECT COUNT( * ) FROM `ticket1` WHERE `status` =10) AS Incomplete,
                        (SELECT COUNT( * ) FROM `ticket1` WHERE `status` =2) AS New")

这是对票进行计数并按状态进行过滤。

我的问题是，如何从ticket1，ticket2，ticket3将所有票据计入完成，未完成和新的总数中。

帮我，谢谢。

Answer 1

尝试一下。

select status, count(1) from 
(select * from ticket1
union all
select * from ticket2
union all
select * from ticket3) group by status;

让我知道您是否遇到任何问题

Answer 2

您可以从联合全部中选择

select  sum(case when status =  9 then  1 else 0 end)  done,
        sum(case when status =  10 then  1 else 0 end)  Incomplete,
        sum(case when status =  2 then  1 else 0 end)  New,  
        sum(case when status  in (9,10,2)  then  1 else 0 end)  deon_incomplete_new,               
        count(*)   tot  
from (
      select ticket_id, status  from ticket1
      union all  
      select ticket_id, status  from ticket2
      union all  
      select ticket_id, status  from ticket2
 ) t

Answer 3

我猜3d表的名称是ticket3，对吧？

SELECT 
  (SELECT COUNT(*) FROM `ticket1` WHERE `status` = 9) +
  (SELECT COUNT(*) FROM `ticket2` WHERE `status` = 9) + 
  (SELECT COUNT(*) FROM `ticket3` WHERE `status` = 9) AS TotalDone,
  (SELECT COUNT(*) FROM `ticket1` WHERE `status` = 10) +
  (SELECT COUNT(*) FROM `ticket2` WHERE `status` = 10) + 
  (SELECT COUNT(*) FROM `ticket3` WHERE `status` = 10) AS TotalIncomplete,
  (SELECT COUNT(*) FROM `ticket1` WHERE `status` = 2) +
  (SELECT COUNT(*) FROM `ticket2` WHERE `status` = 2) + 
  (SELECT COUNT(*) FROM `ticket3` WHERE `status` = 2) AS TotalNew

Answer 4

我建议：

select sum(case when status = 9 then cnt else 0 end) as done,
       sum(case when status = 10 then cnt else 0 end) as incomplete,
       sum(case when status = 2 then cnt else 0 end) as new
from ((select status, count(*) as cnt from ticket1 group by status
      ) union all
      (select status, count(*) as cnt from ticket2 group by status
      ) union all
      (select status, count(*) as cnt from ticket3 group by status
      ) 
     ) t;

或者，您可以考虑使用group by将值放在单独的行中。

如果性能是一个问题，而您还有许多其他故障单状态（或更具体地说，许多具有不同状态的行），则每个子查询中的where status in (2, 9, 10)可能会有所帮助。

从多个表总计

4 个答案: