Question

我有一个定义表-def

event_data

及其交易记录在差异表-event

中

Id        Device            Location
1         GGHY199           USA
12        DFGHY71           India
145       APPHY75           USA

我希望输出为对于每一天，所有三个设备均应显示，如果没有事务发生，则应显示为null，我认为这是日期列中的月份的第一个日期。喜欢，

eventid        deviceid         event                date
123              12             Login              12-01-2019
32               12             Unreachable        18-02-2019
223              145            Unreachable        19-02-2019
334              1              DOWN               01-03-2019

当前我正在做

eventid        deviceid         event                date
null             1              null               01-01-2019
123              12             Login              12-01-2019
null             145            null               01-01-2019
null             1              null               01-02-2019
32               12             Unreachable        18-02-2019
223              145            Unreachable        19-02-2019
334              1              DOWN               01-03-2019
null             12             null               01-03-2019
null             145            null               01-03-2019

我显然没有得到我想要的东西。

谢谢！

Answer 1

假设您有一个有效的日历表来存放所需的所有日期（通过select union all表t模拟），则可以尝试在calendare和其他表之间使用左联接

select t.date, e.eventid, e.device, e.event, d.device
from (
  select   '2019-01-01'  date 
  union all 
  select   '2019-01-02' 
  union all 
  select   '2019-01-03' 
  .....
  union all 
  select   '2019-03-31' 
) t 
left event e on t.date = e.event 
left join device d  on e.device = d.id

Answer 2

您似乎正在追寻类似的东西...

DROP TABLE IF EXISTS my_table;

CREATE TABLE my_table
(event_id SERIAL PRIMARY KEY
,device_id INT NOT NULL
,event VARCHAR(12) NOT NULL
,date DATE NOT NULL
);

INSERT INTO my_table VALUES
(123, 12,'Login','2019-01-12'),
( 32, 12,'Unreachable','2019-02-18'),
(223,145,'Unreachable','2019-02-19'),
(334,  1,'DOWN','2019-03-01');

 SELECT DISTINCT z.event_id
               , x.device_id
               , z.event
               , y.date 
            FROM my_table x 
            JOIN my_table y 
            LEFT 
            JOIN my_table z 
              ON z.device_id = x.device_id 
             AND z.date = y.date 
           ORDER 
              BY date
               , device_id;
+----------+-----------+-------------+------------+
| event_id | device_id | event       | date       |
+----------+-----------+-------------+------------+
|     NULL |         1 | NULL        | 2019-01-12 |
|      123 |        12 | Login       | 2019-01-12 |
|     NULL |       145 | NULL        | 2019-01-12 |
|     NULL |         1 | NULL        | 2019-02-18 |
|       32 |        12 | Unreachable | 2019-02-18 |
|     NULL |       145 | NULL        | 2019-02-18 |
|     NULL |         1 | NULL        | 2019-02-19 |
|     NULL |        12 | NULL        | 2019-02-19 |
|      223 |       145 | Unreachable | 2019-02-19 |
|      334 |         1 | DOWN        | 2019-03-01 |
|     NULL |        12 | NULL        | 2019-03-01 |
|     NULL |       145 | NULL        | 2019-03-01 |
+----------+-----------+-------------+------------+

Answer 3

使用CTE table，我们可以创建日期，然后将它们与事件表关联起来。

with t0 (i)    as (select 0  union all  select 0  union all  select 0),
     t1 (i)    as (select a.i from t0 a ,t0 b ),
     t2 (i)    as (select a.i from t1 a ,t1 b ),
     t3 (srno) as (select row_number()over(order by a.i) from t2 a ,t2 b ),
     tbldt(dt) as (select dateadd(day,t3.srno-1,'01/01/2019') from t3)   
select eventid, deviceid, event, tbldt.dt 
from tbldt
left join event e on e.date = tbldt.dt
left join def d  on e.deviceid = d.id 
where tbldt.dt <= (select max(date) from event)

SQL：列出没有任何事务的行

3 个答案: