PostgreSQL选择具有相同列值的行

如何解决PostgreSQL选择具有相同列值的行

  | location_id  |  lat  |  long  |  speed  |
    ------------- ------- -------- --------- 
      101241        0.12    1.1       0.0    
    ------------- ------- -------- --------- 
      101242        0.12    1.1       0.0
    ------------- ------- -------- --------- 
      101243        0.12    1.1       0.0
    ------------- ------- -------- --------- 
      101244        1.25    0.74      7.4
    ------------- ------- -------- ---------

我想选择speed = 0和lat && long相同的所有位置

因此，从上面的示例答案应该是：

   | location_id  |
    --------------
        101241     
    --------------
        101242     
    --------------
        101243     
    --------------

注意：： 速度恒定为0 ，但纬度和经度取决于前一行的值

解决方法

您可以使用inner join：

select distinct t1.id
from table_name t1
inner join table_name t2
on t1.location_id <> t2.location_id 
and t1.lat = t2.lat
and t1.long = t2.long
where t1.speed = 0
and t2.speed = 0

或存在：

select t.id
from table_name t
where exists (
    select *
    from table_name it
    where t.location_id <> it.location_id 
    and t.lat = it.lat
    and t.long = it.long
    and it.speed = 0
)
and t.speed = 0

,

我实际上将其理解为“间隔与孤岛”问题，在该问题中，您希望相邻的行具有相同的纬度和经度，并且速度为0。

您可以使用窗口函数来实现这一点：行号之间的差为您提供孤岛：然后您可以计算每个孤岛的长度，并对这些长度进行过滤，过滤长度大于1且速度为{{ 1}}：

0

Demo on DB Fiddle

,

另一个解决方案：

SELECT location_id 
FROM device_location 
WHERE (lat,long) IN (
  SELECT lat,long
  FROM device_location
  WHERE speed = 0.0
  GROUP BY lat,long
  HAVING COUNT(*) > 1
);

在SQLize.online上进行测试

,

如果要相邻的行，可以只使用lead()和lag()。 。 。但是通过使用locationid有一点技巧：

select t.*
from (select t.*,lag(locationid) over (order by locationid) as prev_locationid,lead(locationid) over (order by locationid) as next_locationid,lag(locationid) over (partition by lat,long order by locationid) as prev_locationid_ll,lead(locationid) over (partition by lat,long order by locationid) as next_locationid_ll
      from t
     ) t
where speed = 0 and
      (prev_locationid = prev_locationid_ll or
       next_locationid = next_locationid_ll
      );

仅对位置ID进行比较。根据位置ID计算一个。第二个是基于纬度和经度的上一个或下一个。如果这些相同，则相邻行上的值相同。