Question

我在Postgres SQL 8.4中有如下表：

 1   | John Smith | 2011-08-12 12:44:13.125+08
 2   | John Smith | 2011-08-16 08:38:57.968+08
 3   | John Smith | 2011-08-16 08:38:58.062+08
 4   | Kenny Long | 2011-08-16 17:06:35.843+08
 5   | Kenny Long | 2011-08-16 17:06:35.906+08
 6   | Kenny Long | 2011-08-16 17:06:59.281+08
 7   | Kenny Long | 2011-08-16 17:07:00.234+08
 8   | Kenny Long | 2011-08-16 17:07:32.859+08
 9   | Kenny Long | 2011-08-16 17:08:00.437+08
 10  | Kenny Long | 2011-08-16 17:08:22.718+08
 11  | Kenny Long | 2011-08-16 17:08:22.781+08

我想选择基于时间戳的列。对于那些彼此相距2分钟的记录，只需要一行。例如，记录编号 4到9 应仅返回行号 4 ，并忽略其余行。

我怎样才能做到这一点？非常感谢您的帮助。

提前谢谢。

Joe Liew

Answer 1

我用recursive方式试了一下。我不确定这是更好的方法，我很确定我应该研究一些Window操作来减少它。

但它适用于我的测试用例。目标是从每个人的一个最小时间戳开始，然后跟踪要删除的行（在2分钟范围内），以及哪一行是下一个有效行。然后在每次迭代时，我们从这个有效行继续（每个人一个）。

所以这是查询表myschema.mytable，其中包含colums id，name，tm。请注意，level列只是为了跟踪递归和调试，而不是必需的：

WITH RECURSIVE mytmp(id,name,thetime,thelevel) AS (
   -- recursive query: 1st row
   -- starting point, one row of the table for each people
   -- with a subquery to get the min time with id, maybe a better way to do it
   (
    select myschema.mytable.id,myschema.mytable.name,myschema.mytable.tm as thetime,1 as thelevel
    from (
      select name,min(tm) as mintm
      from myschema.mytable
      group by name
     ) q,myschema.mytable
    WHERE myschema.mytable.name=q.name
    AND myschema.mytable.tm=q.mintm
   ORDER BY name ASC) -- end of starting point of recursive query
  UNION ALL 
    -- now the recursive part, starting from the 1st row and then again and again (loop)
    -- get descendants in the 2 minutes interval for every computed row already in mytmp
    --
    -- get from previous iterations targets, one per guy
    -- and track the 1st new valid row (>2min) for that guy
    -- removing bad rows (<2min) is easy, several way to do it
    -- keeping only one valid row (and not all the others is harder, limit and aggregates functions are restricted in recursive terms
    -- we must keep only one, as the future valid rows will depend on the 2 minutes range from this one
    -- maybe some window function could help me, but at least I've a working solution
    select myschema.mytable.id,myschema.mytable.name,myschema.mytable.tm as thetime,q2.thelevel
    FROM myschema.mytable,(
    -- here need to keep 1st true one
    select myschema.mytable.name,MIN(myschema.mytable.tm) as tm,mytmp2.thelevel +1 as thelevel
    FROM myschema.mytable,(
        select id,name,thetime,thelevel
        from mytmp 
    ) mytmp2
    -- hack: mytmp2 is useless, mytmp should have been used
    -- we create this indirection to avoid this message:
    -- "ERROR:  aggregate functions not allowed in a recursive query's recursive term"
    -- on the MIN functions
    -- I do not know why it worked :-)
    WHERE myschema.mytable.name=mytmp2.name
    -- future
    AND myschema.mytable.tm - mytmp2.thetime > INTERVAL '0'
    GROUP BY
        -- hack the group by, to make 2 groups
        -- the first one for rows in the 2 min range and the second one for others
        CASE WHEN ((myschema.mytable.tm - mytmp2.thetime) > INTERVAL '2 minutes') THEN 1 ELSE 2 END,
        myschema.mytable.name,mytmp2.thelevel,mytmp2.thetime
        -- then with the having we keep only the second group, containing the first valid > 2min row
        HAVING ((MIN(myschema.mytable.tm) - mytmp2.thetime) > INTERVAL '2 minutes')=true
    ) q2 -- q2contains 1st true row and all false rows for each people
    -- q2 is used to get the id, that we cannot have in a group by request
    WHERE q2.tm=myschema.mytable.tm
    AND q2.name=myschema.mytable.name
 ) -- end of recursive query
  SELECT *
  FROM mytmp
  ORDER BY name asc, thelevel asc,thetime asc
  -- LIMIT 100 -- to debug, avoid infinite loops

另一个解决方案可能是使用存储过程，在临时表中执行相同的操作（获取有效行，删除2分钟范围内的行，然后获取下一个有效行等），也许更容易维护。

Answer 2

只是一些想法，没有经过测试。窗口函数需要8.4 /晚。

SELECT * FROM 
(SELECT
     name,
     tm, 
     case
         when lagname is NULL OR   -- first row of everything
              (name <> lagname) OR -- we have order by name, this is first row of this name
              (name = lagname AND lagtm + interval '2 minutes' >= tm) 
         then 1
         else 0
     end as flags
  FROM
  ( 
     SELECT name, 
         tm,
         lag(name) over (order by name,tm) as lagname,
         lag(tm) over (order by name,tm) as lagtm
      from "table"."table"
  ) AS lagtable
) AS blar
WHERE "flags" = 1

如何根据最大时间间隔进行选择

2 个答案: