通过增加订单SQL

Question

表选择：

id | year |  score 
-----+------+----------- 
    12 | 2011 |  0.929 
    12 | 2014 |  0.933 
    12 | 2010 |  0.937 
    12 | 2013 |  0.938 
    12 | 2009 |  0.97 
    13 | 2010 |  0.851 
    13 | 2014 |  0.881 
    13 | 2011 |  0.885 
    13 | 2013 |  0.895 
    13 | 2009 |  0.955 
    16 | 2009 |  0.867 
    16 | 2011 |  0.881 
    16 | 2012 |  0.886 
    16 | 2013 |  0.897 
    16 | 2014 |  0.953 

所需的输出：

id | year |  score 
-----+------+----------- 
    16 | 2009 |  0.867 
    16 | 2011 |  0.881 
    16 | 2012 |  0.886 
    16 | 2013 |  0.897 
    16 | 2014 |  0.953 

我有在尝试了在对于今年增产得分困难。 任何帮助将不胜感激。

Answer 1

所以你想选择id = 16，因为它是唯一一个稳步增加值。

许多版本的SQL支持lag()，它可以帮助解决这个问题。您可以确定，对于一个给定的ID，如果所有的值都增加或做递减：

select id, 
     (case when min(score - prev_score) < 0 then 'nonincreasing' else 'increasoing' end) as grp 
from (select t.*, lag(score) over (partition by id order by year) as prev_score 
     from table t 
    ) t 
group by id; 

然后你可以选择全部使用“增加” ID的加入：

select t.* 
from table t join 
    (select id 
     from (select t.*, lag(score) over (partition by id order by year) as prev_score 
      from table t 
      ) t 
     group by id 
     having min(score - prev_score) > 0 
    ) inc 
    on t.id = inc.id;