你可以用 generate_array() 以及 unnest() :

select num_opens, count(t.num_opens)
from (select min(num_opens) as min_no, max(num_opens) as max_no
      from t
     ) x cross join
     unnest(generate_array(t.min_no, t.max_no)) num_opens left join
     t
     on t.num_opens = num_opens
group by num_opens;

首先需要一个参考表。从你的照片上可以看出 users ，但实际上任何一张（足够大的）table都可以。
首先，您将使用 rank() 或者 row_count() 功能。或者如果你的 users.id 没有空隙，使用起来更容易。 SELECT *, rank() OVER (ORDER BY id) as reference_value FROM users 这将生成一个表 1....n 为了 users .
现在加入到该表中，但从加入的表中计数：

SELECT 
    a.reference_value, count(b.num_opens) as total
FROM   
     (SELECT rank() OVER (ORDER BY id) as reference_value from users) a
LEFT JOIN 
     [whatever table] b ON a.reference_value = b.num_opens
GROUP BY 
   a.reference_value

但这行太多了！你肯定有比这些事件计数更多的用户。所以在里面放一个快速过滤器。

SELECT 
    a.reference_value, count(b.num_opens) as total
FROM   
     (SELECT rank() OVER (ORDER BY id) as reference_value from users) a
LEFT JOIN 
     [whatever table] b ON a.reference_value = b.num_opens
WHERE
     a.reference_value <= (SELECT max(num_opens) FROM [whatever table])
GROUP BY 
   a.reference_value