计算SQL中每个唯一列组合的行数

我想从基于两列的表中返回一组唯一记录，以及最近的发布时间和这两列组合在其输出记录出现之前的总次数

因此，我想得到的是以下几点：

select col1, col2, max_posted, count from T
join (
 select col1, col2, max(posted) as posted  from T where groupid = "XXX" 
group by col1, col2) h
on ( T.col1 = h.col1 and
  T.col2 = h.col2 and
  T.max_posted = h.tposted)
where T.groupid = 'XXX'

Count需要是col1和col2的每个组合在输出中每条记录的max_过帐之前发生的次数。我希望我的解释是正确的：

编辑：在尝试以下建议时：

 select dx.*,
   count(*) over (partition by dx.cicd9, dx.cdesc order by dx.tposted) as   cnt
from dx
join (
select cicd9, cdesc, max(tposted) as tposted  from dx where groupid ="XXX" 
group by cicd9, cdesc) h
on ( dx.cicd9 = h.cicd9 and
  dx.cdesc = h.cdesc and
  dx.tposted = h.tposted)
where groupid =  'XXX';

计数始终返回“1”。此外，您如何仅统计tposted之前发生的记录

这也失败了，但我希望你能达到我的目的：

  WITH H AS (
    SELECT cicd9, cdesc, max(tposted) as tposted  from dx where groupid =  'XXX' 
    group by cicd9, cdesc), 
    J AS (
    SELECT  count(*) as cnt
    FROM dx, h
    WHERE dx.cicd9 = h.cicd9
      and dx.cdesc = h.cdesc
      and dx.tposted <= h.tposted
      and dx.groupid = 'XXX'
 )
SELECT H.*,J.cnt
FROM H,J 

---

有人需要帮忙吗？

您只需要累计计数吗

select t.*,
       count(*) over (partition by col1, col2 order by posted) as cnt
from table t
where groupid = 'xxx';

这个怎么样：

SELECT DISTINCT ON (cicd9, cdesc) cicd9, cdesc,
  max(posted) OVER w AS last_post,
  count(*) OVER w AS num_posts
FROM dx
WHERE groupid = 'XXX'
WINDOW w AS (
  PARTITION BY cicd9, cdesc
  RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING
);

---

由于缺少PG版本、表定义、数据和所需的输出，这只是一时兴起，但原则应该是可行的：在groupid='XXX'的两列上做一个分区，然后找到发布列的最大值和窗口框架中的行总数，从而确定范围。。。窗口定义中的子句。

这是我能想到的最好的方法-欢迎提供更好的建议

这将产生我所需要的结果，但要理解，从联接开始，计数将始终至少为1：

  SELECT dx.cicd9, dx.cdesc, max(dx.tposted), count(*)
from dx 
join (
SELECT cicd9, cdesc, max(tposted) as tposted  from dx where groupid   =  'XXX' 
    group by cicd9, cdesc) h
on 
  (dx.cicd9 = h.cicd9 and dx.cdesc = h.cdesc and dx.tposted <= h.tposted 
  and dx.groupid = 'XXX')
group by dx.cicd9, dx.cdesc
order by dx.cdesc;

或

这令人困惑：

Count需要是col1和col1的每次组合的次数 col2发生在输出中每条记录的max_过帐之前

因为，根据定义，每个记录都在最新帖子之前或同时出现，这本质上意味着每个组合的总计数忽略了句子中一个错误的假设

因此，这可以归结为一个简单的组：

---

这与当前接受的答案完全相同。只是更快更简单。

样本数据和期望的结果将有助于澄清问题。是的。只需要对过帐前发生的记录进行累计计数。请参阅我的上述编辑。谢谢。我想补充一点，我正在寻找与输出行匹配的行总数，包括重复的行。您的问题仍然模糊。您应该添加示例数据和所需结果。这里有很多要学习的内容，如果这包括所有行，那么它可能会起作用。。当我没有脑死亡的时候，我需要研究一下。谢谢PG版本9.3如何更改查询以使每个cicd9、cdesc组只生成一行？它当前为dx中找到的每一行复制相同的输出行。谢谢你给我的所有新概念。@AlanWayne:添加了一个独特的ON子句，请参阅更新的答案。是的！成功了。非常感谢您，先生。它是否还您需要的东西？如果没有，它是如何失败的？如果无法用文字说明任务，则应在问题中添加有意义的示例值和所需结果。@ErwinBrandstetter Yes。如果count=0就更好了，因为我希望对以前日期的行进行计数。否则，它工作得很好。

 WITH H AS (
    SELECT cicd9, cdesc, max(tposted) as tposted  from dx where groupid =  'XXX' 
    group by cicd9, cdesc)  
SELECT dx.cicd9, dx.cdesc, max(dx.tposted), count(*)
from dx, H
where dx.cicd9 = h.cicd9 and dx.cdesc = h.cdesc and dx.tposted <= h.tposted 
  and dx.groupid = 'XXX'
group by dx.cicd9, dx.cdesc
order by cdesc;

SELECT cicd9, cdesc
     , max(posted) AS last_posted
     , count(*)    AS ct
FROM   dx
WHERE  groupid = 'XXX'
GROUP  BY 1, 2
ORDER  BY 1, 2;