根据其他表中的匹配选择分区

Question

有下表（conversations）：

 id | record_id  |  is_response  |         text         |
 ---+------------+---------------+----------------------+
 1  |     1      |      false    | in text 1            |
 2  |     1      |      true     | response text 3      |
 3  |     1      |      false    | in text 2            |
 4  |     1      |      true     | response text 2      |
 5  |     1      |      true     | response text 3      |
 6  |     2      |      false    | in text 1            |
 7  |     2      |      true     | response text 1      |
 8  |     2      |      false    | in text 2            |
 9  |     2      |      true     | response text 3      |
 10 |     2      |      true     | response text 4      |

还有另一个帮助表 (responses)：

 id |         text         |
 ---+----------------------+
 1  | response text 1      |
 2  | response text 2      |
 3  | response text 4      |

我正在寻找一个 SQL 查询来输出以下内容：

  record_id |       context
  ----------+-----------------------+---------------------
       1    | in text 1 response text 3 in text 2 response text 2
  ----------+-----------------------+---------------------
       2    | in text 1 response text 1
  ----------+-----------------------+---------------------
       2    | in text 2 response text 3 response text 4

所以每次is_response 是true 并且text 是在响应表中，聚合到目前为止的对话上下文，忽略对话部分不会以池中的响应结束。

在上面的例子中，响应文本 3 在record_id 1 中。

我尝试了以下复杂的 SQL，但有时会错误地聚合文本：

with context as(
    with answers as (

       SELECT record_id, is_response, id as ans_id
        , max(id)
          OVER (PARTITION BY record_id ORDER BY id
          ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING) AS previous_ans_id
       FROM (select * from conversations where text in (select text from responses)) ans
       ),
     lines as (
      select answers.record_id, con.id, COALESCE(previous_ans_id || ',' || ans_id, '0') as block, con.text as text from answers, conversations con where con.engagement_id = answers.record_id and ((previous_ans_id is null and con.id <= ans_id) OR (con.id > previous_ans_id and con.id <= ans_id)) order by engagement_id, id asc
      )

      select record_id, block,replace(trim(both ' ' from string_agg(text, E' ')) ,'  ',' ') ctx from lines group by record_id, block order by record_id,block
      )

select * from context

我相信有更好的方法。

Answer 1

这是我的看法：

SELECT
    record_id,
    string_agg(text, ' ' ORDER BY id) AS context
FROM (
    SELECT
        *,
        coalesce(sum(incl::integer) OVER (ORDER BY id ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING),0) AS grp
    FROM (
        SELECT *, is_response AND text IN (SELECT text FROM responses) as incl
        FROM conversations
         ) c
     ) c1
GROUP BY record_id, grp
HAVING bool_or(incl)
ORDER BY max(id);

这将扫描表conversations 一次，但我不确定它是否会比您的解决方案执行得更好。基本思想是使用窗口函数来计算同一记录中的前行如何结束对话。然后我们可以按该号码和record_id 分组并丢弃不完整的对话。

Answer 2

有一个简单快速的解决方案：

SELECT record_id, string_agg(text, ' ') As context
FROM  (
   SELECT c.*, count(r.text) OVER (PARTITION BY c.record_id ORDER BY c.id DESC) AS grp
   FROM   conversations  c
   LEFT   JOIN responses r ON r.text = c.text AND c.is_response
   ORDER  BY record_id, id
   ) sub
WHERE  grp > 0  -- ignore conversation part that does not end with a response
GROUP  BY record_id, grp
ORDER  BY record_id, grp;

count() 只计算非空值。如果LEFT JOIN 到responses 为空，则r.text 为NULL：

• Select rows which are not present in other table

grp（“组”的缩写）中的值仅在触发新的输出行时才会增加。属于同一输出行的所有行都以相同的grp 编号结束。然后很容易在外部 SELECT 中聚合。

特殊的技巧是以相反的顺序计算会话结束。最后一个 end 之后的所有内容（从末尾开始时首先出现）获取 grp = 0 并在外部 SELECT 中删除。

更多解释的类似案例：

• Row number with reset in PostgreSQL
• Select longest continuous sequence