Postgres COUNT 个带有 INNER JOIN 的列值

Question

我正在 Postgres 9.3 中创建报告。这是我的SQL Fiddle。
基本上我有两张表，responses和questions，结构是：

responses
->id
->question_id
->response

questions
->id
->question
->costperlead

response 列只能有 3 个值，Yes/No/Possbily， 我的报告应该有以下列：

  question_id
, # of Yes Responses
, # of No Responses
, # of Possbily Responses
, Revenue

然后：

# of Yes Responses - count of all Yes values in the response column
# of No Responses - count of all No values in the response column
# of Possbily Responses - count of all 'Possbily' values in the response column

收入是costperlead *（是响应的数量 + 可能响应的数量）。

我不知道如何构建查询，我是新来的，而且我来自 MySQL，所以对于 postgres，有些事情是不同的。在我的 SQL Fiddle 示例中，大多数响应是 Yes 和 Null，最终没关系，会有可能和 No。

到目前为止，我只有：

SELECT a.question_id
FROM responses a
INNER JOIN questions b ON a.question_id = b.id
WHERE a.created_at = '2015-07-17'
GROUP BY a.question_id;

Answer 1

由于唯一的谓词过滤表 responses 中的行，因此首先聚合响应，然后加入问题是最有效的：

SELECT *, q.costperlead * (r.ct_yes + r.ct_maybe) AS revenue
FROM  (
   SELECT question_id
        , count(*) FILTER (WHERE response = 'Yes')      AS ct_yes
        , count(*) FILTER (WHERE response = 'No')       AS ct_no
        , count(*) FILTER (WHERE response = 'Possibly') AS ct_maybe
   FROM   responses
   WHERE  created_at = '2015-07-17'
   GROUP  BY 1
   ) r
JOIN   questions q ON q.id = r.question_id;

db小提琴here

这使用聚合 FILTER 子句（在 Postgres 9.4 或更高版本中）。见：

• Aggregate columns with additional (distinct) filters

另外：考虑将response 实现为boolean 类型和true/false/null。

对于 Postgres 9.3：

SELECT *, q.costperlead * (r.ct_yes + r.ct_maybe) AS revenue
FROM  (
   SELECT question_id
        , count(response = 'Yes' OR NULL)      AS ct_yes
        , count(response = 'No' OR NULL)       AS ct_no
        , count(response = 'Possibly' OR NULL) AS ct_maybe
   FROM   responses
   WHERE  created_at = '2015-07-17'
   GROUP  BY 1
   ) r
JOIN   questions q ON q.id = r.question_id;

_旧sqlfiddle

技术综合对比：

• For absolute performance, is SUM faster or COUNT?

Answer 2

你应该试试：

SELECT a.question_id, 
       SUM(CASE WHEN a.response = 'Yes' THEN 1 ELSE 0 END) AS NumsOfYes, 
       SUM(CASE WHEN a.response = 'No' THEN 1 ELSE 0 END) AS NumsOfNo,
       SUM(CASE WHEN a.response = 'Possibly' THEN 1 ELSE 0 END) AS NumOfPossibly,
       costperlead  * SUM(CASE WHEN a.response = 'Yes' THEN 1 ELSE 0 END) + SUM(CASE WHEN a.response = 'Possibly' THEN 1 ELSE 0 END) AS revenue
FROM responses a 
     INNER JOIN questions b ON a.question_id = b.id 
GROUP BY a.question_id, b.costperlead