Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
1.1k views
in Technique[技术] by (71.8m points)

database - Postgres COUNT number of column values with INNER JOIN

I am creating a report in Postgres 9.3. This is my SQL Fiddle.
Basically I have two tables, responses and questions, the structure is:

responses
->id
->question_id
->response

questions
->id
->question
->costperlead

for the column response there can only be 3 values, Yes/No/Possbily, and my report should have the columns:

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

Then:

# 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

Revenue is the costperlead * (Number of Yes Responses + Number of Possibly Responses).

I don't know how to construct the query, I'm new plus I came from MySQL so some things are different for postgres. In my SQL Fiddle sample most responses are Yes and Null, it's ok eventually, there will be Possibly and No.

So far I have only:

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;
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

You should try:

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

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

2.1m questions

2.1m answers

60 comments

56.6k users

...