5

There are 3 tables from which I need to pull data, and print it in a specific way.

http://sqlfiddle.com/#!15/59481/8

Refer to this question to see how the query was produced. The version in the fiddle working partially, but I need to modify the query to produce a different output.

I have the table project_report which is storing data for keyword:

1. If a keyword has data for a particular day it will store it in `project_report`
2. A keyword might have no data to store for a particular day.
3. A keyword might store multiple rows of data for a particular day (the data is uniq)
4. A keyword might not have data for a particular day, but it might have for a different day.

I believe to solve my problem a DB function will be the best approach, because I would like to be able to specify a date and get all the data from project_reports for each keyword.

I don't know how to write this query, to produces the below results, also I don't know what will be better, to have a view or a db function, I think that with a view in order to select a specific date, I will need for each keyword and each date to produce those outputs. The other problem is that I would also like to use these results in a chart, which will take the data for a range, for that I think a view is better, then calling a function n times.


So if keyword_id: 1 has in the project_report:

ID: 1, keyword_id: 1, project_id: 1, position: 1, created_at: '2014-07-09'
ID: 2, keyword_id: 1, project_id: 1, position: 2, created_at: '2014-07-09'

For the selected date 2014-07-09 I should get (please see that position will be the smallest position instead of [1, 2]):

keyword_id: 1, project_id: 1, position: 1, yesterday_pos: 0, last_week_pos: 0, last_month_pos:0 created_at: '2014-07-09'

If keyword_id: 2 has in the project_report:

ID: 3, keyword_id: 2, project_id: 1, position: 3, created_at: '2014-07-09'
ID: 4, keyword_id: 2, project_id: 1, position: 4, created_at: '2014-07-08'
ID: 5, keyword_id: 2, project_id: 1, position: 5, created_at: '2014-07-08'
ID: 6, keyword_id: 2, project_id: 1, position: 4, created_at: '2014-07-03'

For the selected date 2014-07-09 I should get:

keyword_id: 2, project_id: 1, position: 3, yesterday_pos: 4, last_week_pos: 4, last_month_pos: 0 created_at: '2014-07-09'

If keyword_id: 3 has in the project_report:

ID: 7, keyword_id: 3, project_id: 1, position: 10, created_at: '2014-07-08'
ID: 8, keyword_id: 3, project_id: 1, position: 11, created_at: '2014-07-03'

For the selected date 2014-07-09 I should get:

keyword_id: 3, project_id: 1, position: 0, yesterday_pos: 10, last_week_pos: 11, last_month_pos: 0 created_at: '2014-07-09'

If keyword_id: 4 has in the project_report:

`nothing`

For the selected date 2014-07-09 I should get:

keyword_id: 4, project_id: 1, position: 0, yesterday_pos: 0, last_week_pos: 0, last_month_pos: 0 created_at: '2014-07-09'

1 Answer 1

6

I suggest a radically new approach with crosstab() from the additional module tablefunc. You need to install it once per database. Detailed instructions:

SELECT k.name, keyword_id, project_id, now()::date AS the_date
     , t AS today, y As yesterday, w AS week, m AS month
FROM   crosstab(
   $$
   SELECT rn
        , pk.keyword_id
        , pk.project_id
        , d.created_at
        , COALESCE(pr.pos, 0) AS pos
   FROM  (SELECT *, row_number() OVER () AS rn FROM project_keyword) pk
   CROSS  JOIN (
      VALUES
         (now()::date)
       , (now()::date - 1)
       , (now()::date - 7)
       ,((now() - interval '1 month')::date)
     ) d(created_at)
   LEFT JOIN (
      SELECT keyword_id
           , project_id
           , created_at::date AS created_at
           , min(position) AS pos
      FROM   project_report
      GROUP  BY keyword_id, project_id, created_at::date
      ) pr USING (keyword_id, project_id, created_at)
   ORDER  BY pk.rn, d.created_at
   $$

  ,$$
   VALUES
         (now()::date)
       , (now()::date - 1)
       , (now()::date - 7)
       , ((now() - interval '1 month')::date)
   $$
   ) ct (rn int, keyword_id int, project_id int
       , t int, y int, w int, m int)
JOIN   keyword k USING (keyword_id);

Returns:

name    keyword_id   project_id   the_date     today   yesterday   week   month
Cheese  1            1            2014-07-11   1       1           1      0
Monitor 2            1            2014-07-11   2       2           2      0
Apple   3            1            2014-07-11   9       9           9      0
Apple1  4            1            2014-07-11   0       0           0      0
Iphone  5            1            2014-07-11   0       33          34     0

Explain

  • We need a single unique column per group for crosstab(). Since you don't have that per (keyword_id int, project_id) I am adding a surrogate key with row_number() to project_keyword.

  • JOIN that to a set of dates you are interested in (today, yesterday etc.). One row per date.

  • LEFT JOIN to project_reports to attach the position for each date if available. Default to 0 with COALESCE.

  • Cross-tabulate to get a single row with one position column per given date.

  • Finally JOIN to keyword just to add the name. Add the current date and remove rn from the result while being at it.

Aside: "name" is not a good name. I'd use something descriptive instead.

5
  • (keyword_id, project_id, created_at) here I get common column name "created_at" appears more than once in left table but If I remove it is working fine, It could be because I have on the project_keyword a column created_at?
    – alexsmn
    Commented Jul 11, 2014 at 14:57
  • 1
    @uhn-nohn: Yes, in this case the column name is ambiguous and you need to switch to explicit ON syntax in the join condition instead of the shorter USING. Commented Jul 11, 2014 at 15:24
  • What is the best way of creating a variable instead of now()::date, so I can pass a date into that variable, and get data based on its value?
    – alexsmn
    Commented Jul 14, 2014 at 9:55
  • @uhn-nohn: Create an SQL function with a date parameter and RETURNS TABLE and replace now()::date with $1. Examples: stackoverflow.com/questions/11862936/…, stackoverflow.com/questions/11401749/…. If you are still stuck, write a new question here. Comments are not the place for new questions. Commented Jul 14, 2014 at 14:14
  • I already have a question on here, I tried to the function, but I got stuck not knowing what type to return, dba.stackexchange.com/questions/71429/…
    – alexsmn
    Commented Jul 14, 2014 at 14:17

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