Postgresql – Select row if column data was not selected

postgresqlpostgresql-9.3

How can I selected rows from a table when a column data is different.

Table: id(pk), name, title, book_id, date

I want to get all the rows where book_id is uniq.

Data example:

test1, test2, 1, 2015-01-01
test2, test4, 1, 2015-01-02
test2, test3, 1, 2015-01-02
test2, test2, 4, 2015-01-03
test4, test2, 5, 2015-01-04
test5, test2, 5, 2015-01-05

Expected output:

test1, test2, 1, 2015-01-01
test2, test3, 1, 2015-01-02
test2, test2, 4, 2015-01-03
test4, test2, 5, 2015-01-04

Best Answer

This can easily be done using Postgres' distinct on operator:

select distinct on (book_id) name, title, bookd_id, date
from the_table
order by book_id, date;

This will pick the first row for each book_id based on the date column.

Related Solutions

PostgreSQL 9.2/9.3 – Get Distinct Results Based on Column and Condition

If you want the distinct keywords then you need to count the keyword_id, not the position column.

So your expression:

COUNT(CASE WHEN position BETWEEN 4 AND 10 THEN position ELSE NULL END )

becomes:

COUNT(distinct CASE WHEN position BETWEEN 4 AND 10 THEN keyword_id ELSE NULL END )

So the final statement is:

SELECT
  pr.project_id,
  COUNT(DISTINCT pr.keyword_id) as total_keywords,
  count(distinct CASE WHEN position BETWEEN 4 AND 10 THEN keyword_id ELSE NULL END ) AS pos4
FROM
  project_reports pr
GROUP BY pr.project_id;

SQLFiddle example: http://sqlfiddle.com/#!15/b1015/1

PostgreSQL – Select Data Based on Records Created_at Column

Note the slightly modified schema of my test in the fiddle.
Using actual primary keys and proper column names instead of id.

Also, you seem to be operating with dates exclusively. So I suggest to convert your timestamp columns to date.

Items 1 and 2

SELECT k.keyword_id
     , k.name
     , pr.project_id
     , COALESCE(min(pr.position), 0) AS pos
     , COALESCE(pr.created_at, now()::date) AS created_at
FROM   keyword             k 
LEFT   JOIN project_report pr USING (keyword_id)
GROUP  BY k.keyword_id, pr.project_id, pr.created_at
ORDER  BY keyword_id, created_at
;

In Postgres 9.1 or later the pk column covers the whole table in GROUP BY.
Use COALESCE to replace possible NULL values.

A guess at item 3

WITH cte AS (
   SELECT k.keyword_id
        , k.name
        , pr.project_id
        , COALESCE(min(pr.position), 0) AS pos
        , COALESCE(pr.created_at, now()::date) AS created_at
   FROM   keyword   k 
   LEFT   JOIN project_report pr USING (keyword_id)
   GROUP  BY k.keyword_id, pr.project_id, pr.created_at
   )
, x AS (
   SELECT DISTINCT ON (keyword_id, project_id) *
   FROM   cte
   ORDER  BY keyword_id, project_id, created_at DESC
   )
SELECT x.*
     , COALESCE(y.pos, 0) AS yesterday_pos 
     , COALESCE(w.pos, 0) AS week_pos 
     , COALESCE(m.pos, 0) AS month_pos 
FROM   x
LEFT   JOIN  cte y ON y.keyword_id = x.keyword_id
                  AND y.project_id = x.project_id
                  AND y.created_at = x.created_at - interval '1 day'
LEFT   JOIN  cte w ON w.keyword_id = x.keyword_id
                  AND w.project_id = x.project_id
                  AND w.created_at = x.created_at - interval '1 week'
LEFT   JOIN  cte m ON m.keyword_id = x.keyword_id
                  AND m.project_id = x.project_id
                  AND m.created_at = x.created_at - interval '1 month'
;

Explain

In CTE cte produce daily aggregates per (k.keyword_id, pr.project_id).
In CTE x pick the latest day per (k.keyword_id, pr.project_id).
In the outer query LEFT JOIN the latest day x to cte multiple times to retrieve past values from the same (k.keyword_id, pr.project_id) for 1 day / week / month earlier.

SQL Fiddle.