GROUP BY consecutive dates delimited by gaps

Question

Assume you have (in Postgres 9.1 ) a table like this:

date | value 

which have some gaps in it (I mean: not every possible date between min

Answer 1

Here is a way of solving it.

First, to get the beginning of consecutive series, this query would give you the first date:

SELECT first.date
FROM raw_data first
     LEFT OUTER JOIN raw_data prior_first ON first.date = prior_first + 1
WHERE prior_first IS NULL

likewise for the end of consecutive series,

SELECT last.date
FROM raw_data last
     LEFT OUTER JOIN raw_data after_last ON last.date = after_last - 1
WHERE after_last IS NULL

You might consider making these views, to simplify queries using them.

We only need the first to form group ranges

CREATE VIEW beginings AS
SELECT first.date
FROM raw_data first
     LEFT OUTER JOIN raw_data prior_first ON first.date = prior_first + 1
WHERE prior_first IS NULL

CREATE VIEW endings AS
SELECT last.date
FROM raw_data last
     LEFT OUTER JOIN raw_data after_last ON last.date = after_last - 1
WHERE after_last IS NULL

SELECT MIN(raw.date), MAX(raw.date), SUM(raw.value)
FROM raw_data raw
  INNER JOIN (SELECT lo.date AS lo_date, MIN(hi.date) as hi_date
              FROM beginnings lo, endings hi
              WHERE lo.date < hi.date
              GROUP BY lo.date) range
     ON raw.date >= range.lo_date AND raw.date <= range.hi_date
GROUP BY range.lo_date