PostgreSQL – How to Perform Left Join Without Duplicate Rows

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I have two tables called record and record_history. For each record, there could be more than one history. They can be joined by id and record_id. I want to get all the record's entries with recent record_history data. I have created the query like,

SELECT rec.id, rec.name, rech1.data AS last_history_data
FROM record rec
LEFT OUTER JOIN record_history rech1 ON (rec.id = rech1.record_id)
LEFT OUTER JOIN record_history rech2 ON (rec.id = rech2.record_id AND rech2.ts > rech1.ts)
WHERE rech2.id IS NULL
ORDER BY rec.id DESC

Here, I am getting the latest one by ts. This works as long as there are no duplicate ts entries. If the recent timestamp is repeated in record_history, this query returns more than one row for a record. How can we apply the limit here on the left join to restrict duplicate rows?

Best Answer

Unless you are in a very old version of Postgres, you don't need the double join. You can get the same result by using a LATERAL join.

The duplicate results can be avoided in your method by adding a second condition besides the rec.id = rech2.record_id. With the LATERAL join method, the use of LIMIT is avoiding it anyway. There can be only 1 row returned from the lateral subquery. We can add a second condition so the choice is deterministic (from the two or more rows with same timestamp):

SELECT rec.id, rec.name, rech.data AS last_history_data
FROM record AS rec
     LEFT OUTER JOIN LATERAL
     ( SELECT rech.data
       FROM record_history AS rech
       WHERE rec.id = rech.record_id
       ORDER BY rech.ts DESC
                -- ,rech.id DESC               -- optional
       LIMIT 1 
     ) AS rech
     ON TRUE
ORDER BY rec.id DESC ;

Regarding how to do this with the original method (2 joins and IS NULL check), you could change the ON condition - assuming there is an id column in history table so that (id) or at least (ts, id) is unique:

LEFT OUTER JOIN record_history rech2 
ON rec.id = rech2.record_id 
   AND (rech2.ts > rech1.ts OR rech2.ts = rech1.ts AND rech2.id > rech1.id)

By the way, you could replace that second LEFT join and IS NULL check with a NOT EXISTS subquery with same results and possibly similar efficiency (or even with a NOT IN subquery although that needs extra care for nullable columns, not recommended).

1st case

You seem to forget the valid_during range. As your third case suggests, there can be multiple entries per (rec_id, val), so you must select the right one:

UPDATE master m
SET    valid_on = f_array_sort(m.valid_on || u.valid_on) -- sorted array, see below
FROM   updates u
WHERE  m.rec_id = u.rec_id 
AND    m.valid_during @> u.valid_on  -- additional check
AND    m.val = u.val 
AND    NOT m.valid_on @> ARRAY[u.valid_on];

I assume the whole possible date range is always covered for each existing rec_id and valid_during shall not overlap per rec_id, or you'd have to do more.

After installing the additional module btree_gist, add an exclusion constraint to rule out overlapping date ranges if you don't have one, yet:

ALTER TABLE master ADD CONSTRAINT EXCLUDE
USING gist (rec_id WITH =, valid_during WITH &&)  -- disallow overlap

The GiST index this is implemented with is also a perfect match for the query. Details:

2nd / 3rd case

Assuming that every date range starts with the smallest date in the (now sorted!) array: lower(m.valid_during) = m.valid_on[1]. I would enforce that with a CHECK constraint.

Here we need to create one or two new rows In the 2nd case it is enough to shrink the range of the old row and insert one new row In the 3rd case we update the old row with the left half of array and range, insert the new row and finally insert the with the right half of array and range.

Helper functions

To keep it simple I introduce a new constraint: every array is sorted. Use this helper function

-- sort array
CREATE OR REPLACE FUNCTION f_array_sort(anyarray) 
  RETURNS anyarray LANGUAGE sql IMMUTABLE AS
$$SELECT ARRAY (SELECT unnest($1) ORDER BY 1)$$;

I don't need your helper function arraymin() any more, but it could be simplified to:

CREATE OR REPLACE FUNCTION f_array_min(anyarray) 
  RETURNS anyelement LANGUAGE sql IMMUTABLE AS
$$SELECT min(a) FROM unnest($1) a$$;

Two more to get the left and right half of an array split at a given element:

-- split left array at given element
CREATE OR REPLACE FUNCTION f_array_left(anyarray, anyelement) 
  RETURNS anyarray LANGUAGE sql IMMUTABLE AS
$$SELECT ARRAY (SELECT * FROM unnest($1) a WHERE a < $2 ORDER BY 1)$$;

-- split right array at given element
CREATE OR REPLACE FUNCTION f_array_right(anyarray, anyelement) 
  RETURNS anyarray LANGUAGE sql IMMUTABLE AS
$$SELECT ARRAY (SELECT * FROM unnest($1) a WHERE a >= $2 ORDER BY 1)$$;

Query

This does all the rest:

WITH u AS (  -- identify candidates
   SELECT m.id, rec_id, m.val, m.valid_on, m.valid_during
        , u.val AS u_val, u.valid_on AS u_valid_on
   FROM   master  m
   JOIN   updates u USING (rec_id)
   WHERE  m.val <> u.val
   AND    m.valid_during @> u.valid_on
   FOR    UPDATE  -- lock for update
   )
, upd1 AS (  -- case 2: no overlap, no split
   UPDATE master m  -- shrink old row
   SET    valid_during = daterange(lower(u.valid_during), u.u_valid_on)
   FROM   u
   WHERE  u.id = m.id
   AND    u.u_valid_on > m.valid_on[array_upper(m.valid_on, 1)]
   RETURNING m.id
   )
, ins1 AS (  -- insert new row
   INSERT INTO master (rec_id, val, valid_on, valid_during)
   SELECT u.rec_id, u.u_val, ARRAY[u.u_valid_on]
        , daterange(u.u_valid_on, upper(u.valid_during))
   FROM   upd1
   JOIN   u USING (id)
   )
, upd2 AS (  -- case 3: overlap, need to split row
   UPDATE master m  -- shrink to first half
   SET    valid_during = daterange(lower(u.valid_during), u.u_valid_on)
        , valid_on = f_array_left(u.valid_on, u.u_valid_on)
   FROM   u
   LEFT   JOIN upd1 USING (id)
   WHERE  upd1.id IS NULL  -- all others
   AND    u.id = m.id
   RETURNING m.id, f_array_right(u.valid_on, u.u_valid_on) AS arr_right
   )
INSERT INTO master (rec_id, val, valid_on, valid_during)
          -- new row
SELECT u.rec_id, u.u_val, ARRAY[u.u_valid_on]
     , daterange(u.u_valid_on, upd2.arr_right[1])
FROM   upd2
JOIN   u USING (id)
UNION ALL  -- second half of old row
SELECT u.rec_id, u.val, upd2.arr_right
     , daterange(upd2.arr_right[1], upper(u.valid_during))
FROM   upd2
JOIN   u USING (id);

SQL Fiddle.

Notes

You need to understand the concept of data-modifying CTEs (writeable CTEs), before you touch this. Judging from the code you provided, you know your way around Postgres.
- Are SELECT type queries the only type that can be nested?
FOR UPDATE is to avoid race conditions with concurrent write access. If you are the only user writing to the tables, you don't need it.
I took a piece of paper and drew a timeline so not to get lost in all of this.
Each row is only updated / inserted once, and operations are simple and roughly optimized. No expensive window functions. This should perform well. Much faster than your previous approach in any case.
It would be a bit less confusing if you'd use distinct column names for u.valid_on and m.valid_on, which are related but different things.
I compute the right half of the split array in the RETURNING clause of CTE upd2: f_array_right(u.valid_on, u.u_valid_on) AS arr_right, because I need it several times in the next step. This is a (legal) trick to save one more CTE.
As for solutions that don't involve unnesting the master table: You have to unnest the array valid_on either way, to split it, at least as long as it's not sorted. Also, your helper function arraymin() is already unnesting it anyway.

PostgreSQL – LEFT JOIN with Recurring Dates and Users

You have to do with users the same thing you're doing with days:

WITH
  days AS (
    SELECT generate_series(current_date-7, current_date, '1d')::date AS day
    ),
  eves AS (
    SELECT user_id, created_at::date AS full_day, COUNT(*) as evs
    FROM logged_events
    WHERE logged_events.created_at >= current_date-6
    GROUP BY user_id, full_day
    ),
  users AS (
    SELECT DISTINCT user_id FROM eves
    )
SELECT eves.full_day, days.day, users.user_id, eves.evs
FROM days
CROSS JOIN users
LEFT JOIN eves ON (eves.full_day = days.day AND eves.user_id = users.user_id)
GROUP BY users.user_id, days.day;