PostgreSQL remove duplicates by GROUP BY - sql

Question

I would like to print the last message of a person, but only his latest message should be printed per person. I use PostgreSQL 10.
+-----------+----------+--------------+
| name | body | created_at |
+-----------+----------+--------------+
| Maria | Test3 | 2017-07-07 |
| Paul | Test5 | 2017-06-01 |
+-----------+----------+--------------+
I have tried this with the following SQL query, this gives me exactly that back but unfortunately the people are doubled in it.
SELECT * FROM messages
WHERE receive = 't'
GROUP BY name
ORDER BY MAX(created_at) DESC
+-----------+----------+--------------+
| name | body | created_at |
+-----------+----------+--------------+
| Maria | Test1 | 2016-06-01 |
| Maria | Test2 | 2016-11-01 |
| Maria | Test3 | 2017-07-07 |
| Paul | Test4 | 2017-01-01 |
| Paul | Test5 | 2017-06-01 |
+-----------+----------+--------------+
I tried to remove the duplicates with a DISTINCT, but unfortunately I get this error message:
SELECT DISTINCT ON (name) * FROM messages
WHERE receive = 't'
GROUP BY name
ORDER BY MAX(created_at) DESC
ERROR: SELECT DISTINCT ON expressions must match initial ORDER BY expressions LINE 1: SELECT DISTINCT ON (name) * FROM messages ^ : SELECT DISTINCT ON (name) * FROM messages WHERE receive = 't' GROUP BY name ORDER BY MAX(created_at) DESC
Do you have any ideas how I can solve this ?

Answer 1

You would use DISTINCT ON as follows:
SELECT DISTINCT ON (name) *
FROM messages
WHERE receive = 't'
ORDER BY name, created_at DESC
That is:
no GROUP BY clause is needed
the column(s) listed in DISTINCT ON(...) must appear first in the ORDER BY clause
... followed by the column that should be use to break the group (here, that is created_at)
Note that the results of a distinct on query are always sorted by columns in the clause (because this sort is what is used to identifiy which rows should be kept).
If you want more control over the sort order, then you can use window functions instead:
SELECT *
FROM (
SELECT m.*, ROW_NUMBER() OVER(PARTITION BY name ORDER BY created_at DESC) rn
FROM messages m
WHERE receive = 't'
) t
WHERE rn = 1
ORDER BY created_at DESC

Answer 2

Use DISTINCT ON, but with the right ORDER BY:
SELECT DISTINCT ON (name) m.*
FROM messages m
WHERE receive = 't'
ORDER BY name, created_at DESC;
In general, you don't use DISTINCT ON with GROUP BY. It is used with ORDER BY. The way it works is to that it chooses the first row for each name based on the ORDER BY clause.
You should not be thinking of what you are doing as aggregation. You want to filter based on the created_at. In many databases, you would express this using a correlated subquery:
select m.*
from messages m
where m.created_at = (select max(m2.created_at)
from messages m2
where m2.name = m.name and m2.receive = 't'
) and
m.receive = 't'; -- this condition is probably not needed

Answer 3

SELECT *
FROM messages
WHERE receive = 't' and not exists (
select 1
from messages m
where m.receive = message.receive and messages.name = m.name and m.created_at > messages.created_at
)
ORDER BY created_at DESC
The query above finds the messages which fulfill the following criteria:
receive is 't'
there not exists another message which
has the same value for receive
has the same name
and is newer
Assuming that the same name does not send two messages at exactly the same time this should be enough. Another point to make is that the name might look similar, but be different, if some white characters are present inside the value, so, if you see two records in the result with the same name, but with different created_at in the query above, then it is highly probable that white characters are playing tricks on you.