How to show distinct sender id from conversation - php

Question

Updated Question:
My private_messages database structure is like this:
+----+-------+-------+-------+--------+------------+
| id | text | byuid | touid | unread | timesent |
+----+-------+-------+-------+--------+------------+
| 1 | Hi | 1 | 4 | 1 | 1514764805 |
| 2 | hello | 1 | 4 | 1 | 1514764804 |
| 3 | hlw | 1 | 4 | 1 | 1514764803 |
| 4 | good | 2 | 4 | 1 | 1514764802 |
| 5 | fine | 3 | 4 | 0 | 1514764801 |
+----+-------+-------+-------+--------+------------+
My all_users_table is:
+----+-------+------+
| id | name | pass |
+----+-------+------+
| 1 | user1 | 123 |
| 2 | user2 | 112 |
| 3 | user3 | 124 |
| 4 | user4 | 258 |
| 5 | user5 | 315 |
+----+-------+------+
My current SQL code is
$sql = "SELECT
a.name, b.id, b.byuid, b.unread, b.starred FROM all_users_table a
INNER JOIN private_messages b ON a.id = b.byuid
WHERE b.touid='4' AND starred='0'
ORDER BY b.timesent DESC, b.unread
LIMIT $limit_start, $items_per_page
";
Which Print Like below:
1
1
1
2
3
All I want is to (For touid = 4)
print all unique byuid as a output.
Order By Unread DESC
Order By timesent DESC
Limit 0, 2 (for pagination purpose)
Output should be:
1
2
Can anyone help me please ? I tried Group By but it shows empty results.

Answer 1

$sql = "SELECT
a.name, b.id, b.byuid, b.unread, b.starred FROM all_users_table a
INNER JOIN private_messages b ON a.id = b.byuid
WHERE b.touid='".$myid."' AND starred='0' GROUP BY a.name
ORDER BY b.timesent DESC, b.unread
LIMIT $limit_start, $items_per_page";
Try this one with GROUP BY CLAUSE

Answer 2

"SELECT
a.name, b.id, b.byuid, b.unread, b.starred
FROM all_users_table a
LEFT JOIN private_messages b ON a.id = b.byuid
WHERE b.touid = '".$myid."' AND starred = '0'
ORDER BY b.timesent DESC, b.unread
LIMIT $limit_start, $items_per_page"

Answer 3

If you need not duplicated rows you should use DISTINCT clause
$sql = "SELECT DISTINCT
a.name, b.id, b.byuid, b.unread, b.starred
FROM all_users_table a
INNER JOIN private_messages b ON a.id = b.byuid
WHERE b.touid='".$myid."' AND starred='0'
ORDER BY b.timesent DESC, b.unread
LIMIT $limit_start, $items_per_page";
(the use of group by for avoid duplicated row in absence of aggregation function as SUM() or MAX() is deprecated in sql and in the most recent version of mySQL generate error)
But looking to your updated data you are not looking for the distinct result but at the result for the most recent sent values
and for this you don't need distinct but a join between al the columns you need and a subselect that retrive the sot recent value for byuid
select a.name, b.id, b.byuid, b.unread, b.starred
FROM all_users_table a
INNER JOIN private_messages b ON a.id = b.byuid A
INNER JOIN (
select byuid, max(timesent) max_sent
from private_messages
WHERE b.touid= 4
group by byuid
) t on t.byuid = b.byuid and t.max_sent = b.timesent
ORDER BY b.timesent DESC, b.unread
LIMIT $limit_start, $items_per_page