Is there a [straightforward] way to order results *first*, *then* group by another column, with SQL? - sql

Question

I see that in an SQL query, the GROUP BY has to precede the ORDER BY expression. Does this imply that ordering is done after grouping would have discarded identical rows?
Because I seem to need to order rows by a timestamp first, then discard the rows with identical timestamp. And I don't know how to accomplish this.
I am using MySQL 5.1.41.
Here is the definition of the table expressed with create table:
create table
(
A int,
B timestamp
)
The data could be:
+-----+-----------------------+
| A | B |
+-----+-----------------------+
| 1 | today |
| 1 | yesterday |
| 2 | yesterday |
| 2 | tomorrow |
+-----+-----------------------+
The results of the query on the above table, which I am after, would be:
+-----+-----------------------+
| A | B |
+-----+-----------------------+
| 1 | today |
| 2 | tomorrow |
+-----+-----------------------+
Basically, I want the rows with the latest timestamp in column "B" (hence the mention of ORDER BY), and only one row for each value in column "A" (think DISTINCT or GROUP BY).
The actual problem behind the simplified example above:
In reality, I have two tables - users and payment_receipts:
create table users
(
phone_nr int(10) unsigned not null,
primary key (phone_nr)
)
create table payment_receipts
(
phone_nr int(10) unsigned not null,
payed_ts timestamp default current_timestamp not null,
payed_until_ts timestamp not null,
primary key (phone_nr, payed_ts, payed_until_ts)
)
The tables may include other columns but I omit these as irrelevant. Implementing a payment scheme, I have to send SMS to users across the cellular network, in periodic intervals depending on whether the payment is due or not. The payment is actualized when the SMS is sent as the recipient is taxed for it. I use the payment_receipts table to keep records of all payments done, i.e. for book-keeping. This is intended to model a real shop where both the buyer and the seller get a copy of the receipt of purchase, for reference. This table stores my (seller's) copy [of each receipt]. The customer's receipt is the received SMS itself. Each time an SMS is sent (and thus a payment is accomplished), the table is inserted a receipt record, stating who paid, when and "until when". To explain the latter, imagine a subscription service, but one which spans indefinitely until the user opt-out explicitly, at which point the corresponding user record is removed. A payment is made a month in advance, so as a rule, the difference between the payed_ts and payed_until_ts is 30 days worth of time.
I have a batch job that executes every day and needs to select a list of users that are due monthly payment as part of the automatic subscription renewal described above. To link this to the dummy example earlier, the phone number column phone_nr would be the column "A" and payed_until_ts would be column "B", but in reality there are two tables, which has to do with the following behaviour: when a user record is removed, the receipt must remain, for book-keeping. So not only do I need to group payments by date and discard all but the latest payment receipt date, I also need to watch out not to select receipts for which there no longer is a matching user record.
To solve the problem of selecting required records -- those that are due payment -- I need to find receipts with the latest payed_until_ts timestamp for each phone_nr (there may be several, obviously) and out of those records I further need to select only those phone numbers where payed_until_ts is earlier than the time the batch job executes. I then would send an SMS to each of these numbers, inserting a receipt record for each sent SMS, where payed_ts is now() and payed_until_ts is now() + interval 30 days.
But I can't seem to come up with the query required.

Answer 1

Select a,b from (select a,b from table order by b) as c group by a;

Answer 2

Yes, grouping is done first, and it affects a single select whereas ordering affects all the results from all select statements in a union, such as:
select a, 'max', max(b) from tbl group by a
union all select a, 'min', min(b) from tbl group by a
order by 1, 2
(using field numbers in order by since I couldn't be bothered to name my columns). Each group by affects only its select, the order by affects the combined result set.
It seems that what you're after can be achieved with:
select A, max(B) from tbl group by A
This uses the max aggregation function to basically do your pre-group ordering (it doesn't actually sort it in any decent DBMS, rather it will simply choose the maximum from an suitable index if available).

Answer 3

SELECT DISTINCT a,b
FROM tbl t
WHERE b = (SELECT MAX(b) FROM tbl WHERE tbl.a = t.a);

Answer 4

According to your new rules (tested with PostgreSQL)
Query You'd Want:
SELECT pr.phone_nr, pr.payed_ts, pr.payed_until_ts
FROM payment_receipts pr
JOIN users
ON (pr.phone_nr = users.phone_nr)
JOIN (select phone_nr, max(payed_until_ts) as payed_until_ts
from payment_receipts
group by phone_nr
) sub
ON ( pr.phone_nr = sub.phone_nr
AND pr.payed_until_ts = sub.payed_until_ts)
ORDER BY pr.phone_nr, pr.payed_ts, pr.payed_until_ts;
Original Answer (with updates):
CREATE TABLE foo (a NUMERIC, b TEXT, DATE);
INSERT INTO foo VALUES
(1,'a','2010-07-30'),
(1,'b','2010-07-30'),
(1,'c','2010-07-31'),
(1,'d','2010-07-31'),
(1,'a','2010-07-29'),
(1,'c','2010-07-29'),
(2,'a','2010-07-29'),
(2,'a','2010-08-01');
-- table contents
SELECT * FROM foo ORDER BY c,a,b;
a | b | c
---+---+------------
1 | a | 2010-07-29
1 | c | 2010-07-29
2 | a | 2010-07-29
1 | a | 2010-07-30
1 | b | 2010-07-30
1 | c | 2010-07-31
1 | d | 2010-07-31
2 | a | 2010-08-01
-- The following solutions both retrieve records based on the latest date
-- they both return the same result set, solution 1 is faster, solution 2
-- is easier to read
-- Solution 1:
SELECT foo.a, foo.b, foo.c
FROM foo
JOIN (select a, max(c) as c from foo group by a) bar
ON (foo.a=bar.a and foo.c=bar.c)
ORDER BY foo.a, foo.b, foo.c;
-- Solution 2:
SELECT a, b, MAX(c) AS c
FROM foo main
GROUP BY a, b
HAVING MAX(c) = (select max(c) from foo sub where main.a=sub.a group by a)
ORDER BY a, b;
a | b | c
---+---+------------
1 | c | 2010-07-31
1 | d | 2010-07-31
2 | a | 2010-08-01
(3 rows)
Comment:
1 is returned twice because their are multiple b values. This is acceptable (and advised). Your data should never have this problem, because c is based on b's value.

Answer 5

create table user_payments
(
phone_nr int NOT NULL,
payed_until_ts datetime NOT NULL
)
insert into user_payments
(phone_nr, payed_until_ts)
values
(1, '2016-01-28'), -- today
(1, '2016-01-27'), -- yesterday
(2, '2016-01-27'), -- yesterday
(2, '2016-01-29') -- tomorrow
select phone_nr, MAX(payed_until_ts) as latest_payment
from user_payments
group by phone_nr
-- OUTPUT:
-- phone_nr latest_payment
-- 1 2016-01-28 00:00:00.000
-- 2 2016-01-29 00:00:00.000
In the above example, I have used datetime column but similar query should work for timestamp column.
The MAX function will basically do the "ORDER BY" payed_until_ts column and pick the latest value for each phone_nr.
Also, you will get only one value for each phone_nr due to "GROUP BY" clause.