Select rows with highest value in one column in SQL - sql

in MySQL, I am trying to select one row for each "foreign_id". It must be the row with the highest value in column "time" (which is of type DATETIME). Can you help me how the SQL SELECT statement must look like? Thank you!
This would be really great! I am already trying for hours to find a solution :(
This is my table:
primary_id | foreign_id | name | time
----------------------------------------------------
1 | 3 | a | 2017-05-18 01:02:03
2 | 3 | b | 2017-05-19 01:02:03
3 | 3 | c | 2017-05-20 01:02:03
4 | 5 | d | 2017-07-18 01:02:03
5 | 5 | e | 2017-07-20 01:02:03
6 | 5 | f | 2017-07-18 01:02:03
And this is what the result should look like:
primary_id | foreign_id | name | time
----------------------------------------------------
3 | 3 | c | 2017-05-20 01:02:03
5 | 5 | e | 2017-07-20 01:02:03
I tried to order the intermediate result by time (descending) and then to select only the first row by using LIMIT 1. But like this I cannot get one row for each foreign_id.
Another try was to first order the intermediate result by time (descending) and then to GROUP BY foreign_id. But the GROUP BY statement seems to be executed before the ORDER BY statement (I received the rows with primary_id 1 and 4 as a result, not 3 and 5).

Try this
SELECT DISTINCT *
From my_table A
INNER JOIN (SELECT foreign_id, Max(time) AS time FROM my_table GROUP BY foreign_id) B
ON A.foreign_id = B.foreign_id AND A.time = B.time

Just add some data sample to analyze special case
CREATE TABLE Table1
(`primary_id` int, `foreign_id` int, `name` varchar(1), `time` datetime)
;
INSERT INTO Table1
(`primary_id`, `foreign_id`, `name`, `time`)
VALUES
(1, 3, 'a', '2017-05-18 01:02:03'),
(2, 3, 'b', '2017-05-19 01:02:03'),
(3, 3, 'c', '2017-05-20 01:02:03'),
(7, 3, 'H', '2017-05-20 01:02:03'),
(4, 5, 'd', '2017-07-18 01:02:03'),
(5, 5, 'e', '2017-07-20 01:02:03'),
(6, 5, 'f', '2017-07-18 01:02:03')
;
http://sqlfiddle.com/#!9/38947b/6
select d.primary_id, d.foreign_id, c.name, d.time
from table1 c inner join (
select max(b.primary_id) primary_id, a.foreign_id, a.time
from table1 b inner join
( select foreign_id, max(time) time
from table1
group by foreign_id) a
on a.foreign_id = b.foreign_id and a.time=b.time
group by a.foreign_id, a.time ) d
on c.primary_id=d.primary_id

In days gone by you would code this as a correlated subquery:
SELECT *
FROM Table1 o
WHERE primary_id = (
SELECT min (m.primary_id) FROM Table1 m
WHERE m.time= (
SELECT max (i.time) FROM Table1 i
WHERE o.foreign_id=i.foreign_id
)
)
The extra subquery handles the case of duplicate foreign_id & time values. If you were sure that time was unique for each foreign_id you could omit the middle subquery.

Related

Selecting only top parent table row with all of it's children table rows

So I have two tables:
#ProjectHealthReports
Id | From | SubmittedOn
1 | 2020-01-01 |
2 | 2020-02-01 | 2020-10-23
3 | 2020-03-01 |
4 | 2020-04-01 | 2020-10-23
5 | 2020-05-01 | 2020-10-23
#ProjectHealthReportItems
Id | Note | ProjectHealthReportId
1 | First for 2020-01-01 | 1
2 | Second for 2020-01-01 | 1
3 | First for 2020-02-01 | 2
4 | Second for 2020-02-01 | 2
5 | First for 2020-03-01 | 3
6 | Second for 2020-03-01 | 3
7 | First for 2020-04-01 | 4
8 | Second for 2020-04-01 | 4
9 | (We want this one) First for 2020-05-01 | 5
10 | (We want this one) Second for 2020-05-01 | 5
How can I get all #ProjectHealthReportItems and #ProjectHealthReport details for the last From date which has value for SubmittedOn (so in this case it would be ProjectHealthReport 5 and ProjectHealthReportItems 9, 10).
Basically, I need something like this just, obviously without top 1 as it only returns one row and I need, in this case, to return 2 rows :)
select top 1 phr.Id, phr.[From], phr.SubmittedOn, phri.Note from #ProjectHealthReports phr
inner join #ProjectHealthReportItems phri on phr.Id = phri.ProjectHealthReportId
where phr.SubmittedOn is not null
order by phr.[From] desc
Here is the SQL for creating and seeding the tables
create table #ProjectHealthReports(
Id int primary key,
[From] date not null ,
SubmittedOn date null
)
go
create table #ProjectHealthReportItems(
Id int primary key,
Note nvarchar(max),
ProjectHealthReportId int constraint FK_PHR references #ProjectHealthReports
)
go
insert into #ProjectHealthReports(Id, [From], SubmittedOn)
values (1, '2020-01-01', null),
(2, '2020-02-01', getutcdate()),
(3, '2020-03-01', null),
(4, '2020-04-01', getutcdate()),
(5, '2020-05-01', getutcdate())
go
insert into #ProjectHealthReportItems(Id, Note, ProjectHealthReportId)
values (1, 'First for 2020-01-01', 1),
(2, 'Second for 2020-01-01', 1),
(3, 'First for 2020-02-01', 2),
(4, 'Second for 2020-02-01', 2),
(5, 'First for 2020-03-01', 3),
(6, 'Second for 2020-03-01', 3),
(7, 'First for 2020-04-01', 4),
(8, 'Second for 2020-04-01', 4),
(9, '(We want this one) First for 2020-05-01', 5),
(10, '(We want this one) Second for 2020-05-01', 5)
go
First select top then join
select t.*, phri.Note
from (select top(1) phr.Id phrid, phr.[From], phr.SubmittedOn
from #ProjectHealthReports phr
where phr.SubmittedOn is not null
order by phr.[From] desc) t
inner join #ProjectHealthReportItems phri on t.phrId = phri.ProjectHealthReportId
I would suggest window functions:
select phr.*, phri.*
from #ProjectHealthReports phr left join
(select phri.*,
row_number() over (partition by ProjectHealthReportId order by id desc) as seqnum
from #ProjectHealthReportItems phri
) phri
on phr.Id = phri.ProjectHealthReportId and seqnum = 1
order by phr.[From] desc;
You can also do this using filtering in the where, such as correlated subquery:
select phr.*, phri.*
from #ProjectHealthReports phr join
#ProjectHealthReportItems phri
on phr.Id = phri.ProjectHealthReportId and seqnum = 1
where phri.id = (select max(phri2.id)
from #ProjectHealthReportItems phri2
where phri2.ProjectHealthReportId = phri.ProjectHealthReportId
)
order by phr.[From] desc
An efficient way to do this without a LEFT JOIN would be assign a row number, using the ROW_NUMBER() windowing function, to the #ProjectHealthReports table. Something like this
with lv_cte as (
select *, row_number() over (order by [From] desc) rn
from #ProjectHealthReports)
select l.*, phri.*
from lv_cte l
join #ProjectHealthReportItems phri on l.id=phri.ProjectHealthReportId
where l.rn=1;
Output
Id From SubmittedOn rn Id Note ProjectHealthReportId
5 2020-05-01 2020-10-23 1 9 (We want this one) First for 2020-05-01 5
5 2020-05-01 2020-10-23 1 10 (We want this one) Second for 2020-05-01 5

A very basic SQL issue I'm stuck with [duplicate]

I have a table of player performance:
CREATE TABLE TopTen (
id INT UNSIGNED PRIMARY KEY AUTO_INCREMENT,
home INT UNSIGNED NOT NULL,
`datetime`DATETIME NOT NULL,
player VARCHAR(6) NOT NULL,
resource INT NOT NULL
);
What query will return the rows for each distinct home holding its maximum value of datetime? In other words, how can I filter by the maximum datetime (grouped by home) and still include other non-grouped, non-aggregate columns (such as player) in the result?
For this sample data:
INSERT INTO TopTen
(id, home, `datetime`, player, resource)
VALUES
(1, 10, '04/03/2009', 'john', 399),
(2, 11, '04/03/2009', 'juliet', 244),
(5, 12, '04/03/2009', 'borat', 555),
(3, 10, '03/03/2009', 'john', 300),
(4, 11, '03/03/2009', 'juliet', 200),
(6, 12, '03/03/2009', 'borat', 500),
(7, 13, '24/12/2008', 'borat', 600),
(8, 13, '01/01/2009', 'borat', 700)
;
the result should be:
id
home
datetime
player
resource
1
10
04/03/2009
john
399
2
11
04/03/2009
juliet
244
5
12
04/03/2009
borat
555
8
13
01/01/2009
borat
700
I tried a subquery getting the maximum datetime for each home:
-- 1 ..by the MySQL manual:
SELECT DISTINCT
home,
id,
datetime AS dt,
player,
resource
FROM TopTen t1
WHERE `datetime` = (SELECT
MAX(t2.datetime)
FROM TopTen t2
GROUP BY home)
GROUP BY `datetime`
ORDER BY `datetime` DESC
The result-set has 130 rows although database holds 187, indicating the result includes some duplicates of home.
Then I tried joining to a subquery that gets the maximum datetime for each row id:
-- 2 ..join
SELECT
s1.id,
s1.home,
s1.datetime,
s1.player,
s1.resource
FROM TopTen s1
JOIN (SELECT
id,
MAX(`datetime`) AS dt
FROM TopTen
GROUP BY id) AS s2
ON s1.id = s2.id
ORDER BY `datetime`
Nope. Gives all the records.
I tried various exotic queries, each with various results, but nothing that got me any closer to solving this problem.
You are so close! All you need to do is select BOTH the home and its max date time, then join back to the topten table on BOTH fields:
SELECT tt.*
FROM topten tt
INNER JOIN
(SELECT home, MAX(datetime) AS MaxDateTime
FROM topten
GROUP BY home) groupedtt
ON tt.home = groupedtt.home
AND tt.datetime = groupedtt.MaxDateTime
The fastest MySQL solution, without inner queries and without GROUP BY:
SELECT m.* -- get the row that contains the max value
FROM topten m -- "m" from "max"
LEFT JOIN topten b -- "b" from "bigger"
ON m.home = b.home -- match "max" row with "bigger" row by `home`
AND m.datetime < b.datetime -- want "bigger" than "max"
WHERE b.datetime IS NULL -- keep only if there is no bigger than max
Explanation:
Join the table with itself using the home column. The use of LEFT JOIN ensures all the rows from table m appear in the result set. Those that don't have a match in table b will have NULLs for the columns of b.
The other condition on the JOIN asks to match only the rows from b that have bigger value on the datetime column than the row from m.
Using the data posted in the question, the LEFT JOIN will produce this pairs:
+------------------------------------------+--------------------------------+
| the row from `m` | the matching row from `b` |
|------------------------------------------|--------------------------------|
| id home datetime player resource | id home datetime ... |
|----|-----|------------|--------|---------|------|------|------------|-----|
| 1 | 10 | 04/03/2009 | john | 399 | NULL | NULL | NULL | ... | *
| 2 | 11 | 04/03/2009 | juliet | 244 | NULL | NULL | NULL | ... | *
| 5 | 12 | 04/03/2009 | borat | 555 | NULL | NULL | NULL | ... | *
| 3 | 10 | 03/03/2009 | john | 300 | 1 | 10 | 04/03/2009 | ... |
| 4 | 11 | 03/03/2009 | juliet | 200 | 2 | 11 | 04/03/2009 | ... |
| 6 | 12 | 03/03/2009 | borat | 500 | 5 | 12 | 04/03/2009 | ... |
| 7 | 13 | 24/12/2008 | borat | 600 | 8 | 13 | 01/01/2009 | ... |
| 8 | 13 | 01/01/2009 | borat | 700 | NULL | NULL | NULL | ... | *
+------------------------------------------+--------------------------------+
Finally, the WHERE clause keeps only the pairs that have NULLs in the columns of b (they are marked with * in the table above); this means, due to the second condition from the JOIN clause, the row selected from m has the biggest value in column datetime.
Read the SQL Antipatterns: Avoiding the Pitfalls of Database Programming book for other SQL tips.
Here goes T-SQL version:
-- Test data
DECLARE #TestTable TABLE (id INT, home INT, date DATETIME,
player VARCHAR(20), resource INT)
INSERT INTO #TestTable
SELECT 1, 10, '2009-03-04', 'john', 399 UNION
SELECT 2, 11, '2009-03-04', 'juliet', 244 UNION
SELECT 5, 12, '2009-03-04', 'borat', 555 UNION
SELECT 3, 10, '2009-03-03', 'john', 300 UNION
SELECT 4, 11, '2009-03-03', 'juliet', 200 UNION
SELECT 6, 12, '2009-03-03', 'borat', 500 UNION
SELECT 7, 13, '2008-12-24', 'borat', 600 UNION
SELECT 8, 13, '2009-01-01', 'borat', 700
-- Answer
SELECT id, home, date, player, resource
FROM (SELECT id, home, date, player, resource,
RANK() OVER (PARTITION BY home ORDER BY date DESC) N
FROM #TestTable
)M WHERE N = 1
-- and if you really want only home with max date
SELECT T.id, T.home, T.date, T.player, T.resource
FROM #TestTable T
INNER JOIN
( SELECT TI.id, TI.home, TI.date,
RANK() OVER (PARTITION BY TI.home ORDER BY TI.date) N
FROM #TestTable TI
WHERE TI.date IN (SELECT MAX(TM.date) FROM #TestTable TM)
)TJ ON TJ.N = 1 AND T.id = TJ.id
EDIT
Unfortunately, there are no RANK() OVER function in MySQL.
But it can be emulated, see Emulating Analytic (AKA Ranking) Functions with MySQL.
So this is MySQL version:
SELECT id, home, date, player, resource
FROM TestTable AS t1
WHERE
(SELECT COUNT(*)
FROM TestTable AS t2
WHERE t2.home = t1.home AND t2.date > t1.date
) = 0
This will work even if you have two or more rows for each home with equal DATETIME's:
SELECT id, home, datetime, player, resource
FROM (
SELECT (
SELECT id
FROM topten ti
WHERE ti.home = t1.home
ORDER BY
ti.datetime DESC
LIMIT 1
) lid
FROM (
SELECT DISTINCT home
FROM topten
) t1
) ro, topten t2
WHERE t2.id = ro.lid
I think this will give you the desired result:
SELECT home, MAX(datetime)
FROM my_table
GROUP BY home
BUT if you need other columns as well, just make a join with the original table (check Michael La Voie answer)
Best regards.
Since people seem to keep running into this thread (comment date ranges from 1.5 year) isn't this much simpler:
SELECT * FROM (SELECT * FROM topten ORDER BY datetime DESC) tmp GROUP BY home
No aggregation functions needed...
Cheers.
You can also try this one and for large tables query performance will be better. It works when there no more than two records for each home and their dates are different. Better general MySQL query is one from Michael La Voie above.
SELECT t1.id, t1.home, t1.date, t1.player, t1.resource
FROM t_scores_1 t1
INNER JOIN t_scores_1 t2
ON t1.home = t2.home
WHERE t1.date > t2.date
Or in case of Postgres or those dbs that provide analytic functions try
SELECT t.* FROM
(SELECT t1.id, t1.home, t1.date, t1.player, t1.resource
, row_number() over (partition by t1.home order by t1.date desc) rw
FROM topten t1
INNER JOIN topten t2
ON t1.home = t2.home
WHERE t1.date > t2.date
) t
WHERE t.rw = 1
SELECT tt.*
FROM TestTable tt
INNER JOIN
(
SELECT coord, MAX(datetime) AS MaxDateTime
FROM rapsa
GROUP BY
krd
) groupedtt
ON tt.coord = groupedtt.coord
AND tt.datetime = groupedtt.MaxDateTime
This works on Oracle:
with table_max as(
select id
, home
, datetime
, player
, resource
, max(home) over (partition by home) maxhome
from table
)
select id
, home
, datetime
, player
, resource
from table_max
where home = maxhome
Try this for SQL Server:
WITH cte AS (
SELECT home, MAX(year) AS year FROM Table1 GROUP BY home
)
SELECT * FROM Table1 a INNER JOIN cte ON a.home = cte.home AND a.year = cte.year
Here is MySQL version which prints only one entry where there are duplicates MAX(datetime) in a group.
You could test here http://www.sqlfiddle.com/#!2/0a4ae/1
Sample Data
mysql> SELECT * from topten;
+------+------+---------------------+--------+----------+
| id | home | datetime | player | resource |
+------+------+---------------------+--------+----------+
| 1 | 10 | 2009-04-03 00:00:00 | john | 399 |
| 2 | 11 | 2009-04-03 00:00:00 | juliet | 244 |
| 3 | 10 | 2009-03-03 00:00:00 | john | 300 |
| 4 | 11 | 2009-03-03 00:00:00 | juliet | 200 |
| 5 | 12 | 2009-04-03 00:00:00 | borat | 555 |
| 6 | 12 | 2009-03-03 00:00:00 | borat | 500 |
| 7 | 13 | 2008-12-24 00:00:00 | borat | 600 |
| 8 | 13 | 2009-01-01 00:00:00 | borat | 700 |
| 9 | 10 | 2009-04-03 00:00:00 | borat | 700 |
| 10 | 11 | 2009-04-03 00:00:00 | borat | 700 |
| 12 | 12 | 2009-04-03 00:00:00 | borat | 700 |
+------+------+---------------------+--------+----------+
MySQL Version with User variable
SELECT *
FROM (
SELECT ord.*,
IF (#prev_home = ord.home, 0, 1) AS is_first_appear,
#prev_home := ord.home
FROM (
SELECT t1.id, t1.home, t1.player, t1.resource
FROM topten t1
INNER JOIN (
SELECT home, MAX(datetime) AS mx_dt
FROM topten
GROUP BY home
) x ON t1.home = x.home AND t1.datetime = x.mx_dt
ORDER BY home
) ord, (SELECT #prev_home := 0, #seq := 0) init
) y
WHERE is_first_appear = 1;
+------+------+--------+----------+-----------------+------------------------+
| id | home | player | resource | is_first_appear | #prev_home := ord.home |
+------+------+--------+----------+-----------------+------------------------+
| 9 | 10 | borat | 700 | 1 | 10 |
| 10 | 11 | borat | 700 | 1 | 11 |
| 12 | 12 | borat | 700 | 1 | 12 |
| 8 | 13 | borat | 700 | 1 | 13 |
+------+------+--------+----------+-----------------+------------------------+
4 rows in set (0.00 sec)
Accepted Answers' outout
SELECT tt.*
FROM topten tt
INNER JOIN
(
SELECT home, MAX(datetime) AS MaxDateTime
FROM topten
GROUP BY home
) groupedtt ON tt.home = groupedtt.home AND tt.datetime = groupedtt.MaxDateTime
+------+------+---------------------+--------+----------+
| id | home | datetime | player | resource |
+------+------+---------------------+--------+----------+
| 1 | 10 | 2009-04-03 00:00:00 | john | 399 |
| 2 | 11 | 2009-04-03 00:00:00 | juliet | 244 |
| 5 | 12 | 2009-04-03 00:00:00 | borat | 555 |
| 8 | 13 | 2009-01-01 00:00:00 | borat | 700 |
| 9 | 10 | 2009-04-03 00:00:00 | borat | 700 |
| 10 | 11 | 2009-04-03 00:00:00 | borat | 700 |
| 12 | 12 | 2009-04-03 00:00:00 | borat | 700 |
+------+------+---------------------+--------+----------+
7 rows in set (0.00 sec)
SELECT c1, c2, c3, c4, c5 FROM table1 WHERE c3 = (select max(c3) from table)
SELECT * FROM table1 WHERE c3 = (select max(c3) from table1)
Another way to gt the most recent row per group using a sub query which basically calculates a rank for each row per group and then filter out your most recent rows as with rank = 1
select a.*
from topten a
where (
select count(*)
from topten b
where a.home = b.home
and a.`datetime` < b.`datetime`
) +1 = 1
DEMO
Here is the visual demo for rank no for each row for better understanding
By reading some comments what about if there are two rows which have same 'home' and 'datetime' field values?
Above query will fail and will return more than 1 rows for above situation. To cover up this situation there will be a need of another criteria/parameter/column to decide which row should be taken which falls in above situation. By viewing sample data set i assume there is a primary key column id which should be set to auto increment. So we can use this column to pick the most recent row by tweaking same query with the help of CASE statement like
select a.*
from topten a
where (
select count(*)
from topten b
where a.home = b.home
and case
when a.`datetime` = b.`datetime`
then a.id < b.id
else a.`datetime` < b.`datetime`
end
) + 1 = 1
DEMO
Above query will pick the row with highest id among the same datetime values
visual demo for rank no for each row
Why not using:
SELECT home, MAX(datetime) AS MaxDateTime,player,resource FROM topten GROUP BY home
Did I miss something?
In MySQL 8.0 this can be achieved efficiently by using row_number() window function with common table expression.
(Here row_number() basically generating unique sequence for each row for every player starting with 1 in descending order of resource. So, for every player row with sequence number 1 will be with highest resource value. Now all we need to do is selecting row with sequence number 1 for each player. It can be done by writing an outer query around this query. But we used common table expression instead since it's more readable.)
Schema:
create TABLE TestTable(id INT, home INT, date DATETIME,
player VARCHAR(20), resource INT);
INSERT INTO TestTable
SELECT 1, 10, '2009-03-04', 'john', 399 UNION
SELECT 2, 11, '2009-03-04', 'juliet', 244 UNION
SELECT 5, 12, '2009-03-04', 'borat', 555 UNION
SELECT 3, 10, '2009-03-03', 'john', 300 UNION
SELECT 4, 11, '2009-03-03', 'juliet', 200 UNION
SELECT 6, 12, '2009-03-03', 'borat', 500 UNION
SELECT 7, 13, '2008-12-24', 'borat', 600 UNION
SELECT 8, 13, '2009-01-01', 'borat', 700
Query:
with cte as
(
select id, home, date , player, resource,
Row_Number()Over(Partition by home order by date desc) rownumber from TestTable
)
select id, home, date , player, resource from cte where rownumber=1
Output:
id
home
date
player
resource
1
10
2009-03-04 00:00:00
john
399
2
11
2009-03-04 00:00:00
juliet
244
5
12
2009-03-04 00:00:00
borat
555
8
13
2009-01-01 00:00:00
borat
700
db<>fiddle here
This works in SQLServer, and is the only solution I've seen that doesn't require subqueries or CTEs - I think this is the most elegant way to solve this kind of problem.
SELECT TOP 1 WITH TIES *
FROM TopTen
ORDER BY ROW_NUMBER() OVER (PARTITION BY home
ORDER BY [datetime] DESC)
In the ORDER BY clause, it uses a window function to generate & sort by a ROW_NUMBER - assigning a 1 value to the highest [datetime] for each [home].
SELECT TOP 1 WITH TIES will then select one record with the lowest ROW_NUMBER (which will be 1), as well as all records with a tying ROW_NUMBER (also 1)
As a consequence, you retrieve all data for each of the 1st ranked records - that is, all data for records with the highest [datetime] value with their given [home] value.
Try this
select * from mytable a join
(select home, max(datetime) datetime
from mytable
group by home) b
on a.home = b.home and a.datetime = b.datetime
Regards
K
#Michae The accepted answer will working fine in most of the cases but it fail for one for as below.
In case if there were 2 rows having HomeID and Datetime same the query will return both rows, not distinct HomeID as required, for that add Distinct in query as below.
SELECT DISTINCT tt.home , tt.MaxDateTime
FROM topten tt
INNER JOIN
(SELECT home, MAX(datetime) AS MaxDateTime
FROM topten
GROUP BY home) groupedtt
ON tt.home = groupedtt.home
AND tt.datetime = groupedtt.MaxDateTime
this is the query you need:
SELECT b.id, a.home,b.[datetime],b.player,a.resource FROM
(SELECT home,MAX(resource) AS resource FROM tbl_1 GROUP BY home) AS a
LEFT JOIN
(SELECT id,home,[datetime],player,resource FROM tbl_1) AS b
ON a.resource = b.resource WHERE a.home =b.home;
Hope below query will give the desired output:
Select id, home,datetime,player,resource, row_number() over (Partition by home ORDER by datetime desc) as rownum from tablename where rownum=1
(NOTE: The answer of Michael is perfect for a situation where the target column datetime cannot have duplicate values for each distinct home.)
If your table has duplicate rows for homexdatetime and you need to only select one row for each distinct home column, here is my solution to it:
Your table needs one unique column (like id). If it doesn't, create a view and add a random column to it.
Use this query to select a single row for each unique home value. Selects the lowest id in case of duplicate datetime.
SELECT tt.*
FROM topten tt
INNER JOIN
(
SELECT min(id) as min_id, home from topten tt2
INNER JOIN
(
SELECT home, MAX(datetime) AS MaxDateTime
FROM topten
GROUP BY home) groupedtt2
ON tt2.home = groupedtt2.home
) as groupedtt
ON tt.id = groupedtt.id
Accepted answer doesn't work for me if there are 2 records with same date and home. It will return 2 records after join. While I need to select any (random) of them. This query is used as joined subquery so just limit 1 is not possible there.
Here is how I reached desired result. Don't know about performance however.
select SUBSTRING_INDEX(GROUP_CONCAT(id order by datetime desc separator ','),',',1) as id, home, MAX(datetime) as 'datetime'
from topten
group by (home)

SQL Ignore duplicate primary keys

Imagine you have a string of results from a SELECT statement:
ID (pk) Name Address
1 a b
1 c d
1 e f
2 a b
3 a d
2 a d
Is it possible to alter the SQL statement to get one record ONLY for the record with ID 1?
I have a SELECT statement that displays multiple values which can have the same primary key. I want to only take one of those records, if say, I have 5 records with the same primary key.
SQL: http://pastebin.com/cFCBA2Uy
Screenshot: http://i.imgur.com/UlMBZhC.png
What I want is to show only one file which is for e.g. File Number: 925, 890
You stated that no matter which row to choose when there are more than one row for the same Id, you just want one row for each id.
The following query does what you asked for:
DECLARE #T table
(
id int,
name varchar(50),
address varchar(50)
)
INSERT INTO #T VALUES
(1, 'a', 'b'),
(1, 'c', 'd'),
(1, 'e', 'f'),
(2, 'a', 'b'),
(3, 'a', 'd'),
(2, 'a', 'd');
WITH A AS
(
SELECT
t.id, t.name, t.address,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY (SELECT NULL)) AS RowNumber
FROM
#T t
)
SELECT
A.id, A.name, A.address
FROM
A
WHERE
A.RowNumber = 1
But I think there should be a criteria. If you find one, express your criteria as the ORDER BY inside the OVER clause.
EDIT:
Here you have the result:
+----+------+---------+
| id | name | address |
+----+------+---------+
| 1 | a | b |
| 2 | a | b |
| 3 | a | d |
+----+------+---------+
Disclaimer: the query I wrote is non-deterministic, different conditions (indexes, statistics, etc) might lead to different results.

SQL inner join using multiple in statements on single table

Having a bit of trouble with an SQL query I am trying to create. The table format is as follows,
ID | Data Identifier | Date Added | Data Column
1 | 1001 | 15400 | Newest Value
1 | 1001 | 15000 | Oldest Value
1 | 1001 | 15200 | Older Value
1 | 1002 | 16000 | Newest Value
2 | 1001 | 16000 | Newest Value
What I am trying to do is, for each ID in a list (1,2) , and for each Data Identifier id in (1001,1002) return just the rows with the first matching field id and date nearest and below 16001.
So the results would be :
1 | 1001 | 15400 | Newest Value
1 | 1002 | 16000 | Newest Value
2 | 1001 | 16000 | Newest Value
I have tried several manner of joins but I keep returning duplicate records. Any advice or help would be appreciated.
It seems as if you want to GROUP BY and maybe a self join onto the table.
I have the following code for you:
-- Preparing a test table
INSERT INTO #tmpTable(ID, Identifier, DateAdded, DataColumn)
SELECT 1, 1001, 15400, 'Newest Value'
UNION
SELECT 1, 1001, 15000, 'Oldest Value'
UNION
SELECT 1, 1001, 15200, 'Older Value'
UNION
SELECT 1, 1002, 16000, 'Newest Value'
UNION
SELECT 2, 1001, 16000, 'Newest Value'
-- Actual Select
SELECT b.ID, b.Identifier, b.DateAdded, DataColumn
FROM
(SELECT ID, Identifier, MAX(DateAdded) AS DateAdded
FROM #tmpTable
WHERE DateAdded < 16001
GROUP BY ID, Identifier) a
INNER JOIN #tmpTable b ON a.DateAdded = b.DateAdded
AND a.ID = b.ID
AND a.Identifier = b.Identifier
You need to create a primary key column on your table that will not be used as an aggregate. Then you can create a CTE to select the rows required and then use it to select the data.
The aggregate function MIN(ABS(15500 - DateAdded)) will return the closest value to 15500.
WITH g AS
(
SELECT MAX(UniqueKey) AS UniqueKey, ID, DataIdentifier, MIN(ABS(15500 - DateAdded)) AS "DateTest"
FROM test
GROUP BY ID, DataIdentifier
)
SELECT test.ID, test.DataIdentifier, test.DateAdded, test.DataColumn
FROM g
INNER JOIN test
ON g.UniqueKey = test.UniqueKey
EDIT:
Screenshot of working example:
I think in this case self-join would be the best, but I still don't get the nearest and below value... (may be 15400)

Query for missing elements

I have a table with the following structure:
timestamp | name | value
0 | john | 5
1 | NULL | 3
8 | NULL | 12
12 | john | 3
33 | NULL | 4
54 | pete | 1
180 | NULL | 4
400 | john | 3
401 | NULL | 4
592 | anna | 2
Now what I am looking for is a query that will give me the sum of the values for each name, and treats the nulls in between (orderd by the timestamp) as the first non-null name down the list, as if the table were as follows:
timestamp | name | value
0 | john | 5
1 | john | 3
8 | john | 12
12 | john | 3
33 | pete | 4
54 | pete | 1
180 | john | 4
400 | john | 3
401 | anna | 4
592 | anna | 2
and I would query SUM(value), name from this table group by name. I have thought and tried, but I can't come up with a proper solution. I have looked at recursive common table expressions, and think the answer may lie in there, but I haven't been able to properly understand those.
These tables are just examples, and I don't know the timestamp values in advance.
Could someone give me a hand? Help would be very much appreciated.
With Inputs As
(
Select 0 As [timestamp], 'john' As Name, 5 As value
Union All Select 1, NULL, 3
Union All Select 8, NULL, 12
Union All Select 12, 'john', 3
Union All Select 33, NULL, 4
Union All Select 54, 'pete', 1
Union All Select 180, NULL, 4
Union All Select 400, 'john', 3
Union All Select 401, NULL, 4
Union All Select 592, 'anna', 2
)
, NamedInputs As
(
Select I.timestamp
, Coalesce (I.Name
, (
Select I3.Name
From Inputs As I3
Where I3.timestamp = (
Select Max(I2.timestamp)
From Inputs As I2
Where I2.timestamp < I.timestamp
And I2.Name Is not Null
)
)) As name
, I.value
From Inputs As I
)
Select NI.name, Sum(NI.Value) As Total
From NamedInputs As NI
Group By NI.name
Btw, what would be orders of magnitude faster than any query would be to first correct the data. I.e., update the name column to have the proper value, make it non-nullable and then run a simple Group By to get your totals.
Additional Solution
Select Coalesce(I.Name, I2.Name), Sum(I.value) As Total
From Inputs As I
Left Join (
Select I1.timestamp, MAX(I2.Timestamp) As LastNameTimestamp
From Inputs As I1
Left Join Inputs As I2
On I2.timestamp < I1.timestamp
And I2.Name Is Not Null
Group By I1.timestamp
) As Z
On Z.timestamp = I.timestamp
Left Join Inputs As I2
On I2.timestamp = Z.LastNameTimestamp
Group By Coalesce(I.Name, I2.Name)
You don't need CTE, just a simple subquery.
select t.timestamp, ISNULL(t.name, (
select top(1) i.name
from inputs i
where i.timestamp < t.timestamp
and i.name is not null
order by i.timestamp desc
)), t.value
from inputs t
And summing from here
select name, SUM(value) as totalValue
from
(
select t.timestamp, ISNULL(t.name, (
select top(1) i.name
from inputs i
where i.timestamp < t.timestamp
and i.name is not null
order by i.timestamp desc
)) as name, t.value
from inputs t
) N
group by name
I hope I'm not going to be embarassed by offering you this little recursive CTE query of mine as a solution to your problem.
;WITH
numbered_table AS (
SELECT
timestamp, name, value,
rownum = ROW_NUMBER() OVER (ORDER BY timestamp)
FROM your_table
),
filled_table AS (
SELECT
timestamp,
name,
value
FROM numbered_table
WHERE rownum = 1
UNION ALL
SELECT
nt.timestamp,
name = ISNULL(nt.name, ft.name),
nt.value
FROM numbered_table nt
INNER JOIN filled_table ft ON nt.rownum = ft.rownum + 1
)
SELECT *
FROM filled_table
/* or go ahead aggregating instead */