Hi apologies for formatting but im stumped and frustrated and i just need some help.
I've got two tables. I have made a good faith attempt to follow community standards but just in case it doesnt work, Table A has 3 columns 'ID', to identify a sales rep, 'Start' to indicate what company term they started, and 'Sales' to indicate their sales in that first term. Table B is just an expansion of Table A where it lists all terms (i marked it as quarters) a sales person was there and their sales.
Table A
+----+---------+-------+
| ID | Quarter | Sales |
+----+---------+-------+
| 1 | 141 | 30 |
| 2 | 151 | 50 |
| 3 | 151 | 80 |
+----+---------+-------+
Table B
+----+---------+-------+
| ID | Quarter | Sales |
+----+---------+-------+
| 1 | 141 | 30 |
| 1 | 142 | 25 |
| 1 | 143 | 45 |
| 2 | 151 | 50 |
| 2 | 152 | 60 |
| 2 | 153 | 75 |
| 3 | 151 | 80 |
| 3 | 152 | 50 |
| 3 | 153 | 70 |
+----+---------+-------+
My desired output is a table with ID, start term, sales from that term, second term, sales from that term, etc. for the first 6 terms an employee is there
my code is this
select a.id, start, a.sales,
case when a.start+1 = b.quarter then sales end as secondquartersales,
case when a.start+2 = b.quarter then sales end as thridquartersales,.....
from tablea a
left join tableb b
on a.id = b.id;
it gives nulls for all case when statements. please help
maybe try GROUP BY
create table a ( id number, strt number, sales number);
create table b (id number, quarter number , sales number);
insert into a values (1,141,30);
insert into a values (2,151,50);
insert into a values (3,151,80);
insert into b values ( 1,141,30);
insert into b values ( 1,142,25);
insert into b values ( 1,143,45);
insert into b values ( 2,151,50);
insert into b values ( 2,152,60);
insert into b values ( 2,153,75);
insert into b values ( 3,151,80);
insert into b values ( 3,152,50);
insert into b values ( 3,153,70);
select a.id, a.strt, a.sales,
max(case when a.strt+1 = b.quarter then b.sales end ) as secondquartersales,
max(case when a.strt+2 = b.quarter then b.sales end ) as thridquartersales
from a, b
where a.id = b.id
group by a.id, a.strt, a.sales;
OR PIVOT
select * from (
select a.id,
case when a.strt+1 = b.quarter then 'Q2'
when a.strt+2 = b.quarter then 'Q3'
when a.strt+3 = b.quarter then 'Q4'
when a.strt = b.quarter then 'Q1'end q,
b.sales sales
from a, b
where a.id = b.id)
pivot ( max(nvl(sales,0)) for Q in ('Q1', 'Q2', 'Q3', 'Q4'));
This is valid ANSI 92 SQL, as it is an inner join. The whole ANSI style version is just syntax candy.
I have two tables:
Table 'bookings':
id | date | hours
--------------------------
1 | 06/01/2016 | 2
1 | 06/02/2016 | 1
2 | 06/03/2016 | 2
3 | 06/03/2016 | 4
Table 'lookupCalendar':
date
-----
06/01/2016
06/02/2016
06/03/2016
I want to join them together so that I have a date for each booking so that the results look like this:
Table 'results':
id | date | hours
--------------------------
1 | 06/01/2016 | 2
1 | 06/02/2016 | 1
1 | 06/03/2016 | 0 <-- Added by query
2 | 06/01/2016 | 0 <-- Added by query
2 | 06/02/2016 | 0 <-- Added by query
2 | 06/03/2016 | 2
3 | 06/01/2016 | 0 <-- Added by query
3 | 06/02/2016 | 0 <-- Added by query
3 | 06/03/2016 | 4
I have tried doing a cross-apply, but that doesn't get me there, neither does a full join. The FULL JOIN just gives me nulls in the id column and the cross-apply gives me too much data.
Is there a query that can give me the results table above?
More Information
It might be beneficial to note that I am doing this so that I can calculate an average hours booked over a period of time, not just the number of records in the table.
Ideally, I'd be able to do
SELECT AVG(hours) AS my_average, id
FROM bookings
GROUP BY id
But since that would just give me a count of the records instead of the count of the days I want to cross apply it with the dates. Then I think I can just do the query above with the results table.
select i.id, c.date, coalesce(b.hours, 0) as hours
from lookupCalendar c
cross join (select distinct id from bookings) i
left join bookings b
on b.id = i.id
and b.date = c.date
order by i.id, c.date
Try this:
select c.date, b.id, isnull(b.hours, 0)
from lookupCalendar c
left join bookings b on b.date = c.date
LookupCalendar is your main table because you want the bookings against each date, irrespective of whether there was a booking on that date or not, so a left join is required.
I am not sure if you need to include b.id to solve your actual problem though. Wouldn't you just want to get the total number of hours booked against each date like this, to then calculate the average?:
select c.date, sum(isnull(b.hours, 0))
from lookupCalendar c
left join bookings b on b.date = c.date
group by c.date
You can try joining all the combinations of IDs and dates and left joining the data;
WITH Booking AS (SELECT *
FROM (VALUES
( 1 , '06/01/2016', 2 )
, ( 1 , '06/02/2016', 1 )
, ( 2 , '06/03/2016', 2 )
, ( 3 , '06/03/2016', 4 )
) x (id, date, hours)
)
, lookupid AS (
SELECT DISTINCT id FROM Booking
)
, lookupCalender AS (
SELECT DISTINCT date FROM Booking
)
SELECT ID.id, Cal.Date, ISNULL(B.Hours,0) AS hours
FROM lookupid id
INNER JOIN lookupCalender Cal
ON 1 = 1
LEFT JOIN Booking B
ON id.id = B.id
AND Cal.date = B.Date
ORDER BY ID.id, Cal.Date
Having some trouble figuring out how to make these query.
In general I have a table with
sales_ID
Employee_ID
sale_date
sale_price
what I want to do is have a view that shows for each sales item how much the employee on average sells for 1 year previous of the sale_date.
example: Suppose I have this in the sales table
sales_ID employee_id sale_date sale_price
1 Bob 2016/06/10 100
2 Bob 2016/01/01 75
3 Bob 2014/01/01 475
4 Bob 2015/12/01 100
5 Bob 2016/05/01 200
6 Fred 2016/01/01 30
7 Fred 2015/05/01 50
for sales_id 1 record I want to pull all sales from Bob by 1 year up to the month of the sale (so 2015-05-01 to 2016-05-31 which has 3 sales for 75, 100, 200) so the final output would be
sales_ID employee_id sale_date sale_price avg_sale
1 Bob 2016/06/10 100 125
2 Bob 2016/01/01 75 275
3 Bob 2014/01/01 475 null
4 Bob 2015/12/01 100 475
5 Bob 2016/05/01 200 87.5
6 Fred 2016/01/01 30 50
7 Fred 2015/05/01 50 null
What I tried doing is something like this
select a.sales_ID, a.sale_price, a.employee_ID, a.sale_date, b.avg_price
from sales a
left join (
select employee_id, avg(sale_price) as avg_price
from sales
where sale_date between Date(VARCHAR(YEAR(a.sale_date)-1) ||'-'|| VARCHAR(MONTH(a.sale_date)-1) || '-01')
and Date(VARCHAR(YEAR(a.sale_date)) ||'-'|| VARCHAR(MONTH(a.sale_date)) || '-01') -1 day
group by employee_id
) b on a.employee_id = b.employee_id
which DB2 doesn't like using the parent table a in the sub query, but I can't think of how to properly write this query. any thoughts?
Ok. I think I figured it out. Please note 3 things.
I couldn't test it in DB2, so I used Oracle. But syntax would be more or less same.
I didn't use your 1 year logic exactly. I am counting current_date minus 365 days, but you can change the between part in where clause in inner query, as you mentioned in the question.
The expected output you mentioned is incorrect. So for every sale_id, I took the date, found the employee_id, took all the sales of that employee for last 1 year, excluding the current date, and then took average. If you want to change it, you can change the where clause in subquery.
select t1.*,t2.avg_sale
from
sales t1
left join
(
select a.sales_id
,avg(b.sale_price) as avg_sale
from sales a
inner join
sales b
on a.employee_id=b.employee_id
where b.sale_date between a.sale_date - 365 and a.sale_date -1
group by a.sales_id
) t2
on t1.sales_id=t2.sales_id
order by t1.sales_id
Output
+----------+-------------+-------------+------------+----------+
| SALES_ID | EMPLOYEE_ID | SALE_DATE | SALE_PRICE | AVG_SALE |
+----------+-------------+-------------+------------+----------+
| 1 | Bob | 10-JUN-2016 | 100 | 125 |
| 2 | Bob | 01-JAN-2016 | 75 | 100 |
| 3 | Bob | 01-JAN-2014 | 475 | |
| 4 | Bob | 01-DEC-2015 | 100 | |
| 5 | Bob | 01-MAY-2016 | 200 | 87.5 |
| 6 | Fred | 01-JAN-2016 | 30 | 50 |
| 7 | Fred | 01-MAY-2015 | 50 | |
+----------+-------------+-------------+------------+----------+
You can almost fix your original query by doing a LATERAL join. Lateral allows you to reference previously declared tables as in:
select a.sales_ID, a.sale_price, a.employee_ID, a.sale_date, b.avg_price
from sales a
left join LATERAL (
select employee_id, avg(sale_price) as avg_price
from sales
where sale_date between Date(VARCHAR(YEAR(a.sale_date)-1) ||'-'|| VARCHAR(MONTH(a.sale_date)-1) || '-01')
and Date(VARCHAR(YEAR(a.sale_date)) ||'-'|| VARCHAR(MONTH(a.sale_date)) || '-01') -1 day
group by employee_id
) b on a.employee_id = b.employee_id
However, I get an syntax error from your date arithmetic, so using #Utsav solution for this yields:
select a.sales_ID, a.sale_price, a.employee_ID, a.sale_date, b.avg_price
from sales a
left join lateral (
select employee_id, avg(sale_price) as avg_price
from sales b
where a.employee_id = b.employee_id
and b.sale_date between a.sale_date - 365 and a.sale_date -1
group by employee_id
) b on a.employee_id = b.employee_id
Since we already pushed the predicate inside the LATERAL join, it is strictly speaking not necessary to use the on clause:
select a.sales_ID, a.sale_price, a.employee_ID, a.sale_date, b.avg_price
from sales a
left join lateral (
select employee_id, avg(sale_price) as avg_price
from sales b
where a.employee_id = b.employee_id
and b.sale_date between a.sale_date - 365 and a.sale_date -1
group by employee_id
) b on 1=1
By using a LATERAL join we removed one access against the sales table. A comparison of the plans show:
No LATERAL Join
Access Plan:
Total Cost: 20,4571
Query Degree: 1
Rows
RETURN
( 1)
Cost
I/O
|
7
>MSJOIN
( 2)
20,4565
3
/---+----\
7 0,388889
TBSCAN FILTER
( 3) ( 6)
6,81572 13,6402
1 2
| |
7 2,72222
SORT GRPBY
( 4) ( 7)
6,81552 13,6397
1 2
| |
7 2,72222
TBSCAN TBSCAN
( 5) ( 8)
6,81488 13,6395
1 2
| |
7 2,72222
TABLE: LELLE SORT
SALES ( 9)
Q6 13,6391
2
|
2,72222
HSJOIN
( 10)
13,6385
2
/-----+------\
7 7
TBSCAN TBSCAN
( 11) ( 12)
6,81488 6,81488
1 1
| |
7 7
TABLE: LELLE TABLE: LELLE
SALES SALES
Q2 Q1
LATERAL Join
Access Plan:
Total Cost: 13,6565
Query Degree: 1
Rows
RETURN
( 1)
Cost
I/O
|
7
>^NLJOIN
( 2)
13,6559
2
/---+----\
7 0,35
TBSCAN GRPBY
( 3) ( 4)
6,81488 6,81662
1 1
| |
7 0,35
TABLE: LELLE TBSCAN
SALES ( 5)
Q5 6,81656
1
|
7
TABLE: LELLE
SALES
Q1
Window functions with framing
DB2 does not yet support range frames over dates, but by using a clever trick by #mustaccio in:
https://dba.stackexchange.com/questions/141263/what-is-the-meaning-of-order-by-x-range-between-n-preceding-if-x-is-a-dat
we can actually use only one table access and solve the problem:
select a.sales_ID, a.sale_price, a.employee_ID, a.sale_date
, avg(sale_price) over (partition by employee_id
order by julian_day(a.sale_date)
range between 365 preceding
and 1 preceding
) as avg_price
from sales a
Access Plan:
Total Cost: 6.8197
Query Degree: 1
Rows
RETURN
( 1)
Cost
I/O
|
7
TBSCAN
( 2)
6.81753
1
|
7
SORT
( 3)
6.81703
1
|
7
TBSCAN
( 4)
6.81488
1
|
7
TABLE: LELLE
SALES
Q1
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)