Related
Here's what I'm trying to do. Let's say I have this table t:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
2 | 18 | 2012-05-19 | y
3 | 18 | 2012-08-09 | z
4 | 19 | 2009-06-01 | a
5 | 19 | 2011-04-03 | b
6 | 19 | 2011-10-25 | c
7 | 19 | 2012-08-09 | d
For each id, I want to select the row containing the minimum record_date. So I'd get:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
The only solutions I've seen to this problem assume that all record_date entries are distinct, but that is not this case in my data. Using a subquery and an inner join with two conditions would give me duplicate rows for some ids, which I don't want:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
5 | 19 | 2011-04-03 | b
4 | 19 | 2009-06-01 | a
How about something like:
SELECT mt.*
FROM MyTable mt INNER JOIN
(
SELECT id, MIN(record_date) AS MinDate
FROM MyTable
GROUP BY id
) t ON mt.id = t.id AND mt.record_date = t.MinDate
This gets the minimum date per ID, and then gets the values based on those values. The only time you would have duplicates is if there are duplicate minimum record_dates for the same ID.
I could get to your expected result just by doing this in mysql:
SELECT id, min(record_date), other_cols
FROM mytable
GROUP BY id
Does this work for you?
To get the cheapest product in each category, you use the MIN() function in a correlated subquery as follows:
SELECT categoryid,
productid,
productName,
unitprice
FROM products a WHERE unitprice = (
SELECT MIN(unitprice)
FROM products b
WHERE b.categoryid = a.categoryid)
The outer query scans all rows in the products table and returns the products that have unit prices match with the lowest price in each category returned by the correlated subquery.
I would like to add to some of the other answers here, if you don't need the first item but say the second number for example you can use rownumber in a subquery and base your result set off of that.
SELECT * FROM
(
SELECT
ROW_NUM() OVER (PARTITION BY Id ORDER BY record_date, other_cols) as rownum,
*
FROM products P
) INNER
WHERE rownum = 2
This also allows you to order off multiple columns in the subquery which may help if two record_dates have identical values. You can also partition off of multiple columns if needed by delimiting them with a comma
This does it simply:
select t2.id,t2.record_date,t2.other_cols
from (select ROW_NUMBER() over(partition by id order by record_date)as rownum,id,record_date,other_cols from MyTable)t2
where t2.rownum = 1
If record_date has no duplicates within a group:
think of it as of filtering. Simpliy get (WHERE) one (MIN(record_date)) row from the current group:
SELECT * FROM t t1 WHERE record_date = (
select MIN(record_date)
from t t2 where t2.group_id = t1.group_id)
If there could be 2+ min record_date within a group:
filter out non-min rows (see above)
then (AND) pick only one from the 2+ min record_date rows, within the given group_id. E.g. pick the one with the min unique key:
AND key_id = (select MIN(key_id)
from t t3 where t3.record_date = t1.record_date
and t3.group_id = t1.group_id)
so
key_id | group_id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
8 | 19 | 2009-06-01 | e
will select key_ids: #1 and #4
SELECT p.* FROM tbl p
INNER JOIN(
SELECT t.id, MIN(record_date) AS MinDate
FROM tbl t
GROUP BY t.id
) t ON p.id = t.id AND p.record_date = t.MinDate
GROUP BY p.id
This code eliminates duplicate record_date in case there are same ids with same record_date.
If you want duplicates, remove the last line GROUP BY p.id.
This a old question, but this can useful for someone
In my case i can't using a sub query because i have a big query and i need using min() on my result, if i use sub query the db need reexecute my big query. i'm using Mysql
select t.*
from (select m.*, #g := 0
from MyTable m --here i have a big query
order by id, record_date) t
where (1 = case when #g = 0 or #g <> id then 1 else 0 end )
and (#g := id) IS NOT NULL
Basically I ordered the result and then put a variable in order to get only the first record in each group.
The below query takes the first date for each work order (in a table of showing all status changes):
SELECT
WORKORDERNUM,
MIN(DATE)
FROM
WORKORDERS
WHERE
DATE >= to_date('2015-01-01','YYYY-MM-DD')
GROUP BY
WORKORDERNUM
select
department,
min_salary,
(select s1.last_name from staff s1 where s1.salary=s3.min_salary ) lastname
from
(select department, min (salary) min_salary from staff s2 group by s2.department) s3
I have 2 tables - TC and T, with columns specified below. TC maps to T on column T_ID.
TC
----
T_ID,
TC_ID
T
-----
T_ID,
V_ID,
Datetime,
Count
My current result set is:
V_ID TC_ID Datetime Count
----|-----|------------|--------|
2 | 1 | 2013-09-26 | 450600 |
2 | 1 | 2013-12-09 | 14700 |
2 | 1 | 2014-01-22 | 15000 |
2 | 1 | 2014-01-22 | 15000 |
2 | 1 | 2014-01-22 | 7500 |
4 | 1 | 2014-01-22 | 1000 |
4 | 1 | 2013-12-05 | 0 |
4 | 2 | 2013-12-05 | 0 |
Using the following query:
select T.V_ID,
TC.TC_ID,
T.Datetime,
T.Count
from T
inner join TC
on TC.T_ID = T.T_ID
Result set I want:
V_ID TC_ID Datetime Count
----|-----|------------|--------|
2 | 1 | 2014-01-22 | 15000 |
4 | 1 | 2014-01-22 | 1000 |
4 | 2 | 2013-12-05 | 0 |
I want to write a query to select each distinct V_ID + TC_ID combination, but only with the maximum datetime, and for that datetime the maximum count. E.g. for the distinct combination of V_ID = 2 and TC_ID = 1, '2014-01-22' is the maximum datetime, and for that datetime, 15000 is the maximum count, so select this record for the new table. Any ideas? I don't know if this is too ambitious for a query and I should just handle the result set in code instead.
One method uses row_number():
select v_id, tc_id, datetime, count
from (select T.V_ID, TC.TC_ID, T.Datetime, T.Count,
row_number() over (partition by t.V_ID, tc.tc_id
order by datetime desc, count desc
) as seqnum
from t join
tc
on tc.t_id = t._id
) tt
where seqnum = 1;
The only issue is that some rows have the same maximum datetime value. SQL tables represent unordered sets, so there is no way to determine which is really the maximum -- unless the datetime really has a time component or another column specifies the ordering within a day.
It is possible to solve this using CTEs. First, extracting the data from your query. Second, get the maxdates. Third, get the highest count for each maxdate.:
;WITH Dataset AS
(
select T.V_ID,
TC.TC_ID,
T.[Datetime],
T.[Count]
from T
inner join TC
on TC.T_ID = T._ID
),
MaxDates AS
(
SELECT V_ID, TC_ID, MAX(t.[Datetime]) AS MaxDate
FROM Dataset t
GROUP BY t.V_ID, t.TC_ID
)
SELECT t.V_ID, t.TC_ID, t.[Datetime], MAX(t.[Count]) AS [Count]
FROM Dataset t
INNER JOIN MaxDates m ON t.V_ID = m.V_ID AND t.TC_ID = m.TC_ID AND m.MaxDate = t.[Datetime]
GROUP BY t.V_ID, t.TC_ID, t.[Datetime]
Just to keep it simple:
You need to group by T.V_ID,TC.TC_ID,
with selecting the max of date and then to get the maximum count, you must use a sub query as follows,
select T.V_ID,
TC.TC_ID,
max(T.Datetime) as Date_Time,
(select max(Count) from T as tb where v_ID = T.v_ID and DateTime = max(T.DateTime)) as Count
from T
inner join TC
on TC.T_ID = T._ID
group by T.V_ID,TC.TC_ID,
Here's what I'm trying to do. Let's say I have this table t:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
2 | 18 | 2012-05-19 | y
3 | 18 | 2012-08-09 | z
4 | 19 | 2009-06-01 | a
5 | 19 | 2011-04-03 | b
6 | 19 | 2011-10-25 | c
7 | 19 | 2012-08-09 | d
For each id, I want to select the row containing the minimum record_date. So I'd get:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
The only solutions I've seen to this problem assume that all record_date entries are distinct, but that is not this case in my data. Using a subquery and an inner join with two conditions would give me duplicate rows for some ids, which I don't want:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
5 | 19 | 2011-04-03 | b
4 | 19 | 2009-06-01 | a
How about something like:
SELECT mt.*
FROM MyTable mt INNER JOIN
(
SELECT id, MIN(record_date) AS MinDate
FROM MyTable
GROUP BY id
) t ON mt.id = t.id AND mt.record_date = t.MinDate
This gets the minimum date per ID, and then gets the values based on those values. The only time you would have duplicates is if there are duplicate minimum record_dates for the same ID.
I could get to your expected result just by doing this in mysql:
SELECT id, min(record_date), other_cols
FROM mytable
GROUP BY id
Does this work for you?
To get the cheapest product in each category, you use the MIN() function in a correlated subquery as follows:
SELECT categoryid,
productid,
productName,
unitprice
FROM products a WHERE unitprice = (
SELECT MIN(unitprice)
FROM products b
WHERE b.categoryid = a.categoryid)
The outer query scans all rows in the products table and returns the products that have unit prices match with the lowest price in each category returned by the correlated subquery.
I would like to add to some of the other answers here, if you don't need the first item but say the second number for example you can use rownumber in a subquery and base your result set off of that.
SELECT * FROM
(
SELECT
ROW_NUM() OVER (PARTITION BY Id ORDER BY record_date, other_cols) as rownum,
*
FROM products P
) INNER
WHERE rownum = 2
This also allows you to order off multiple columns in the subquery which may help if two record_dates have identical values. You can also partition off of multiple columns if needed by delimiting them with a comma
This does it simply:
select t2.id,t2.record_date,t2.other_cols
from (select ROW_NUMBER() over(partition by id order by record_date)as rownum,id,record_date,other_cols from MyTable)t2
where t2.rownum = 1
If record_date has no duplicates within a group:
think of it as of filtering. Simpliy get (WHERE) one (MIN(record_date)) row from the current group:
SELECT * FROM t t1 WHERE record_date = (
select MIN(record_date)
from t t2 where t2.group_id = t1.group_id)
If there could be 2+ min record_date within a group:
filter out non-min rows (see above)
then (AND) pick only one from the 2+ min record_date rows, within the given group_id. E.g. pick the one with the min unique key:
AND key_id = (select MIN(key_id)
from t t3 where t3.record_date = t1.record_date
and t3.group_id = t1.group_id)
so
key_id | group_id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
8 | 19 | 2009-06-01 | e
will select key_ids: #1 and #4
SELECT p.* FROM tbl p
INNER JOIN(
SELECT t.id, MIN(record_date) AS MinDate
FROM tbl t
GROUP BY t.id
) t ON p.id = t.id AND p.record_date = t.MinDate
GROUP BY p.id
This code eliminates duplicate record_date in case there are same ids with same record_date.
If you want duplicates, remove the last line GROUP BY p.id.
This a old question, but this can useful for someone
In my case i can't using a sub query because i have a big query and i need using min() on my result, if i use sub query the db need reexecute my big query. i'm using Mysql
select t.*
from (select m.*, #g := 0
from MyTable m --here i have a big query
order by id, record_date) t
where (1 = case when #g = 0 or #g <> id then 1 else 0 end )
and (#g := id) IS NOT NULL
Basically I ordered the result and then put a variable in order to get only the first record in each group.
The below query takes the first date for each work order (in a table of showing all status changes):
SELECT
WORKORDERNUM,
MIN(DATE)
FROM
WORKORDERS
WHERE
DATE >= to_date('2015-01-01','YYYY-MM-DD')
GROUP BY
WORKORDERNUM
select
department,
min_salary,
(select s1.last_name from staff s1 where s1.salary=s3.min_salary ) lastname
from
(select department, min (salary) min_salary from staff s2 group by s2.department) s3
Raw Data
| ID | STATUS |
| 1 | A |
| 2 | A |
| 3 | B |
| 4 | B |
| 5 | B |
| 6 | A |
| 7 | A |
| 8 | A |
| 9 | C |
Result
| START | END |
| 1 | 2 |
| 6 | 8 |
Range of STATUS A
How to query ?
This should give you the correct ranges:
SELECT
STATUS,
MIN(ID),
max_id
FROM (
SELECT
t1.STATUS,
t1.ID,
COALESCE(MAX(t2.ID), t1.ID) max_id
FROM
yourtable t1 LEFT JOIN yourtable t2
ON t1.STATUS=t2.STATUS AND t1.ID<t2.ID
WHERE
NOT EXISTS (SELECT NULL
FROM yourtable t3
WHERE
t3.STATUS!=t1.STATUS
AND t3.ID>t1.ID AND t3.ID<t2.ID)
GROUP BY
t1.ID,
t1.STATUS
) s
WHERE
status = 'A'
GROUP BY
STATUS,
max_id
Please see fiddle here.
You are probably better off with a cursor-based solution or a client-side function.
However, if you were using Oracle - the following would work.
WITH LOWER_VALS AS
( -- All the Ids with no immediate predecessor
SELECT ROWNUM AS RN, STATUS, ID AS LOWER FROM
(
SELECT STATUS, ID
FROM RAWDATA RD1
WHERE RD1.ID -1 NOT IN
(SELECT ID FROM RAWDATA PRED_TABLE WHERE PRED_TABLE.STATUS = RD1.STATUS)
ORDER BY STATUS, ID
)
) ,
UPPER_VALS AS
( -- All the Ids with no immediate successor
SELECT ROWNUM AS RN, STATUS, ID AS UPPER FROM
(
SELECT STATUS, ID
FROM RAWDATA RD2
WHERE RD2.ID +1 NOT IN
(SELECT ID FROM RAWDATA SUCC_TABLE WHERE SUCC_TABLE.STATUS = RD2.STATUS)
ORDER BY STATUS, ID
)
)
SELECT
L.STATUS, L.LOWER, U.UPPER
FROM
LOWER_VALS L
JOIN UPPER_VALS U ON
U.RN = L.RN;
Results in the set
A 1 2
A 6 8
B 3 5
C 9 9
http://sqlfiddle.com/#!4/10184/2
There is not a lot to go on from what you put, but I think this might work. I am using T-SQL because I don't know what you are using?
SELECT
min(ID)
, max(ID)
FROM RawData
WHERE [Status] = 'A'
Following / copying computhomas's question, but adding some twists...
I have the following table in MSSQL2008
id | business_key | result | date
1 | 1 | 0 | 9
2 | 1 | 1 | 8
3 | 2 | 1 | 7
4 | 3 | n | 6
5 | 4 | 1 | 5
6 | 4 | 0 | 4
And now i want to group based on the business_key returning the complete entry with the newest date.
So my expected result is:
id | business_key | result | date
1 | 1 | 0 | 9
3 | 2 | 1 | 7
4 | 3 | n | 6
5 | 4 | 1 | 5
I also bet that there is a way to achieve that, i just can't find / see / think of it at the moment.
edit: sorry about this, I actually meant something else from original question I did. I felt like editing this might be better than accepting a solution and making another question. my original problem was that I am not filtering by id.
SELECT t.*
FROM
(
SELECT *, ROW_NUMBER() OVER
(
PARTITION BY [business_key]
ORDER BY [date] DESC
) AS [RowNum]
FROM yourTable
) AS t
WHERE t.[RowNum] = 1
SELECT
*
FROM
mytable
WHERE
ID IN (SELECT MAX(ID) FROM mytable GROUP BY business_key)
SELECT
MAX(T1.id) AS [id],
T1.business_key,
T1.result
FROM
dbo.My_Table T1
LEFT OUTER JOIN dbo.My_Table T2 ON
T2.business_key = T1.business_key AND
T2.id > T1.id
WHERE
T2.id IS NULL
GROUP BY T1.business_key,
T1.result
ORDER BY MAX(T1.id)
Edited based on clarifications
SELECT M1.*
FROM My_Table M1
INNER JOIN
(
SELECT [business_key], MAX([date]) as MaxDate
FROM My_Table
GROUP BY [business_key]
) M2 ON M1.business_key = M2.business_key AND M1.[date] = M2.MaxDate
ORDER BY M1.[id]
Assuming the combination of business_key & date is unique then....
Working example (3rd time is a charm):
declare #src as table(id int, business_key int,result int,[date] int)
insert into #src
SELECT 1,1,0,9
UNION SELECT 2,1,1,8
UNION SELECT 3,2,1,7
UNION SELECT 4,3,1,6
UNION SELECT 5,4,1,5
UNION SELECT 6,4,0,4
;with bkdate(business_key,[date])
AS
(
select business_key,MAX([date])
from #src
group by business_key
)
select src.* from #src src
inner join bkdate
ON src.[date] = bkdate.date
and src.business_key = bkdate.business_key
order by id
How about (edited after question change):
with latestdate as (
select business_key, maxdate=max(date)
from the_table
group by business_key
), latest as (
select ID = max(id)
from the_table
inner join latestdate
on the_table.business_key=latestdate.business_key
and the_table.date=latestdate.maxdate
group by the_table.business_key
)
select the_table.*
from the_table
inner join latest
on latest.id=the_table.id