Select info from table where row has max date - sql

My table looks something like this:
group date cash checks
1 1/1/2013 0 0
2 1/1/2013 0 800
1 1/3/2013 0 700
3 1/1/2013 0 600
1 1/2/2013 0 400
3 1/5/2013 0 200
-- Do not need cash just demonstrating that table has more information in it
I want to get the each unique group where date is max and checks is greater than 0. So the return would look something like:
group date checks
2 1/1/2013 800
1 1/3/2013 700
3 1/5/2013 200
attempted code:
SELECT group,MAX(date),checks
FROM table
WHERE checks>0
GROUP BY group
ORDER BY group DESC
problem with that though is it gives me all the dates and checks rather than just the max date row.
using ms sql server 2005

SELECT group,MAX(date) as max_date
FROM table
WHERE checks>0
GROUP BY group
That works to get the max date..join it back to your data to get the other columns:
Select group,max_date,checks
from table t
inner join
(SELECT group,MAX(date) as max_date
FROM table
WHERE checks>0
GROUP BY group)a
on a.group = t.group and a.max_date = date
Inner join functions as the filter to get the max record only.
FYI, your column names are horrid, don't use reserved words for columns (group, date, table).

You can use a window MAX() like this:
SELECT
*,
max_date = MAX(date) OVER (PARTITION BY group)
FROM table
to get max dates per group alongside other data:
group date cash checks max_date
----- -------- ---- ------ --------
1 1/1/2013 0 0 1/3/2013
2 1/1/2013 0 800 1/1/2013
1 1/3/2013 0 700 1/3/2013
3 1/1/2013 0 600 1/5/2013
1 1/2/2013 0 400 1/3/2013
3 1/5/2013 0 200 1/5/2013
Using the above output as a derived table, you can then get only rows where date matches max_date:
SELECT
group,
date,
checks
FROM (
SELECT
*,
max_date = MAX(date) OVER (PARTITION BY group)
FROM table
) AS s
WHERE date = max_date
;
to get the desired result.
Basically, this is similar to #Twelfth's suggestion but avoids a join and may thus be more efficient.
You can try the method at SQL Fiddle.

Using an in can have a performance impact. Joining two subqueries will not have the same performance impact and can be accomplished like this:
SELECT *
FROM (SELECT msisdn
,callid
,Change_color
,play_file_name
,date_played
FROM insert_log
WHERE play_file_name NOT IN('Prompt1','Conclusion_Prompt_1','silent')
ORDER BY callid ASC) t1
JOIN (SELECT MAX(date_played) AS date_played
FROM insert_log GROUP BY callid) t2
ON t1.date_played = t2.date_played

SELECT distinct
group,
max_date = MAX(date) OVER (PARTITION BY group), checks
FROM table
Should work.

Related

How to select rows where values changed for an ID

I have a table that looks like the following
id effective_date number_of_int_customers
123 10/01/19 0
123 02/01/20 3
456 10/01/19 6
456 02/01/20 6
789 10/01/19 5
789 02/01/20 4
999 10/01/19 0
999 02/01/20 1
I want to write a query that looks at each ID to see if the salespeople have newly started working internationally between October 1st and February 1st.
The result I am looking for is the following:
id effective_date number_of_int_customers
123 02/01/20 3
999 02/01/20 1
The result would return only the salespeople who originally had 0 international customers and now have at least 1.
I have seen similar posts here that use nested queries to pull records where the first date and last have different values. But I only want to pull records where the original value was 0. Is there a way to do this in one query in SQL?
In your case, a simple aggregation would do -- assuming that 0 is the earliest value:
select id, max(number_of_int_customers)
from t
where effective_date in ('2019-10-01', '2020-02-01')
group by id
having min(number_of_int_customers) = 0;
Obviously, this is not correct if the values can decrease to zero. But this having clause fixes that problem:
having min(case when number_of_int_customers = 0 then effective_date end) = min(effective_date)
An alternative is to use window functions, such asfirst_value():
select distinct id, last_noic
from (select t.*,
first_value(number_of_int_customers) over (partition by id order by effective_date) as first_noic,
first_value(number_of_int_customers) over (partition by id order by effective_date desc) as last_noic,
from t
where effective_date in ('2019-10-01', '2020-02-01')
) t
where first_noic = 0;
Hmmm, on second thought, I like lag() better:
select id, number_of_int_customers
from (select t.*,
lag(number_of_int_customers) over (partition by id order by effective_date) as prev_noic
from t
where effective_date in ('2019-10-01', '2020-02-01')
) t
where prev_noic = 0;

SQL retrieval from a table

Sales table
ItemID Date Code
-------------------------------------------------
12345 2019-02-17 1
12345 2019-02-17 2
12345 2019-02-17 3
-------------------------------------------------
12344 2019-02-18 1
12344 2019-02-18 3
12344 2019-02-18 2
-------------------------------------------------
12443 2019-02-19 1
12443 2019-02-19 2
12443 2019-02-19 3
I want to retrieve those item ids from Sales table where Code 2 is updated after code 3. So my query should return 12344 as result... I cannot think of any functions to achieve this as I have limited knowledge on SQL. Can someone please provide me some suggestion on achieving this? Thanks
Here is the query that you are looking for
select sales1.itemId from sales sales1
inner join sales sales2
on sales1.itemId = sales2.itemId
where sales1.date > sales2.date and sales1.code = 2 and sales2.code = 3;
Check if it helps you.
You can try using lag()
select itemid from
(
select sales1.itemId,sales1.code,
sales1.code-lag(code) over(partition by itemid order by dates) as diff
from sales sales1
where code in (2,3)
)A where diff>=1
You can include a column
update_seq_no
Each time you update you can increase the update_seq_no and then you can achieve your requirement you want using the below sql.
alter table sales add (update_seq_no number);
create SEQUENCE IFSQ_UPDATE_SEQNO MINVALUE 1000 MAXVALUE 9999999999999999 INCREMENT BY 1 START WITH 1 NOCYCLE ;
-- for existing rows (just a work around)
BEGIN
for rec in (select rowid row_id,itemid,date from sales where update_Seq_no is null order by 2,3 asc)
loop
update sales set update_Seq_no=IFSQ_UPDATE_SEQNO.NEXTVAL
where rowid=rec.row_id;
commit;
end loop;
end;
--for new rows
insert into sales values(p_itemid,p_date,3,IFSQ_UPDATE_SEQNO.NEXTVAL);
insert into sales values(p_itemid,p_date,2,IFSQ_UPDATE_SEQNO.NEXTVAL);
with sales1 as (select row_num1,a.* from (select rownum row_num1,b.*
from sales b order by update_seq_no asc)a)
select a.*
from sales1 a,sales1 b
where a.row_num1-1=b.row_num1
and a.code=2 and b.code=3
and a.itemid=b.itemid
and a.Date=b.Date;
If there is only one row per itemid and code, then you can simply aggregate:
select itemid
from mytable
group by itemid
having max(case when code = 2 then date end) > max(case when code = 3 then date end);
With your sample data, however, no itemid will be returned, as all codes per itemid have the same date. None comes before the other.

SQL select specific group from table

I have a table named trades like this:
id trade_date trade_price trade_status seller_name
1 2015-01-02 150 open Alex
2 2015-03-04 500 close John
3 2015-04-02 850 close Otabek
4 2015-05-02 150 close Alex
5 2015-06-02 100 open Otabek
6 2015-07-02 200 open John
I want to sum up trade_price grouped by seller_name when last (by trade_date) trade_status was 'open'. That is:
sum_trade_price seller_name
700 John
950 Otabek
The rows where seller_name is Alex are skipped because the last trade_status was 'close'.
Although I can get desirable output result with the help of nested select
SELECT SUM(t1.trade_price), t1.seller_name
WHERE t1.seller_name NOT IN
(SELECT t2.seller_name FROM trades t2
WHERE t2.seller_name = t1.seller_name AND t2.trade_status = 'close'
ORDER BY t2.trade_date DESC LIMIT 1)
from trades t1
group by t1.seller_name
But it takes more than 1 minute to execute above query (I have approximately 100K rows).
Is there another way to handle it?
I am using PostgreSQL.
I would approach this with window functions:
SELECT SUM(t.trade_price), t.seller_name
FROM (SELECT t.*,
FIRST_VALUE(trade_status) OVER (PARTITION BY seller_name ORDER BY trade_date desc) as last_trade_status
FROM trades t
) t
WHERE last_trade_status <> 'close;
GROUP BY t.seller_name;
This should perform reasonably with an index on seller_name
select
sum(trade_price) as sum_trade_price,
seller_name
from
trades
inner join
(
select distinct on (seller_name) seller_name, trade_status
from trades
order by seller_name, trade_date desc
) s using (seller_name)
where s.trade_status = 'open'
group by seller_name

Get count of two different values but not same values

I have the following table format.
**ID Name Start Date End Date**
1 ABC 1/1/2015 12/31/2015
1 XYZ 4/1/2015 8/31//2015
1 DEF 1/1/2012 12/31/2012
2 ABC 1/23/2011 1/23/2012
2 ABC 1/31/2012 1/31/2013
3 DEF 2/12/2015 5/30/2015
3 XYZ 4/1/2015 6/01/2015
4 DEF 3/1/2015 12/31/2015
4 DEF 4/1/2015 6/30/2015
I need the count of ID's having Different Name which lies in date range of May 2015
Expected Results
ID COUNT
1 2
3 2
P.S - ID 4 also lies in the date range of MAY 2015, but the Name is same i.e DEF. So I need only ID's 1 and 3 but not 4.
Thank You in advance and appreciated for your efforts.
I imagine your sample data doesn't match your desired results, but I think this is what you're looking for using conditional aggregation:
select id, count(*)
from yourtable
group by id
having sum(case when '5/1/2015' between startdate and enddate then 1 else 0 end) > 1
and count(distinct name) = count(name)
SQL Fiddle Demo
The sum aggregation in the having clause is making sure there are multiple records in between that date. The count clause in the having clause is making sure there aren't any duplicates.
declare
#startdate datetime = '20150501',
#enddate datetime = '20150531'
select t.id, count(distinct t.name)
from mytable t
where t.startdate <= #enddate and t.enddate >= #startdate
group by t.id
having count(distinct t.name) > 1

SQL query to group by data but with order by clause

I have table booking in which I have data
GUEST_NO HOTEL_NO DATE_FROM DATE_TO ROOM_NO
1 1 2015-05-07 2015-05-08 103
1 1 2015-05-11 2015-05-12 104
1 1 2015-05-14 2015-05-15 103
1 1 2015-05-17 2015-05-20 101
2 2 2015-05-01 2015-05-02 204
2 2 2015-05-04 2015-05-05 203
2 2 2015-05-17 2015-05-22 202
What I want is to get the result as.
1 ) It should show output as Guest_no, Hotel_no, Room_no, and column with count as number of time previous three column combination repeated.
So OutPut should like
GUEST_NO HOTEL_NO ROOM_NO Count
1 1 103 2
1 1 104 1
1 1 101 1
2 2 204 1
etc. But I want result to in ordered way e.g.: The output should be order by bk.date_to desc
My query is as below its showing me count but if I use order by its not working
select bk.guest_no, bk.hotel_no, bk.room_no,
count(bk.guest_no+bk.hotel_no+bk.room_no) as noOfTimesRoomBooked
from booking bk
group by bk.guest_no, bk.hotel_no, bk.room_no, bk.date_to
order by bk.date_to desc
So with adding order by result is showing different , because as I added order by date_to column so i have to add this column is group by clause too which will end up in different result as below
GUEST_NO HOTEL_NO ROOM_NO Count
1 1 103 1
1 1 104 1
1 1 103 1
1 1 101 1
2 2 204 1
Which is not the output I want.
I want these four column but with order by desc of date_to column and count as no of repetition of first 3 columns
I think a good way to do this would be grouping by guest_no, hotel_no and room_no, and sorting by the maximum (i.e. most recent) booking date in each group.
SELECT
guest_no,
hotel_no,
room_no,
COUNT(1) AS BookingCount
FROM
booking
GROUP BY
guest_no,
hotel_no,
room_no
ORDER BY
MAX(date_to) DESC;
Maybe this is what you're looking for?
select
guest_no,
hotel_no,
room_no,
count(*) as Count
from
booking
group by
guest_no,
hotel_no,
room_no
order by
min(date_to) desc
Or maybe max() instead of min(). SQL Fiddle: http://sqlfiddle.com/#!6/e684c/3
You could try this.
select t.* from
(
select bk.guest_no, bk.hotel_no, bk.room_no, bk.date_to,
count(*) as noOfTimesBooked from booking bk
group by bk.guest_no, bk.hotel_no, bk.room_no, bk.date_to
) t
order by t.date_to
You will also have to select date_to and then group the result by it.
If you use 'group by' clause, SQL Server doesn't allow you to use 'order by'. So you can make a sub query and use 'order by' in the outer query.
SELECT * FROM
(select bk.guest_no,bk.hotel_no,bk.room_no
,count(bk.guest_no+bk.hotel_no+bk.room_no) as noOfTimesRoomBooked,
(SELECT MAX(date_to) FROM booking CK
WHERE CK.guest_no=BK.guest_no AND bk.hotel_no=CK.bk.hotel_no
bk.room_no=CK.ROOM_NO ) AS DATEBOOK
from booking bk
group by bk.guest_no,bk.hotel_no,bk.room_no,bk.date_to) A
ORDER BY DATEBOOK
IT MIGHT HELP YOU