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
Related
I have a table of customer transactions where each item purchased by a customer is stored as one row. So, for a single transaction there can be multiple rows in the table. I have another col called visit_date.
There is a category column called cal_month_nbr which ranges from 1 to 12 based on which month transaction occurred.
The data looks like below
Id visit_date Cal_month_nbr
---- ------ ------
1 01/01/2020 1
1 01/02/2020 1
1 01/01/2020 1
2 02/01/2020 2
1 02/01/2020 2
1 03/01/2020 3
3 03/01/2020 3
first
I want to know how many times customer visits per month using their visit_date
i.e i want below output
id cal_month_nbr visit_per_month
--- --------- ----
1 1 2
1 2 1
1 3 1
2 2 1
3 3 1
and what is the avg frequency of visit per ids
ie.
id Avg_freq_per_month
---- -------------
1 1.33
2 1
3 1
I tried with below query but it counts each item as one transaction
select avg(count_e) as num_visits_per_month,individual_id
from
(
select r.individual_id, cal_month_nbr, count(*) as count_e
from
ww_customer_dl_secure.cust_scan
GROUP by
r.individual_id, cal_month_nbr
order by count_e desc
) as t
group by individual_id
I would appreciate any help, guidance or suggestions
You can divide the total visits by the number of months:
select individual_id,
count(*) / count(distinct cal_month_nbr)
from ww_customer_dl_secure.cust_scan c
group by individual_id;
If you want the average number of days per month, then:
select individual_id,
count(distinct visit_date) / count(distinct cal_month_nbr)
from ww_customer_dl_secure.cust_scan c
group by individual_id;
Actually, Hive may not be efficient at calculating count(distinct), so multiple levels of aggregation might be faster:
select individual_id, avg(num_visit_days)
from (select individual_id, cal_month_nbr, count(*) as num_visit_days
from (select distinct individual_id, visit_date, cal_month_nbr
from ww_customer_dl_secure.cust_scan c
) iv
group by individual_id, cal_month_nbr
) ic
group by individual_id;
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
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
I need to count a value (M_Id) at each change of a date (RS_Date) and create a column grouped by the RS_Date that has an active total from that date.
So the table is:
Ep_Id Oa_Id M_Id M_StartDate RS_Date
--------------------------------------------
1 2001 5 1/1/2014 1/1/2014
1 2001 9 1/1/2014 1/1/2014
1 2001 3 1/1/2014 1/1/2014
1 2001 11 1/1/2014 1/1/2014
1 2001 2 1/1/2014 1/1/2014
1 2067 7 1/1/2014 1/5/2014
1 2067 1 1/1/2014 1/5/2014
1 3099 12 1/1/2014 3/2/2014
1 3099 14 2/14/2014 3/2/2014
1 3099 4 2/14/2014 3/2/2014
So my goal is like
RS_Date Active
-----------------
1/1/2014 5
1/5/2014 7
3/2/2014 10
If the M_startDate = RS_Date I need to count the M_id and then for
each RS_Date that is not equal to the start date I need to count the M_Id and then add that to the M_StartDate count and then count the next RS_Date and add that to the last active count.
I can get the basic counts with something like
(Case when M_StartDate <= RS_Date
then [m_Id] end) as Test.
But I am stuck as how to get to the result I want.
Any help would be greatly appreciated.
Brian
-added in response to comments
I am using Server Ver 10
If using SQL SERVER 2012+ you can use ROWS with your the analytic/window functions:
;with cte AS (SELECT RS_Date
,COUNT(DISTINCT M_ID) AS CT
FROM Table1
GROUP BY RS_Date
)
SELECT *,SUM(CT) OVER(ORDER BY RS_Date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Run_CT
FROM cte
Demo: SQL Fiddle
If stuck using something prior to 2012 you can use:
;with cte AS (SELECT RS_Date
,COUNT(DISTINCT M_ID) AS CT
FROM Table1
GROUP BY RS_Date
)
SELECT a.RS_Date
,SUM(b.CT)
FROM cte a
LEFT JOIN cte b
ON a.RS_DAte >= b.RS_Date
GROUP BY a.RS_Date
Demo: SQL Fiddle
You need a cumulative sum, easy in SQL Server 2012 using Windowed Aggregate Functions. Based on your description this will return the expected result
SELECT p_id, RS_Date,
SUM(COUNT(*))
OVER (PARTITION BY p_id
ORDER BY RS_Date
ROWS UNBOUNDED PRECEDING)
FROM tab
GROUP BY p_id, RS_Date
It looks like you want something like this:
SELECT
RS_Date,
SUM(c) OVER (PARTITION BY M_StartDate ORDER BY RS_Date ROWS UNBOUNDED PRECEEDING)
FROM
(
SELECT M_StartDate, RS_Date, COUNT(DISTINCT M_Id) AS c
FROM my_table
GROUP BY M_StartDate, RS_Date
) counts
The inline view computes the counts of distinct M_Id values within each (M_StartDate, RS_Date) group (distinctness enforced only within the group), and the outer query uses the analytic version of SUM() to add up the counts within each M_StartDate.
Note that this particular query will not exactly reproduce your example results. It will instead produce:
RS_Date Active
-----------------
1/1/2014 5
1/5/2014 7
3/2/2014 8
3/2/2014 2
This is on account of some rows in your example data with RS_Date 3/2/2014 having a later M_StartDate than others. If this is not what you want then you need to clarify the question, which currently seems a bit inconsistent.
Unfortunately, analytic functions are not available until SQL Server 2012. In SQL Server 2010, the job is messier. It could be done like this:
WITH gc AS (
SELECT M_StartDate, RS_Date, COUNT(DISTINCT M_Id) AS c
FROM my_table
GROUP BY M_StartDate, RS_Date
)
SELECT
RS_Date,
(
SELECT SUM(c)
FROM gc2
WHERE gc2.M_StartDate = gc.M_StartDate AND gc2.RS_Date <= gc.RS_Date
) AS Active
FROM gc
If you are using SQL 2012 or newer you can use LAG to produce a running total.
https://msdn.microsoft.com/en-us/library/hh231256(v=sql.110).aspx
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.