customer price date
123 100 1-Jan-15
321 200 2-Jan-15
123 10 3-Jan-15
123 50 4-Jan-15
321 150 5-Jan-15
123 100 6-Jan-15
123 300 7-Jan-15
321 500 8-Jan-15
123 700 9-Jan-15
I would like to take see SUM(Price) between dates 3-Jan and 7-Jan group by customer
Something like:
select customer,sum(price) from table group by customer
having date between 3-Jan-2015 and 8-Jan-2015
This prompts me to have date either in the select or group by clause. When I include it, it groups by Date also.
Desired output is:
123 460
The HAVING clause filters results after the grouping and aggregation, so in order to do this any fields you want in the HAVING clause must be in the GROUP BY or SELECT lists. The WHERE clause filters the rows before doing the grouping and aggregation.
To get the results you want, simply move the date between... condition into a WHERE clause:
SELECT customer,SUM(price)
FROM table
WHERE date BETWEEN 3-Jan-2015 AND 8-Jan-2015
GROUP BY customer
Related
Table:1
Date Customer Amount
12-Dec ABC 200
15-Dec ABC 300
Output:
I need to group the data by Customer and need to take the latest date for that unique record.
Date Customer Amount
15-Dec ABC 500
You seems to want aggregation :
select max(to_date(date, 'DD-MON-YYYY')), cust, sum(amount)
from table t
group by cust;
I have the following View in PostgreSQL:
idshipment idorder quantity_order date quantity_in_shipment percent_sent
50 1 1020 1.1.16 432 42
51 1 1020 17.1.16 299 71
51 1 1020 20.1.16 144 85
51 1 1020 45.1.16 145 100
52 2 1 3.1.17 5 100
This View shows shipments per order.
For example:
idorder=1 was sent by 4 shipments:
quantity in first shipment is 432 which means 42% of order was sent
quantity in second shipment is 299 which means 71% of order was sent
quantity in third shipment is 144 which means 85% of order was sent
quantity in forth shipment is 145 which means 100% of order was sent
I need a query which will show me the first date where each order was sent above 75%. meaning each order shows only one row.
For the above data I should see:
idorder date
1 20.1.16 (cause its 85% first time above 75%)
2 3.1.17 (cause its 100% first time above 75%)
How can i do that?
You can use distinct on:
select distinct on (t.idshipment) t.*
from t
where t.percent_sent >= 75
order by t.idshipment, t.percent_sent asc;
Try something like this:
SELECT iorder, MIN("date") AS "date"
FROM your_view
WHERE percent_sent >= 75
GROUP BY iorder
use group by to get only one record per idorder and MIN() to aggregate date by selecting the earliest date
I created a table call shipment that has data like you provided:
and execute this query
SELECT s.idorder, MIN(s.date) as date
FROM shipment s
WHERE percent_sent >= 75
GROUP BY s.idorder
result:
idorder date
----------- ----------
1 2016-01-20
2 2017-03-01
Hello I am very new to SQL programming, started last week. I am trying to select a userID and Maxdate from a table that looks like this for example:
Key USERID Date
1 111 12/1/2014
2 202 4/1/2014
3 111 3/8/2014
4 111 2/5/2014
5 202 2/10/2014
I want to make a query that would end up with the following results:
USERID DATE
111 12/1/2014
202 4/1/2014
Simply use GROUP BY clause with aggregate function MAX to achieve this:
Try this:
SELECT USERID, MAX(Date) AS Date
FROM tableA
GROUP BY USERID
I've been trying to build an sql query that finds from (table) the most recent date for selected id's that fulfill the condition where 'type' is in hierarchy 'vegetables'. My goal is to be able to get the whole row once max(date) and hierarchy conditions are met for each id.
Example values
ID DATE PREFERENCE AGE
123 1/3/2013 carrot 14
123 1/3/2013 apple 12
123 1/2/2013 carrot 14
124 1/5/2013 carrot 13
124 1/3/2013 apple 13
124 1/2/2013 carrot 14
125 1/4/2013 carrot 13
125 1/3/2013 apple 14
125 1/2/2013 carrot 13
I tried the following
SELECT *
FROM table
WHERE date in
(SELECT max(date) FROM (table) WHERE id in (123,124,125))
and preference in
(SELECT preference FROM (hierarchy_table)
WHERE hierarchy = vegetables))
and id in (123,24,125)
but it doesn't give me the most recent date for each id that meets the hierarchy conditions. (ex. in this scenario I would only get id 124)
Thank you in advance!
SELECT max(date) FROM (table) WHERE id in (123,124,125)
is giving you the max date from all dates, you need to group them.
Try replacing with:
SELECT max(date) FROM (table) GROUP BY id
This way you will get the max date for each id
I figured this out. Please see the query below as an example:
SELECT * FROM (table) t
WHERE t.date in
(SELECT max(date) FROM table sub_t where t.ID = sub_t.ID and (date !> (currentdate))
and preference in
(SELECT preference FROM (hierarchy_table) WHERE hierarchy ='vegetables')
and ID in ('124')
Change:
max(date)
To:
-- if your date data is in mm/dd/yyyy
max( str_to_date( date, '%m/%d/%Y' ) )
OR
-- if your date data is in dd/mm/yyyy
max( str_to_date( date, '%d/%m/%Y' ) )
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.