I'm trying to do a simple Select query by getting the country based on the MAX Last update from the other table.
Order#
1
2
3
4
The other table contains the country and the last update:
Order# Cntry Last Update
1 12/21/2019 9:19 PM
1 US 1/10/2020 1:07 AM
2 JP 7/29/2020 12:15 PM
3 CA 4/12/1992 2:04 PM
3 GB 11/6/2001 9:26 AM
3 DK 2/1/2005 3:04 AM
4 CN 8/20/2013 12:04 AM
4 10/1/2015 4:04 PM
My desired result:
Order# Country
1 US
2 JP
3 DK
4
Not sure the right solution for this. So far i'm stuck with this:
SELECT Main.[Order#], tempTable.Cntry
FROM Main
LEFT JOIN (
SELECT [Order#], Cntry, Max([Last Update]) as LatestDate FROM Country
GROUP BY [Order#], Cntry
) as tempTable ON Main.[Order#] = tempTable.[Order#];
Thanks in advance!
If needs only number of order and country,maybe don't need two tables:
SELECT distinct order, country
FROM
(
SELECT order, LAST_VALUE (country) OVER (PARTITION by [order] order by last_update) country FROM Country
) X
In SQL Server, you can use a correlated subquery:
update main
set country = (select top (1) s.country
from secondtable s
where s.order# = main.order#
order by s.lastupdate desc
);
EDIT:
A select would look quite simimilar:
select m.*,
(select top (1) country
from secondtable s
where s.order# = main.order#
order by s.lastupdate desc
)
from main m
I don't have time to try it with sample data, but is that what you are looking for?
select order orde, cntry
from table
where last_update =
(select max(last_update) from table where order = orde)
Related
I have two tables Patient and PatientStatusLog using SQL tables
1.Patient
sno patientid status createdby Reegion
1 481910 D 1222 India
2 476795 D 1222 India
2.PatientStatusLog
sno patientid status comments createdby CreatedDate(dd/mm/yyyy)
1 481910 A mycommnet 1222 01/01/2000
2 481910 A mycommnet 1222 02/01/2000
3 481910 B mycommnet 1222 01/01/2000
4 481910 C mycommnet 1222 01/01/2000
I need output like below who have status A and createddate should recent pass date from PatientStatusLog table using patientid of both tables
Region status CreatedDate
India A 02/01/2000
You can use window functions:
select . . . -- whatever columns you want
from Patient p join
(select psl.*,
rank() over (partition by patientid order by createddate desc) as seqnum
from PatientStatusLog psl
) psl
on p.patientid = psl.patientid
where psl.seqnum = 1 and psl.status = 'A';
Note that this uses rank() because the creation date appears to have duplicates.
Try this:
SELECT TOP 1
p.Region,
psl.Status,
psl.CreatedDate
FROM
Patient [p]
JOIN PatientStatusLog [psl] ON psl.patientid = p.patientid
AND psl.status = 'A'
ORDER BY
psl.CreatedDate DESC
For the subquery request:
SELECT
p.Region,
x.status,
x.CreatedDate
FROM
(
SELECT TOP 1
PatientId,
Status,
CreatedDate
FROM
PatientStatusLog
WHERE
status = 'A'
ORDER BY
CreatedDate DESC
) AS x
JOIN Patient [p] ON p.PatientId = x.PatientId
Note that this is definitely not the best way to handle this, both from a code readability and optimization perspective.
I have one database and time to time i change some part of query as per requirement.
i want to keep record of results of both before and after result of these queries in one table and want to show queries which generate difference.
For Example,
Consider following table
emp_id country salary
---------------------
1 usa 1000
2 uk 2500
3 uk 1200
4 usa 3500
5 usa 4000
6 uk 1100
Now, my before query is :
Before Query:
select count(emp_id) as count,country from table where salary>2000 group by country;
Before Result:
count country
2 usa
1 uk
After Query:
select count(emp_id) as count,country from table where salary<2000 group by country;
After Query Result:
count country
2 uk
1 usa
My Final Result or Table I want is:
column 1 | column 2 | column 3 | column 4 |
2 usa 2 uk
1 uk 1 usa
...... but if query results are same than it shouldn't show in this table.
Thanks in advance.
I believe that you can use the same approach as here.
select t1.*, t2.* -- if you need specific columns without rn than you have to list them here
from
(
select t.*, row_number() over (order by count) rn
from
(
-- query #1
select count(emp_id) as count,country from table where salary>2000 group by country;
) t
) t1
full join
(
select t.*, row_number() over (order by count) rn
from
(
-- query #2
select count(emp_id) as count,country from table where salary<2000 group by country;
) t
) t2 on t1.rn = t2.rn
In a firebird database with a table "Sales", I need to select the first sale of all customers. See below a sample that show the table and desired result of query.
---------------------------------------
SALES
---------------------------------------
ID CUSTOMERID DTHRSALE
1 25 01/04/16 09:32
2 30 02/04/16 11:22
3 25 05/04/16 08:10
4 31 07/03/16 10:22
5 22 01/02/16 12:30
6 22 10/01/16 08:45
Result: only first sale, based on sale date.
ID CUSTOMERID DTHRSALE
1 25 01/04/16 09:32
2 30 02/04/16 11:22
4 31 07/03/16 10:22
6 22 10/01/16 08:45
I've already tested following code "Select first row in each GROUP BY group?", but it did not work.
In Firebird 2.5 you can do this with the following query; this is a minor modification of the second part of the accepted answer of the question you linked to tailored to your schema and requirements:
select x.id,
x.customerid,
x.dthrsale
from sales x
join (select customerid,
min(dthrsale) as first_sale
from sales
group by customerid) p on p.customerid = x.customerid
and p.first_sale = x.dthrsale
order by x.id
The order by is not necessary, I just added it to make it give the order as shown in your question.
With Firebird 3 you can use the window function ROW_NUMBER which is also described in the linked answer. The linked answer incorrectly said the first solution would work on Firebird 2.1 and higher. I have now edited it.
Search for the sales with no earlier sales:
SELECT S1.*
FROM SALES S1
LEFT JOIN SALES S2 ON S2.CUSTOMERID = S1.CUSTOMERID AND S2.DTHRSALE < S1.DTHRSALE
WHERE S2.ID IS NULL
Define an index over (customerid, dthrsale) to make it fast.
in Firebird 3 , get first row foreach customer by min sales_date :
SELECT id, customer_id, total, sales_date
FROM (
SELECT id, customer_id, total, sales_date
, row_number() OVER(PARTITION BY customer_id ORDER BY sales_date ASC ) AS rn
FROM SALES
) sub
WHERE rn = 1;
İf you want to get other related columns, This is where your self-answer fails.
select customer_id , min(sales_date)
, id, total --what about other colums
from SALES
group by customer_id
So simple as:
select CUSTOMERID min(DTHRSALE) from SALES group by CUSTOMERID
I have a SQL Server 2008 database. In this database, I have a result set that looks like the following:
ID Name Department LastOrderDate
-- ---- ---------- -------------
1 Golf Balls Sports 01/01/2015
2 Compact Disc Electronics 02/01/2015
3 Tires Automotive 01/15/2015
4 T-Shirt Clothing 01/10/2015
5 DVD Electronics 01/07/2015
6 Tennis Balls Sports 01/09/2015
7 Sweatshirt Clothing 01/04/2015
...
For some reason, my users want to get the results ordered by department, then last order date. However, not by department name. Instead, the departments will be in a specific order. For example, they want to see the results ordered by Electronics, Automotive, Sports, then Clothing. To throw another kink in works, I cannot update the table schema.
Is there a way to do this with a SQL Query? If so, how? Currently, I'm stuck at
SELECT *
FROM
vOrders o
ORDER BY
o.LastOrderDate
Thank you!
You can use case expression ;
order by case when department = 'Electronics' then 1
when department = 'Automotive' then 2
when department = 'Sports' then 3
when department = 'Clothing' then 4
else 5 end
create a table for the departments that has the name (or better id) of the department and the display order. then join to that table and order by the display order column.
alternatively you can do a order by case:
ORDER BY CASE WHEN Department = 'Electronics' THEN 1
WHEN Department = 'Automotive' THEN 2
...
END
(that is not recommended for larger tables)
Here solution with CTE
with c (iOrder, dept)
as (
Select 1, 'Electronics'
Union
Select 2, 'Automotive'
Union
Select 3, 'Sports'
Union
Select 4, 'Clothing'
)
Select * from c
SELECT o.*
FROM
vOrders o join c
on c.dept = o.Department
ORDER BY
c.iOrder
These are my tables:
USER:
id_user name email last_access id_company
1 jhonatan abc#abc.com 2014-12-15 1
2 cesar cef#cef.com 2014-12-31 1
3 john 123#123.com 2015-01-09 2
4 steven 897#asdd.cpom 2015-01-02 2
5 greg sd#touch.com 2014-12-07 1
6 kyle fb#fb.com 2014-11-20 1
COMPANY:
id_company company
1 Facebook
2 Appslovers
I need to know, what are the users which has the MIN last_access per company (just one). It could be like this:
id_user name last_access company
6 kyle 2014-11-20 Facebook
4 steven 2015-01-02 Appslovers
Is it possible ?
Use window function
SELECT id_user,
NAME,
last_access,
company
FROM (SELECT id_user,
NAME,
last_access,
company,
Row_number()OVER(partition BY company ORDER BY last_access) rn
FROM users u
JOIN company c
ON u.id_company = c.id_company) a
WHERE rn = 1
or join both the tables find the min last_access date per company then join the result back to the users table to get the result
SELECT id_user,
NAME,
a.last_access,
a.company
FROM users u
JOIN(SELECT u.id_company,
Min(last_access) last_access,
company
FROM users u
JOIN company c
ON u.id_company = c.id_company
GROUP BY u.id_company,
company) a
ON a.id_company = u.id_company
AND u.last_access = a.last_access
This can be done in many ways, for example by using a window function like row_number to partition the data and then selecting the top rows from each group like this:
;with cte (id_user, name, last_access, company, seq) as (
select
id_user,
name,
last_access,
company,
seq = row_number() over (partition by u.id_company order by last_access)
from [user] u
inner join [company] c on u.id_company = c.id_company
)
select id_user, name, last_access, company
from cte where seq = 1