i'm a beginner on sql, and i hope my post will be clear.
So i have 1 table (see pictures) with my SalesMen linked to a contract.
And i want to be able to rank each SalesMen on a contract.
So if today on a contract 001 i have Martin as the only SalesMan he should have '1' as rank.
But if one day Martin is fired and the new guy is Frank. Then Martin will be rank 2 and Frank '1'.
example:
idContract idSalesMan DateLoad
001 Jean 1900-01-01
001 Jean 1900-02-02
001 Jean 1900-03-03
002 Martin 1900-01-01
002 Martin 1900-02-02
002 Frank 1900-03-03
The result i want :
idContract idSalesMan DateLoad RankedSalesMan
001 Jean 1900-01-01 1
001 Jean 1900-02-02 1
001 Jean 1900-03-03 1
002 Martin 1900-01-01 2
002 Martin 1900-02-02 2
002 Frank 1900-03-03 1
What i did:
1st try :
select idContract,IdSalesMan,
Row_number() over (partition by IDContract , IDsALESmAN order by IDsALESmAN )
But it gave me a regular 1,2,3,4
Then i tried
select
x.*
,case when x.PrecedingSalesMan = x.SalesMan then
Row_number() over (partition by IDContract , IDsALESmAN order by IDsALESmAN )
from(
select idContract,IdSalesMan,
,LAG(CodeAgent,1) over (order by NoContract, idCustomers, kCode_Agent) as PrecedingSalesMan) x
But it doesn't work neither :(
I don,t know what to do
it seems you just need dense_rank()
select *,dense_rank()over(partition by idContract order by idSalesMan) as rn
from table_name
You need DENSE_RANK() :
SELECT t.*,
DENSE_RANK() OVER (PARTITION BY t.idContract ORDER BY t.idSalesMan) AS RankedSalesMan
FROM table t;
You need only idContract in partition clause.
Related
Table A has ID and date and name. Each time the record is changed the first 11 digits of the Id remain the same but the final digit would increase by 1. For example
123456789110 01-01-2020 John smith
119876543210 01-01-2020 Peter Griffin
119876543211 05-01-2020 Peter Griffin
How could I write a statement that shows The iD associated with John smith as well as the most recent Id of Peter Griffin? Thanks
Yet another option is using WITH TIES
Select top 1 with ties *
From YourTable
Order by row_number() over (partition by left(id,11) order by date desc)
Why not just use max()?
select name, max(id)
from t
group by name;
[Data]
[Emp] [Emp_group] [Date_purchase]
1 001 12-jan-2016
1 001 13-jan-2016
1 001 19-jan-2016
1 003 14-jan-2016
2 004 21-feb-2016
2 004 22-feb-2016
2 004 23-feb-2016
3 005 01-apr-2016
Need SQL to find consecutive purchase dates. Emp (1) of emp group (001) has purchased consecutively on 12 and 13 of January. Emp and Emp group partition must be considered.
Just use lag()/lead():
select t.*
from (select t.*,
lag(date_purchase) over (partition by emp, emp_group order by date_purchase) as prev_dp,
lead(date_purchase) over (partition by emp, emp_group order by date_purchase) as next_dp
from t
) t
where date_purchase in (prev_dp, next_dp);
player team start_date end_date points
John Jacob SportsBallers 2015-01-01 2015-03-31 100
John Jacob SportsKings 2015-04-01 2015-12-01 115
Joe Smith PointScorers 2014-01-01 2016-12-31 125
Bill Johnson SportsKings 2015-01-01 2015-06-31 175
Bill Johnson AllStarTeam 2015-07-01 2016-12-31 200
The above table has many more rows. I was asked the below questions in an interview.
1.)For each player, which team were they play for on 2015-01-01?
I could not answer this one.
2.)For each player, how can we get the team for whom they scored the most points?
select team from Players
where points in (select max(points) from players group by player).
Please, solutions for both.
1
select *
from PlayerTeams
where startdate <='2015-01-01' and enddate >= '2015-01-01'
2
Select player, team, points
from(
Select *, row_number() over (partition by player order by points desc) as rank
From PlayerTeams) as player
where rank = 1
For #1:
Select Player
,Team
From table
Where '2015-01-01' between start_date and end_date
For #2:
select t.Player
,t.Team
from table t
inner join (select Player
,Max(points)
from table
group by Player) m
on t.Player = m.Player
and t.points = m.points
I'm trying to get the SUM(Values) for each Acct, but my issue is trying to get at least one entire row for a DISTINCT Acct with the SUM(Values).
I have some sample data for example:
Acct Values Name Street
123456789 100.20 John 66 Main Street
123456789 200.80 John 22 Main Avenue
222222222 50.25 Jane 1 Blvd
333333333 25.00 Joe 55 Test Ave
333333333 50.00 Joe 8 Douglas Road
555555555 75.00 Tim 12 Clark Ave
666666666 500.00 Tim 12 Clark Street
666666666 500.00 Tim 3 Main Rd.
My query consisted of:
SELECT DISTINCT Acct, SUM(Value) AS [TOTAL]
FROM TABLE_NAME
GROUP BY Acct
The above query gets me close to what I need, but I need the entire row.
Example below of what I am looking for:
Acct Total Name Addr1
123456789 301.00 John 66 Main Street
222222222 50.25 Jane 1 Blvd
333333333 75.00 Joe 55 Test Ave
555555555 75.00 Tim 12 Clark Ave
666666666 1000.00 Tim 12 Clark Street
Thanks.
If it does not matter what address you return, then you can apply and aggregate to the other columns:
SELECT Acct,
SUM(Value) AS [TOTAL],
max(name) name,
max(Street) addr1
FROM TABLE_NAME
GROUP BY Acct;
See SQL Fiddle with Demo
You can do this using window functions such as row_number() in most databases:
select acct, total, name, addr1
from (select t.*, row_number() over (partition by acct order by acct) as seqnum,
sum(value) over (partition by acct) as Total
from table_name
) t
where seqnum = 1;
I would use Windowing Functions (the OVER clause) to solve this.
SELECT DISTINCT
Acct
,SUM([Values]) OVER (PARTITION BY Acct) AS 'Total'
,Name
,FIRST_VALUE(Street) OVER (PARTITION BY Acct ORDER BY Street DESC) AS 'Addr1'
FROM TABLE_NAME
;
The nice thing about Windowing Functions is that you do not add things to a grouping that you do not need in your functions (e.g. SUM), instead you can focus on describing what you are looking for.
In the SQL above, we are saying we want the SUM of Values grouped by (or PARTITION BY as it is called in the OVER clause) Acct. The FIRST_VALUE allows use to return the first value of the street address. The same did not have a DATETIME column so it is hard to say what the order should be for the first value. There is also a LAST_VALUE windowing function. Assuming you do have a DATETIME column you would want to ORDER BY that column value, if not you can just pick some value like I did with Street (MAX might also be a good option then too, but having some type of DATETIME value would be the best way to do it).
Check out this SQL Fiddle: http://sqlfiddle.com/#!6/a474c/8
Here is the BOL about SUM using the OVER clause: http://msdn.microsoft.com/en-us/library/ms187810.aspx
Here is more info on FIRST_VALUE: http://blog.sqlauthority.com/2011/11/09/sql-server-introduction-to-first-_value-and-last_value-analytic-functions-introduced-in-sql-server-2012/
Here is a blog post I've done on the Windowing Functions: http://comp-phil.blogspot.com/2013/03/higher-order-functions.html
I have:
Table1
ID date amt
-------------------
001 21/01/2012 1200
001 25/02/2012 1400
001 24/03/2012 1500
001 21/04/2012 1000
002 21/03/2012 1200
002 01/01/2012 0500
002 08/09/2012 1000
.....
I want to select the first two rows from each group of ID ordered by date DESC from Table1.
Query looks like this:
SELECT TOP 2 DATE, ID, AMT FROM TABLE1 GROUP BY ID, AMT --(NOT WORKING)
Expected output:
ID date amt
-------------------
001 21/01/2012 1200
001 25/02/2012 1400
002 21/03/2012 1200
002 01/01/2012 0500
.....
you can take advantage of using Common table Expression and Window Function
WITH recordList
AS
(
SELECT ID, DATE, Amt,
DENSE_RANK() OVER (PARTITION BY ID ORDER BY DATE ASC) rn
FROM tableName
)
SELECT ID, DATE, Amt
FROM recordList
WHERE rn <= 2
SQLFiddle Demo
based on your desired result above, you are ordering the date by ASCENDING.
Ok, You can either use DENSER_RANK() or ROW_NUMBER() but in my answer, I've used DENSE_RANK() because I'm thinking of the duplicates. Anyway, it's the choice of the OP to use ROW_NUMBER() instead of DENSE_RANK().
TSQL Ranking Functions