I have a query that calculates points based on multiple criteria, and then orders the result set based on those points.
SELECT * FROM (
SELECT
dbo.afunctionthatcalculates(Something, Something) AS Points1
,dbo.anotherone(Something, Something) AS Points2
,dbo.anotherone(Something, Something) AS Points3
,[TotalPoints] = dbo.function(something) + dbo.function(something)
) AS MyData
ORDER BY MyData.TotalPoints
So my first stab at adding placement, rankings.. was this:
SELECT ROW_NUMBER() OVER(MyData.TotalPoints) AS Ranking, * FROM (
SELECT same as above
) AS MyData
ORDER BY MyData.TotalPoints
This adds the Rankings column, but doesn't work when the points are tied.
Rank | TotalPoints
--------------------
1 100
2 90
3 90
4 80
Should be:
Rank | TotalPoints
--------------------
1 100
2 90
2 90
3 80
Not really sure about how to resolve this.
Thank you for your help.
You should use the DENSE_RANK() function which takes the ties into account, as described here: http://msdn.microsoft.com/en-us/library/ms173825.aspx
DENSE_RANK() instead of ROW_NUMBER()
Related
Reaching out for help. I've seen plenty of answers on how to use DUPLICATE, but not quite how I need it. Let's say I have the result of query that looks like the following.
query result
Incident_No Open_Approval_Step Approval_ID
------------- -------------------- -------------------
1 3 Tech
1 4 Cust_Serv
2 1 Incident_Recorder
2 2 Estimation
2 3 Tech
3 4 Cust_Serv
3 5 Mgmt
3 6 Closure
And I need one row for each incident number with the smallest numbered approval step. So the result should look like this.
filtered query result
Incident_No Open_Approval_Step Approval_ID
------------- -------------------- -------------------
1 3 Tech
2 1 Incident_Recorder
3 4 Cust_Serv
Edit This is what I came up with in the end
SELECT DISTINCT
MIN(OPEN_APPROVAL_STEP) OVER(PARTITION BY INCIDENT_NO ORDER BY OPEN_APPROVAL_STEP ASC) AS CUR_APP_STEP,
INCIDENT_NO
FROM T
You can use row_number():
select *
from (
select
t.*,
row_number() over(partition by incident_no order by open_approval_step) rn
from mytable t
) t
where rn = 1
With just one extra column appart from the incident number and approval step, another option is aggregation and Oracle's keep syntax:
select
incident_no,
min(open_approval_step) open_approval_step,
min(approval_id) keep(dense_rank first order by open_approval_step) approval_id
from mytable
group by incident_no
If you have just three columns, you can easily use aggregation:
select incident_no, min(open_approval_step),
min(approval_id) keep (dense_rank first order by open_approval_step)
from t
group by incident_no;
I have a table like so:
id device group
-----------------
1 a 1000
2 a 1000
3 b 1001
4 b 1001
5 b 1001
6 b 1002
8 a 1003
9 a 1003
10 a 1003
11 a 1003
12 b 1004
13 b 1004
All id's and groups are sequential. What I would like is to select id and device based on groups and devices. Think of it as a pagination type selection. Getting the last group is a simple inner selection, but how do I select the second last group, or the third last group - etc.
I tried the row number function like this:
SELECT * FROM
( SELECT *, ROW_NUMBER() OVER (PARTITION BY device ORDER BY group DESC) rn FROM data) tmp
WHERE rn = 1;
.. but changing rn is giving me the previous id, not the previous group.
I would like to end up with a selection that could accomodate these results:
device = a, group = latest:
id device group
10 a 1003
11 a 1003
device = a, group = latest - 1:
id device group
1 a 1000
2 a 1000
Any one know how to accomplish this?
Edit:
Use case is a GPS enabled device in a car, sending data every 30 seconds. Imagine going on a drive today. First you go to the shops, then you go home. the first trip is you driving to the shop. The second trip is you driving back. I want to show those trips on a map, but it means I need to identify your last trip, and then the trip before it - ad infinitum, until you run out of trips.
You can try this approach:
`with x as (
select distinct page
from test_table),
y as (
select x.page
,row_number() over (order by page desc) as row_num
from x)
select test_table.* from test_table join y on y.page = test_table.page
where y.row_num =2`
I will try to explain what I have did here.
The first block(x) returns the distinct groups(pages in my case).
The second block(y) assigns row numbers to the groups in terms of their rank. In this case the ranking is in descending order of the pages.
Finally the third block, selects the desired value for the desired page. In case you want the pen-ultimate page , type rouw_num=2, if third from last use row_num =3 and likewise.
You can play around with the values [here]: http://sqlfiddle.com/#!15/190c06/26
Use dense_rank():
select d.*
from (select d.*, dense_rank() over (order by group_id desc) as seqnum
from data d
where device = 'a'
) d
where seqnum = 2;
Sorry if this is a basic question.
I have a table
store-ProdCode
13p I10x
13p I20x
13p I30x
14a K38z
17a K38y
my data set has nearly 100,000 records.
What I'm trying to do is, for every store find the top 10 prodCode.
I am unsure of how to do this but what I tried was:
select s_code as store, prod_code,count (prod_code)
from top10_secondary
where prod_code is not null
group by store,prod_code
order by count(prod_code) desc limit 10
this is giving me something completely different and i'm unsure on how I go about achieving my final result.
The expected output should be: for every store(s_code) display the top 10 prodcode (top 10 calculated by the count)
so:
store--prodcode--result
1a abc 5
1a abd 4
1a xxx 7
--- this will be done until top 10 prodcodes for 1a are done--
2a dgf 1
2a ldk 6
--process completes until end of data is reached and top 10 prodcodes are displayed for each store
All help is appreciated. What is the best way to do this?
Thanks
One method uses row_number(), something like this:
select s.*
from (select s_code as store, prod_code, count(prod_code),
row_number() over (partition by s_code order by count(prod_code) desc) as seqnum
from top10_secondary
where prod_code is not null
group by s_code, prod_code
) s
where seqnum <= 10;
You can use window functions directly in an aggregation query. The subquery is needed only to reference the sequence number for filtering.
I looked at other examples, but I don't know enough about SQL to adapt it to my needs. I have a table that looks like this:
ID Month NAME COUNT First LAST TOTAL
------------------------------------------------------
1 JAN2013 fred 4
2 MAR2013 fred 5
3 APR2014 fred 1
4 JAN2013 Tom 6
5 MAR2014 Tom 1
6 APR2014 Tom 1
This could be in separate queries, but I need 'First' to equal the first month that a particular name is used, so every row with fred would have JAN2013 in the first field for example. I need the 'Last" column to equal the month of the last record of each name, and finally I need the 'total' column to be the sum of all the counts for each name, so in each row that had fred the total would be 10 in this sample data. This is over my head. Can one of you assist?
This is crude but should do the trick. I renamed your fields a bit because you are using a bunch of "RESERVED" sql words and that is bad form.
;WITH cte as
(
Select
[NAME]
,[nmCOUNT]
,ROW_NUMBER() over (partition by NAME order by txtMONTH ASC) as 'FirstMonth'
,ROW_NUMBER() over (partition by NAME order by txtMONTH DESC) as 'LastMonth'
,SUM([nmCOUNT]) as 'TotNameCount'
From Table
Group by NAME, [nmCOUNT]
)
,cteFirst as
(
Select
NAME
,[nmCOUNT]
,[TotNameCount]
,[txtMONTH] as 'ansFirst'
From cte
Where FirstMonth = 1
)
,cteLast as
(
Select
NAME
,[txtMONTH] as 'ansLast'
From cte
Where LastMonth = 1
Select c.NAME, c.nmCount, c.ansFirst, l.ansLast, c.TotNameCount
From cteFirst c
LEFT JOIN cteLast l on c.NAME = l.NAME
I am looking for the AVERAGE (overall) of the MINIMUM number (grouped by person).
My table looks like this:
Rank Name
1 Amy
2 Amy
3 Amy
2 Bart
1 Charlie
2 David
5 David
1 Ed
2 Frank
4 Frank
5 Frank
I want to know the AVERAGE of the lowest scores. For these people, the lowest scores are:
Rank Name
1 Amy
2 Bart
1 Charlie
2 David
1 Ed
2 Frank
Giving me a final answer of 1.5 - because three people have a MIN(Rank) of 1 and the other three have a MIN(Rank) of 2. That's what I'm looking for - a single number.
My real data has a couple hundred rows, so it's not terribly big. But I can't figure out how to do this in a single, simple statement. Thank you for any help.
Try this:
;WITH MinScores
AS
(
SELECT
"Rank",
Name,
ROW_NUMBER() OVER(PARTITION BY Name ORDER BY "Rank") row_num
FROM Table1
)
SELECT
CAST(SUM("Rank") AS DECIMAL(10, 2)) /
COUNT("Rank")
FROM MinScores
WHERE row_num = 1;
SQL Fiddle Demo
Selecting the set of minimum values is straightforward. The cast() is necessary to avoid integer division later. You could also avoid integer division by casting to float instead of decimal. (But you should be aware that floats are "useful approximations".)
select name, cast(min(rank) as decimal) as min_rank
from Table1
group by name
Now you can use the minimums as a common table expression, and select from it.
with minimums as (
select name, cast(min(rank) as decimal) as min_rank
from Table1
group by name
)
select avg(min_rank) avg_min_rank
from minimums
If you happen to need to do the same thing on a platform that doesn't support common table expressions, you can a) create a view of minimums, and select from that view, or b) use the minimums as a derived table.
You might try using a derived table to get the minimums, then get the average minimum in the outer query, as in:
-- Get the avg min rank as a decimal
select avg(MinRank * 1.0) as AvgRank
from (
-- Get everyone's min rank
select min([Rank]) as MinRank
from MyTable
group by Name
) as a
I think the easiest one will be
for max
select name , max_rank = max(rank)
from table
group by name;
for average
select name , avg_rank = avg(rank)
from table
cgroup by name;