I've 2 CTE. The first counts the number of users. The second does the same. It is necessary to calculate the percentage ratio between them.
Prompt how it can be done?
WITH count AS ( SELECT user_id
from users u
where u.status = 'Over'),
users as (Select user_id
from users u
where u.status LIKE 'LR'
and user_id IN (select * from count))
Select COUNT(*) From users
WITH count AS ( SELECT user_id
from users u
where u.description = 'Track'),
users as (Select user_id
from from users u
where u.status NOT LIKE 'LR'
and user_id IN (select * from count))
Select COUNT(*) From users
You can do it without CTE, just simple select with 2 counts:
SELECT count( CASE WHEN description = 'Over' AND status LIKE 'LR' THEN 1 END )
/
count( CASE WHEN description = 'Track' AND status NOT LIKE 'LR' THEN 1 END )
As Ratio
FROM users
With minimal changes, you can just do one bigger CTE:
WITH count_1 AS
(
SELECT user_id
FROM users u
WHERE u.status = 'Over'
),
users_1 AS
(
SELECT user_id
FROM users u
WHERE u.status LIKE 'LR'
AND user_id IN (SELECT user_id FROM count_1)
),
count_2 AS
(
SELECT user_id
FROM users u
WHERE u.description = 'Track'
),
users_2 AS
(
SELECT user_id
FROM users u
WHERE u.status NOT LIKE 'LR'
AND user_id IN (select user_id from count_2)
)
SELECT
CAST( (SELECT count(*) FROM users_1) AS FLOAT) /
(SELECT count(*) FROM users_2) AS ratio
NOTE 1: The query doesn't make any sense, so I guess there is some misspelling, or some columns messed up. The count_1 will choose users with a status = 'Over', the users_1 will choose the ones which have also a status = 'LR' (the result is already ZERO).
NOTE 2: You wouldn't make queries this way... The following query means exactly the same, and is much simpler (and faster):
WITH
count_1 AS
(
SELECT count(user_id) AS c
FROM users u
WHERE u.description = 'Over'
AND u.status = 'LR'
),
count_2 AS
(
SELECT count(user_id) AS c
FROM users u
WHERE u.description = 'Track'
AND u.status <> 'LR'
)
SELECT
(count_1.c + 0.0) / count_2.c AS ratio
FROM
count_1, count_2 ;
Yet another version:
SELECT count(*) FILTER (WHERE description = 'Over' AND status LIKE 'LR')
/
count(*) FILTER (WHERE description = 'Track' AND status NOT LIKE 'LR')
As Ratio
FROM users
Related
I'm using SQL Server 2000 (80). So, it's not possible to use the LAG function.
I have a code a data set with four columns:
Purchase_Date
Facility_no
Seller_id
Sale_id
I need to identify missing Sale_ids. So every sale_id is a 100% sequential, so the should not be any gaps in order.
This code works for a specific date and store if specified. But i need to work on entire data set looping looping through every facility_id and every seller_id for ever purchase_date
declare #MAXCOUNT int
set #MAXCOUNT =
(
select MAX(Sale_Id)
from #table
where
Facility_no in (124) and
Purchase_date = '2/7/2020'
and Seller_id = 1
)
;WITH TRX_COUNT AS
(
SELECT 1 AS Number
union all
select Number + 1 from TRX_COUNT
where Number < #MAXCOUNT
)
select * from TRX_COUNT
where
Number NOT IN
(
select Sale_Id
from #table
where
Facility_no in (124)
and Purchase_Date = '2/7/2020'
and seller_id = 1
)
order by Number
OPTION (maxrecursion 0)
My Dataset
This column:
case when
Sale_Id=0 or 1=Sale_Id-LAG(Sale_Id) over (partition by Facility_no, Purchase_Date, Seller_id)
then 'OK' else 'Previous Missing' end
will tell you which Seller_Ids have some sale missing. If you want to go a step further and have exactly your desired output, then filter out and distinct the 'Previous Missing' ones, and join with a tally table on not exists.
Edit: OP mentions in comments they can't use LAG(). My suggestion, then, would be:
Make a temp table that that has the max(sale_id) group by facility/seller_id
Then you can get your missing results by this pseudocode query:
Select ...
from temptable t
inner join tally N on t.maxsale <=N.num
where not exists( select ... from sourcetable s where s.facility=t.facility and s.seller=t.seller and s.sale=N.num)
> because the only way to "construct" nonexisting combinations is to construct them all and just remove the existing ones.
This one worked out
; WITH cte_Rn AS (
SELECT *, ROW_NUMBER() OVER(PARTITION BY Facility_no, Purchase_Date, Seller_id ORDER BY Purchase_Date) AS [Rn_Num]
FROM (
SELECT
Facility_no,
Purchase_Date,
Seller_id,
Sale_id
FROM MyTable WITH (NOLOCK)
) a
)
, cte_Rn_0 as (
SELECT
Facility_no,
Purchase_Date,
Seller_id,
Sale_id,
-- [Rn_Num] AS 'Skipped Sale'
-- , case when Sale_id = 0 Then [Rn_Num] - 1 Else [Rn_Num] End AS 'Skipped Sale for 0'
, [Rn_Num] - 1 AS 'Skipped Sale for 0'
FROM cte_Rn a
)
SELECT
Facility_no,
Purchase_Date,
Seller_id,
Sale_id,
-- [Skipped Sale],
[Skipped Sale for 0]
FROM cte_Rn_0 a
WHERE NOT EXISTS
(
select * from cte_Rn_0 b
where b.Sale_id = a.[Skipped Sale for 0]
and a.Facility_no = b.Facility_no
and a.Purchase_Date = b.Purchase_Date
and a.Seller_id = b.Seller_id
)
--ORDER BY Purchase_Date ASC
Name Value AnotherColumn
-----------
Pump 1 8000.0 Something1
Pump 1 10000.0 Something2
Pump 1 10000.0 Something3
Pump 2 3043 Something4
Pump 2 4594 Something5
Pump 2 6165 Something6
My table looks something like this. I would like to know how to select max value for each pump.
select a.name, value from out_pumptable as a,
(select name, max(value) as value from out_pumptable where group by posnumber)g where and g.value = value
this code does the job, but i get two entries of Pump 1 since it has two entries with same value.
select name, max(value)
from out_pumptable
group by name
select name, value
from( select name, value, ROW_NUMBER() OVER(PARTITION BY name ORDER BY value desc) as rn
from out_pumptable ) as a
where rn = 1
SELECT
b.name,
MAX(b.value) as MaxValue,
MAX(b.Anothercolumn) as AnotherColumn
FROM out_pumptabl
INNER JOIN (SELECT
name,
MAX(value) as MaxValue
FROM out_pumptabl
GROUP BY Name) a ON
a.name = b.name AND a.maxValue = b.value
GROUP BY b.Name
Note this would be far easier if you had a primary key. Here is an Example
SELECT * FROM out_pumptabl c
WHERE PK in
(SELECT
MAX(PK) as MaxPK
FROM out_pumptabl b
INNER JOIN (SELECT
name,
MAX(value) as MaxValue
FROM out_pumptabl
GROUP BY Name) a ON
a.name = b.name AND a.maxValue = b.value)
select Name, Value, AnotherColumn
from out_pumptable
where Value =
(
select Max(Value)
from out_pumptable as f where f.Name=out_pumptable.Name
)
group by Name, Value, AnotherColumn
Try like this, It works.
select * from (select * from table order by value desc limit 999999999) v group by v.name
Using analytic function is the easy way to find max value of every group.
Documentation : https://learn.microsoft.com/en-us/sql/t-sql/functions/row-number-transact-sql?view=sql-server-ver15
Select name,
value,
AnotherColumn
From(
SELECT Row_Number() over(partition by name order by value desc)as
row_number, *
FROM students
)
Where row_number = 1
SELECT t1.name, t1.Value, t1.AnotherColumn
FROM mytable t1
JOIN (SELECT name AS nameMax, MAX(Value) as valueMax
FROM mytable
GROUP BY name) AS t2
ON t2.nameMax = t1.name AND t2.valueMax = t1.Value
WHERE 1 OR <anything you would like>
GROUP BY t1.name;
SELECT DISTINCT (t1.ProdId), t1.Quantity FROM Dummy t1 INNER JOIN
(SELECT ProdId, MAX(Quantity) as MaxQuantity FROM Dummy GROUP BY ProdId) t2
ON t1.ProdId = t2.ProdId
AND t1.Quantity = t2.MaxQuantity
ORDER BY t1.ProdId
this will give you the idea.
Given a table of timestamp,user_id,country,site_id.
How do you find the number of users whose first/last visits are the same website?
/* unique users first site*/
SELECT ts,SWE.site_id, SWE.user_id
FROM SWE
WHERE SWE.ts = (
SELECT MIN(t.timestamp)
FROM SWE t
WHERE
t.user_id = SWE.user_id
)
/* unique users last site*/
SELECT ts,SWE.site_id, SWE.user_id
FROM SWE
WHERE SWE.ts = (
SELECT max(t.timestamp)
FROM SWE t
WHERE
t.user_id = SWE.user_id
)
I am not sure how to count when these are equal?
I'd use the DISTINCT ON operator to pick out the first/last visits for each user, then aggregate over these to check if they're different. something like:
WITH first_visits AS (
SELECT DISTINCT ON (user_id) * FROM user_visits
ORDER BY user_id, timestamp
), last_visits AS (
SELECT DISTINCT ON (user_id) * FROM user_visits
ORDER BY user_id, timestamp DESC
)
SELECT user_id,
array_to_string(array_agg(DISTINCT site_id), ', ') AS sites,
MIN(timestamp) AS first_visit, MAX(timestamp) as last_visit
FROM (
SELECT * FROM first_visits
UNION ALL
SELECT * FROM last_visits) x
GROUP BY user_id
HAVING COUNT(DISTINCT site_id) = 1;
Wonder if someone could give me a quick hand. I have 2 select queries (as shown below) and I want to compare the results of both and only return the result that has the most recent date.
So say I have the following 2 results from the queries:-
--------- ---------- ----------------------- --------------- ------ --
COMPANY A EMPLOYEE A 2007-10-16 17:10:21.000 E-mail 6D29D6D5 SYSTEM 1
COMPANY A EMPLOYEE A 2007-10-15 17:10:21.000 E-mail 6D29D6D5 SYSTEM 1
I only want to return the result with the latest date (so the first one). I thought about putting the results into a temporary table and then querying that but just wondering if there's a simpler, more efficient way?
SELECT * FROM (
SELECT fc.accountidname, fc.owneridname, fap.actualend, fap.activitytypecodename, fap.createdby, fap.createdbyname,
ROW_NUMBER() OVER (PARTITION BY fc.accountidname ORDER BY fap.actualend DESC) AS RN
FROM FilteredContact fc
INNER JOIN FilteredActivityPointer fap ON fc.parentcustomerid = fap.regardingobjectid
WHERE fc.statecodename = 'Active'
AND fap.ownerid = '0F995BDC'
AND fap.createdon < getdate()
) tmp WHERE RN = 1
SELECT * FROM (
SELECT fa.name, fa.owneridname, fa.new_technicalaccountmanageridname, fa.new_customerid, fa.new_riskstatusname,
fa.new_numberofopencases, fa.new_numberofurgentopencases, fap.actualend, fap.activitytypecodename, fap.createdby, fap.createdbyname,
ROW_NUMBER() OVER (PARTITION BY fa.name ORDER BY fap.actualend DESC) AS RN
FROM FilteredAccount fa
INNER JOIN FilteredActivityPointer fap ON fa.accountid = fap.regardingobjectid
WHERE fa.statecodename = 'Active'
AND fap.ownerid = '0F995BDC'
AND fap.createdon < getdate()
) tmp2 WHERE RN = 1
if the tables have the same structure (column count and column types to match), then you could just union the results of the two queries, then order by the date desc and then select the top 1.
select top 1 * from
(
-- your first query
union all
-- your second query.
) T
order by YourDateColumn1 desc
You should GROUP BY and use MAX(createdon)
Given an example table 'Users', which has an int column named 'UserID' (and some arbitrary number of other columns), what is the best way to select all rows from which UserID appears more than once?
So far I've come up with
select * from Users where UserID in
(select UserID from Users group by UserID having COUNT(UserID) > 1)
This seems like quite an innefficient way to do this though, is there a better way?
In SQL Server 2005+ you could use this approach:
;WITH UsersNumbered AS (
SELECT
UserID,
rownum = ROW_NUMBER() OVER (PARTITION BY UserID ORDER BY UserID)
FROM Users
)
SELECT u.*
FROM Users u
INNER JOIN UsersNumbered n ON u.UserID = n.UserID AND n.rownum = 2
Provided there exists a non-clustered index on UserID, this yields a slightly worse execution plan than your approach. To make it better (actually, same as yours), you'll need to use... a subquery, however counter-intuitive it may seem:
;WITH UsersNumbered AS (
SELECT
UserID,
rownum = ROW_NUMBER() OVER (PARTITION BY UserID ORDER BY UserID)
FROM Users
)
SELECT u.*
FROM Users u
WHERE EXISTS (
SELECT *
FROM UsersNumbered n
WHERE u.UserID = n.UserID AND n.rownum = 2
);
In case of a clustered index on UserID all three solutions give the same plan.
This would do the same thing but evaluate the performance and it would likely be faster/more efficient. Of course there should be an index on this UserID column.
select u.*
from Users u
join (select UserID,count(UserID) as CUserID from Users group by UserID) u1 on u1.UserID = u.UserID
where CUserID > 1