Select Different Column Value for Row with Max Value - sql-server

I'm hoping for a cleaner way to do something that I know how to do one way. I want to retrieve the UserId for the MAX ID value as well as that MAX ID value. Let's say I have a table with data like this:
ID UserId Value
1 10 'Foo'
2 15 'Blah'
3 10 'Blech'
4 20 'Qwerty'
I want to retrieve:
ID UserId
4 20
I know I could do this like so:
SELECT
t.ID,
t.UserID
FROM
(
SELECT MAX(ID) as [MaxID]
FROM table
) as m
JOIN table as t ON m.MaxID = t.ID
I'm only vaguely familiar with the ROW_NUMBER(), RANK() and other similar methods and I can't help believing that this scenario could benefit from some such method to get rid of joining back to the table.

You can definitely use ROW_NUMBER for something like this:
with t1Rank as
(
select *
, t1Rank = row_number() over (order by ID desc)
from t1
)
select ID, UserID
from t1Rank
where t1Rank = 1
SQL Fiddle with demo.
The advantage with this approach is you can bring Value (or other fields as required) into the result set, too. Plus you can tweak the ordering/grouping as required.

You could also just do it with a sub-query like this:
SELECT ID ,
UserID
FROM table
WHERE ID = ( SELECT MAX(ID)
FROM table
);

SELECT TOP 1 ID, UserID FROM <table> ORDER BY ID DESC

Related

Max Value with unique values in more than one column

I feel like I'm missing something really obvious here.
Using T-SQL/SQL-Server:
I have unique values in more than one column but want to select the max version based on one particular column.
Dataset:
Example
ID | Name| Version | Code
------------------------
1 | Car | 3 | NULL
1 | Car | 2 | 1000
1 | Car | 1 | 2000
Target status: I want my query to only select the row with the highest version value. Running a MAX on the version column pulls all three because of the distinct values in the 'Code' column:
SELECT ID
,Name
,MAX(Version)
,Code
FROM Table
GROUP BY ID, Name, Code
The net result is that I get all three entries as per the data set due to the unique values in the Code column, but I only want the top row (Version 3).
Any help would be appreciated.
You need to identify the row with the highest version as 1 query and use another outer query to pull out all the fields for that row. Like so:
SELECT t.ID, t.Name, GRP.Version, t.Code
FROM (
SELECT ID
,Name
,MAX(Version) as Version
FROM Table
GROUP BY ID, Name
) GRP
INNER JOIN Table t on GRP.ID = t.ID and GRP.Name = t.Name and GRP.Version = t.Version
You can also use row_number() to do this kind of logic, for example like this:
select ID, Name, Version, Code
from (
select *, row_number() over (order by Version desc) as RN
from Table1
) X where RN = 1
Example in SQL Fiddle
add the top statment to force the return of a single row. Also add the order by notation
SELECT top 1 ID
,Name
,MAX(Version)
,Code
FROM Table
GROUP BY ID, Name, Code
order by max(version) desc

select statement with "Group by" on specific columns but displaying other columns along with group by columns

I want to get all data based on group by of only encounter,medicationname
column data..
select encounter,medicationname,count(*) as freq,labdate,result
from Medications where (labdate between #admitdate and DATEDIFF(dd,24,#admitdate))
group by encounter,medicationname having count(*)>2
I have records like
encounter medicationname freq
8604261 ACC 3
Now based on this data ,I want to get
This is my desired output
encounter medicationname labtime result
8604261 ACC 2015-05-22 18
8604261 ACC 2015-07-23 23
8604261 ACC 2015-09-09 27
You can use COUNT() as a window function, something like this:
;With Counted as (
SELECT encounter,medicationname,labdate,result,
COUNT(*) OVER (PARTITION BY encounter,medicationname) as cnt
from Medications
where (labdate between #admitdate
and DATEDIFF(dd,24,#admitdate))
)
select encounter,medicationname,labdate,result
from Counted
where cnt > 2
I would note that I think DATEDIFF1 is probably wrong also but since I don't have your data, inputs and an actual spec, I've left it as is for now.
1DATEDIFF returns an int, but you're using it in a comparison against a column which is apparently a date. DATEADD would be the more probably desired function here, but as I say, I don't have full information to go on.
If I understand you question correctly what you need is this
;WITH CTE AS
(
select encounter,medicationname,count(*) as freq,labdate,result
from Medications where (labdate between #admitdate and DATEDIFF(dd,24,#admitdate))
group by encounter,medicationname having count(*) > 2
)
select encounter,medicationname,labdate,result
from Medications M
INNER JOIN CTE C
ON M.encounter = C.encounter
AND M.medicationname = C.medicationname
where (labdate between #admitdate and DATEDIFF(dd,24,#admitdate))
or better yet using COUNT()OVER()
;WITH CTE AS
(
SELECT encounter,medicationname,COUNT(*) OVER(PARTITION BY encounter,medicationname)as freq,labdate,result
FROM Medications
WHERE (labdate between #admitdate and DATEDIFF(dd,24,#admitdate))
)
SELECT * FROM CTE
WHERE freq > 2
select encounter,medicationname,count(*) as freq,labdate,result
from Medications
where (labdate between #admitdate and DATEDIFF(dd,24,#admitdate))
group by encounter,medicationname having count(*) > 2

T-SQL order by, based on other column value

I'm stuck with a query which should be pretty simple but, for reasons unknown, my brain is not playing ball here ...
Table:
id(int) | strategy (varchar) | value (whatever)
1 "ABC" whatevs
2 "ABC" yeah
3 "DEF" hello
4 "DEF" kitty
5 "QQQ" hurrr
The query should select ALL rows grouped on strategy but only one row per strategy - the one with the higest id.
In the case above, it should return rows with id 2, 4 and 5
SELECT id, strategy , value
FROM (
SELECT id, strategy , value
,ROW_NUMBER() OVER (PARTITION BY strategy ORDER BY ID DESC) rn
FROM Table_Name
) Sub
WHERE rn = 1
Working SQL FIDDLE
You can use window function to get the solution you want. Fiddle here
with cte as
(
select
rank()over(partition by strategy order by id desc) as rnk,
id, strategy, value from myT
)
select id, strategy, value from
cte where rnk = 1;
Try this:
SELECT T2.id,T1.strategy,T1.value
FROM TableName T1
INNER JOIN
(SELECT MAX(id) as id,strategy
FROM TableName
GROUP BY strategy) T2
ON T1.id=T2.id
Result:
ID STRATEGY VALUE
2 ABC yeah
4 DEF kitty
5 QQQ hurrr
See result in SQL Fiddle.
SELECT id, strategy , value
FROM (
SELECT id, strategy , value
,MAX(id) OVER (PARTITION BY strategy) MaxId
FROM YourTable
) Sub
WHERE id=MaxId
You may try this one as well:
SELECT id, strategy, value FROM TableName WHERE id IN (
SELECT MAX(id) FROM TableName GROUP BY strategy
)
Bit depends on your data, you might get results faster with it as it does not do sorting, but by the other hand it uses IN, which can slow you down if there is many 'strategies'

Problem with unique SQL query

I want to select all records, but have the query only return a single record per Product Name. My table looks similar to:
SellId ProductName Comment
1 Cake dasd
2 Cake dasdasd
3 Bread dasdasdd
where the Product Name is not unique. I want the query to return a single record per ProductName with results like:
SellId ProductName Comment
1 Cake dasd
3 Bread dasdasdd
I have tried this query,
Select distict ProductName,Comment ,SellId from TBL#Sells
but it is returning multiple records with the same ProductName. My table is not realy as simple as this, this is just a sample. What is the solution? Is it clear?
Select ProductName,
min(Comment) , min(SellId) from TBL#Sells
group by ProductName
If y ou only want one record per productname, you ofcourse have to choose what value you want for the other fields.
If you aggregate (using group by) you can choose an aggregate function,
htat's a function that takes a list of values and return only one : here I have chosen MIN : that is the smallest walue for each field.
NOTE : comment and sellid can come from different records, since MIN is taken...
Othter aggregates you might find useful :
FIRST : first record encountered
LAST : last record encoutered
AVG : average
COUNT : number of records
first/last have the advantage that all fields are from the same record.
SELECT S.ProductName, S.Comment, S.SellId
FROM
Sells S
JOIN (SELECT MAX(SellId)
FROM Sells
GROUP BY ProductName) AS TopSell ON TopSell.SellId = S.SellId
This will get the latest comment as your selected comment assuming that SellId is an auto-incremented identity that goes up.
I know, you've got an answer already, I'd like to offer a way that was fastest in terms of performance for me, in a similar situation. I'm assuming that SellId is Primary Key and identity. You'd want an index on ProductName for best performance.
select
Sells.*
from
(
select
distinct ProductName
from
Sells
) x
join
Sells
on
Sells.ProductName = x.ProductName
and Sells.SellId =
(
select
top 1 s2.SellId
from
Sells s2
where
x.ProductName = s2.ProductName
Order By SellId
)
A slower method, (but still better than Group By and MIN on a long char column) is this:
select
*
from
(
select
*,ROW_NUMBER() over (PARTITION BY ProductName order by SellId) OccurenceId
from sells
) x
where
OccurenceId = 1
An advantage of this one is that it's much easier to read.
create table Sale
(
SaleId int not null
constraint PK_Sale primary key,
ProductName varchar(100) not null,
Comment varchar(100) not null
)
insert Sale
values
(1, 'Cake', 'dasd'),
(2, 'Cake', 'dasdasd'),
(3, 'Bread', 'dasdasdd')
-- Option #1 with over()
select *
from Sale
where SaleId in
(
select SaleId
from
(
select SaleId, row_number() over(partition by ProductName order by SaleId) RowNumber
from Sale
) tt
where RowNumber = 1
)
order by SaleId
-- Option #2
select *
from Sale
where SaleId in
(
select min(SaleId)
from Sale
group by ProductName
)
order by SaleId
drop table Sale

How do I select last 5 rows in a table without sorting?

I want to select the last 5 records from a table in SQL Server without arranging the table in ascending or descending order.
This is just about the most bizarre query I've ever written, but I'm pretty sure it gets the "last 5" rows from a table without ordering:
select *
from issues
where issueid not in (
select top (
(select count(*) from issues) - 5
) issueid
from issues
)
Note that this makes use of SQL Server 2005's ability to pass a value into the "top" clause - it doesn't work on SQL Server 2000.
Suppose you have an index on id, this will be lightning fast:
SELECT * FROM [MyTable] WHERE [id] > (SELECT MAX([id]) - 5 FROM [MyTable])
The way your question is phrased makes it sound like you think you have to physically resort the data in the table in order to get it back in the order you want. If so, this is not the case, the ORDER BY clause exists for this purpose. The physical order in which the records are stored remains unchanged when using ORDER BY. The records are sorted in memory (or in temporary disk space) before they are returned.
Note that the order that records get returned is not guaranteed without using an ORDER BY clause. So, while any of the the suggestions here may work, there is no reason to think they will continue to work, nor can you prove that they work in all cases with your current database. This is by design - I am assuming it is to give the database engine the freedom do as it will with the records in order to obtain best performance in the case where there is no explicit order specified.
Assuming you wanted the last 5 records sorted by the field Name in ascending order, you could do something like this, which should work in either SQL 2000 or 2005:
select Name
from (
select top 5 Name
from MyTable
order by Name desc
) a
order by Name asc
You need to count number of rows inside table ( say we have 12 rows )
then subtract 5 rows from them ( we are now in 7 )
select * where index_column > 7
select * from users
where user_id >
( (select COUNT(*) from users) - 5)
you can order them ASC or DESC
But when using this code
select TOP 5 from users order by user_id DESC
it will not be ordered easily.
select * from table limit 5 offset (select count(*) from table) - 5;
Without an order, this is impossible. What defines the "bottom"? The following will select 5 rows according to how they are stored in the database.
SELECT TOP 5 * FROM [TableName]
Well, the "last five rows" are actually the last five rows depending on your clustered index. Your clustered index, by definition, is the way that he rows are ordered. So you really can't get the "last five rows" without some order. You can, however, get the last five rows as it pertains to the clustered index.
SELECT TOP 5 * FROM MyTable
ORDER BY MyCLusteredIndexColumn1, MyCLusteredIndexColumnq, ..., MyCLusteredIndexColumnN DESC
Search 5 records from last records you can use this,
SELECT *
FROM Table Name
WHERE ID <= IDENT_CURRENT('Table Name')
AND ID >= IDENT_CURRENT('Table Name') - 5
If you know how many rows there will be in total you can use the ROW_NUMBER() function.
Here's an examble from MSDN (http://msdn.microsoft.com/en-us/library/ms186734.aspx)
USE AdventureWorks;
GO
WITH OrderedOrders AS
(
SELECT SalesOrderID, OrderDate,
ROW_NUMBER() OVER (ORDER BY OrderDate) AS 'RowNumber'
FROM Sales.SalesOrderHeader
)
SELECT *
FROM OrderedOrders
WHERE RowNumber BETWEEN 50 AND 60;
In SQL Server 2012 you can do this :
Declare #Count1 int ;
Select #Count1 = Count(*)
FROM [Log] AS L
SELECT
*
FROM [Log] AS L
ORDER BY L.id
OFFSET #Count - 5 ROWS
FETCH NEXT 5 ROWS ONLY;
Try this, if you don't have a primary key or identical column:
select [Stu_Id],[Student_Name] ,[City] ,[Registered],
RowNum = row_number() OVER (ORDER BY (SELECT 0))
from student
ORDER BY RowNum desc
You can retrieve them from memory.
So first you get the rows in a DataSet, and then get the last 5 out of the DataSet.
There is a handy trick that works in some databases for ordering in database order,
SELECT * FROM TableName ORDER BY true
Apparently, this can work in conjunction with any of the other suggestions posted here to leave the results in "order they came out of the database" order, which in some databases, is the order they were last modified in.
select *
from table
order by empno(primary key) desc
fetch first 5 rows only
Last 5 rows retrieve in mysql
This query working perfectly
SELECT * FROM (SELECT * FROM recharge ORDER BY sno DESC LIMIT 5)sub ORDER BY sno ASC
or
select sno from(select sno from recharge order by sno desc limit 5) as t where t.sno order by t.sno asc
When number of rows in table is less than 5 the answers of Matt Hamilton and msuvajac is Incorrect.
Because a TOP N rowcount value may not be negative.
A great example can be found Here.
i am using this code:
select * from tweets where placeID = '$placeID' and id > (
(select count(*) from tweets where placeID = '$placeID')-2)
In SQL Server, it does not seem possible without using ordering in the query.
This is what I have used.
SELECT *
FROM
(
SELECT TOP 5 *
FROM [MyTable]
ORDER BY Id DESC /*Primary Key*/
) AS T
ORDER BY T.Id ASC; /*Primary Key*/
DECLARE #MYVAR NVARCHAR(100)
DECLARE #step int
SET #step = 0;
DECLARE MYTESTCURSOR CURSOR
DYNAMIC
FOR
SELECT col FROM [dbo].[table]
OPEN MYTESTCURSOR
FETCH LAST FROM MYTESTCURSOR INTO #MYVAR
print #MYVAR;
WHILE #step < 10
BEGIN
FETCH PRIOR FROM MYTESTCURSOR INTO #MYVAR
print #MYVAR;
SET #step = #step + 1;
END
CLOSE MYTESTCURSOR
DEALLOCATE MYTESTCURSOR
Thanks to #Apps Tawale , Based on his answer, here's a bit of another (my) version,
To select last 5 records without an identity column,
select top 5 *,
RowNum = row_number() OVER (ORDER BY (SELECT 0))
from [dbo].[ViewEmployeeMaster]
ORDER BY RowNum desc
Nevertheless, it has an order by, but on RowNum :)
Note(1): The above query will reverse the order of what we get when we run the main select query.
So to maintain the order, we can slightly go like:
select *, RowNum2 = row_number() OVER (ORDER BY (SELECT 0))
from (
select top 5 *, RowNum = row_number() OVER (ORDER BY (SELECT 0))
from [dbo].[ViewEmployeeMaster]
ORDER BY RowNum desc
) as t1
order by RowNum2 desc
Note(2): Without an identity column, the query takes a bit of time in case of large data
Get the count of that table
select count(*) from TABLE
select top count * from TABLE where 'primary key row' NOT IN (select top (count-5) 'primary key row' from TABLE)
If you do not want to arrange the table in ascending or descending order. Use this.
select * from table limit 5 offset (select count(*) from table) - 5;

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