How to find Max value in a column in SQL Server 2012 - sql

I want to find the max value in a column
ID CName Tot_Val PName
--------------------------------
1 1 100 P1
2 1 10 P2
3 2 50 P2
4 2 80 P1
Above is my table structure. I just want to find the max total value only from the table. In that four row ID 1 and 2 have same value in CName but total val and PName has different values. What I am expecting is have to find the max value in ID 1 and 2
Expected result:
ID CName Tot_Val PName
--------------------------------
1 1 100 P1
4 2 80 P1
I need result same as like mention above
select Max(Tot_Val), CName
from table1
where PName in ('P1', 'P2')
group by CName
This is query I have tried but my problem is that I am not able to bring PName in this table. If I add PName in the select list means it will showing the rows doubled e.g. Result is 100 rows but when I add PName in selected list and group by list it showing 600 rows. That is the problem.
Can someone please help me to resolve this.

One possible option is to use a subquery. Give each row a number within each CName group ordered by Tot_Val. Then select the rows with a row number equal to one.
select x.*
from ( select mt.ID,
mt.CName,
mt.Tot_Val,
mt.PName,
row_number() over(partition by mt.CName order by mt.Tot_Val desc) as No
from MyTable mt ) x
where x.No = 1;
An alternative would be to use a common table expression (CTE) instead of a subquery to isolate the first result set.
with x as
(
select mt.ID,
mt.CName,
mt.Tot_Val,
mt.PName,
row_number() over(partition by mt.CName order by mt.Tot_Val desc) as No
from MyTable mt
)
select x.*
from x
where x.No = 1;
See both solutions in action in this fiddle.

You can search top-n-per-group for this kind of a query.
There are two common ways to do it. The most efficient method depends on your indexes and data distribution and whether you already have another table with the list of all CName values.
Using ROW_NUMBER
WITH
CTE
AS
(
SELECT
ID, CName, Tot_Val, PName,
ROW_NUMBER() OVER (PARTITION BY CName ORDER BY Tot_Val DESC) AS rn
FROM table1
)
SELECT
ID, CName, Tot_Val, PName
FROM CTE
WHERE rn=1
;
Using CROSS APPLY
WITH
CTE
AS
(
SELECT CName
FROM table1
GROUP BY CName
)
SELECT
A.ID
,A.CName
,A.Tot_Val
,A.PName
FROM
CTE
CROSS APPLY
(
SELECT TOP(1)
table1.ID
,table1.CName
,table1.Tot_Val
,table1.PName
FROM table1
WHERE
table1.CName = CTE.CName
ORDER BY
table1.Tot_Val DESC
) AS A
;
See a very detailed answer on dba.se Retrieving n rows per group
, or here Get top 1 row of each group
.

CROSS APPLY might be as fast as a correlated subquery, but this often has very good performance (and better than ROW_NUMBER():
select t.*
from t
where t.tot_val = (select max(t2.tot_val)
from t t2
where t2.cname = t.cname
);
Note: The performance depends on having an index on (cname, tot_val).

Related

Replace nested query to single select query

Consider the table fields as follows.
Appid Client_name is_real RTT
100 C1 1 1
200 C1 1 6
200 C2 1 7
100 C1 1 9
200 C1 0 7
Now I need total number of unique real Appid's in the table. We can say one appid record is real by if 'is_real' is 1.
In above table, we have only 3 real Appid's. Which are (100,C1), (200,C1) and (200, C2).
Postgesql command:
Select sum(r)
from (select count(is_real) as r from table group by Appid, Client_name) as t;
I don't want any recursive query. If you can fetch with single select query, it would be helpful.
Since you seem to define a unique id by (Appid, Client_name) (which is confusing, since you are mixing terms):
SELECT COUNT(DISTINCT (Appid, Client_name)) AS ct
FROM tbl
WHERE is_real = 1;
(Appid, Client_name) is a row-type expression, short for ROW(Appid, Client_name). Only distinct combinations are counted.
Another trick to get this done without subquery is to use a window function:
SELECT DISTINCT count(*) OVER () AS ct
FROM tbl
WHERE is_real = 1
GROUP BY Appid, Client_name;
But neither is going to be faster than using a subquery (which is not a recursive query):
SELECT count(*) AS ct
FROM (
SELECT 1
FROM tbl
WHERE is_real = 1
GROUP BY Appid, Client_name
) sub;
That's what I would use.
It's essential to understand the sequence of events in a SELECT query:
Best way to get result count before LIMIT was applied
total number of unique real Appid's in the table
I assume is_real is 1 = true, 0 = false.
SELECT COUNT(DISTINCT Appid)
FROM table
WHERE is_real = 1;

Ensuring only distinct records are returned with DISTINCT

Given the following table:
date_field_one date_field_two arbitrary_value
---------------- ---------------- -----------------
1/1/11 1/3/11 cheese
1/1/11 1/4/11 the color orange
2/2/11 2/3/11 1
2/2/11 2/4/11 2
My problem: I'm not sure how to go about structuring a query using a set based approach that yields the following results:
for each distinct date, the record with the earliest
date_field_two value is returned
Any ideas?
Edit for new response! The solution posted by M.Ali may be the best fit for your specific case as it will ensure you only ever get one row result from your base data, even if there exist multiple candidate rows for your answer ( as in, date_field_one, date_field_two combinations are not distinct ). The following will return multiple results per date_field_one, date_field_two combination in the not-distinct scenario:
SELECT t.date_field_one, t.date_field_two, t.arbitrary_value
FROM ( SELECT date_field_one,
date_field_two = MIN( date_field_two )
FROM dbo.[table]
GROUP BY date_field_one ) dl
LEFT JOIN dbo.[table] t
ON dl.date_field_one = t.date_field_one
AND dl.date_field_two = t.date_field_two;
;WITH CTE
AS
(
SELECT *, rn = ROW_NUMBER() OVER (PARTITION BY date_field_one ORDER BY date_field_two
ASC)
FROM TableName
)
SELECT * FROM CTE
WHERE rn = 1
Something like this:
select date_field_one, min(date_field_two)
from yourtable
group by date_field_one
select date_field_one, min(date_fileld_two)
from table
group by date_field_one
try this for latest...........
select date_field_one ,min(date_field_two) date_field_two
from table group by date_field_one

SQL. Is there any efficient way to find second lowest value?

I have the following table:
ItemID Price
1 10
2 20
3 12
4 10
5 11
I need to find the second lowest price. So far, I have a query that works, but i am not sure it is the most efficient query:
select min(price)
from table
where itemid not in
(select itemid
from table
where price=
(select min(price)
from table));
What if I have to find third OR fourth minimum price? I am not even mentioning other attributes and conditions... Is there any more efficient way to do this?
PS: note that minimum is not a unique value. For example, items 1 and 4 are both minimums. Simple ordering won't do.
SELECT MIN( price )
FROM table
WHERE price > ( SELECT MIN( price )
FROM table )
select price from table where price in (
select
distinct price
from
(select t.price,rownumber() over () as rownum from table t) as x
where x.rownum = 2 --or 3, 4, 5, etc
)
Not sure if this would be the fastest, but it would make it easier to select the second, third, etc... Just change the TOP value.
UPDATED
SELECT MIN(price)
FROM table
WHERE price NOT IN (SELECT DISTINCT TOP 1 price FROM table ORDER BY price)
To find out second minimum salary of an employee, you can use following:
select min(salary)
from table
where salary > (select min(salary) from table);
This is a good answer:
SELECT MIN( price )
FROM table
WHERE price > ( SELECT MIN( price )
FROM table )
Make sure when you do this that there is only 1 row in the subquery! (the part in brackets at the end).
For example if you want to use GROUP BY you will have to define even further using:
SELECT MIN( price )
FROM table te1
WHERE price > ( SELECT MIN( price )
FROM table te2 WHERE te1.brand = te2.brand)
GROUP BY brand
Because GROUP BY will give you multiple rows, otherwise you will get the error:
SQL Error [21000]: ERROR: more than one row returned by a subquery used as an expression
I guess a simplest way to do is using offset-fetch filter from standard sql, distinct is not necessary if you don't have repeat values in your column.
select distinct(price) from table
order by price
offset 1 row fetch first 1 row only;
no need to write complex subqueries....
In amazon redshift use limit-fetch instead for ex...
Select distinct(price) from table
order by price
limit 1
offset 1;
You can either use one of the following:-
select min(your_field) from your_table where your_field NOT IN (select distinct TOP 1 your_field from your_table ORDER BY your_field DESC)
OR
select top 1 ColumnName from TableName where ColumnName not in (select top 1 ColumnName from TableName order by ColumnName asc)
I think you can find the second minimum using LIMIT and ORDER BY
select max(price) as minimum from (select distinct(price) from tableName order by price asc limit 2 ) --or 3, 4, 5, etc
if you want to find third or fourth minimum and so on... you can find out by changing minimum number in limit. you can find using this statement.
You can use RANK functions,
it may seem complex query but similar results like other answers can be achieved with the same,
WITH Temp_table AS (SELECT ITEM_ID,PRICE,RANK() OVER (ORDER BY PRICE) AS
Rnk
FROM YOUR_TABLE_NAME)
SELECT ITEM_ID FROM Temp_table
WHERE Rnk=2;
Maybe u can check the min value first and then place a not or greater than the operator. This will eliminate the usage of a subquery but will require a two-step process
select min(price)
from table
where min(price) <> -- "the min price you previously got"

SQL query to select distinct row with minimum value

I want an SQL statement to get the row with a minimum value.
Consider this table:
id game point
1 x 5
1 z 4
2 y 6
3 x 2
3 y 5
3 z 8
How do I select the ids that have the minimum value in the point column, grouped by game? Like the following:
id game point
1 z 4
2 y 5
3 x 2
Use:
SELECT tbl.*
FROM TableName tbl
INNER JOIN
(
SELECT Id, MIN(Point) MinPoint
FROM TableName
GROUP BY Id
) tbl1
ON tbl1.id = tbl.id
WHERE tbl1.MinPoint = tbl.Point
This is another way of doing the same thing, which would allow you to do interesting things like select the top 5 winning games, etc.
SELECT *
FROM
(
SELECT ROW_NUMBER() OVER (PARTITION BY ID ORDER BY Point) as RowNum, *
FROM Table
) X
WHERE RowNum = 1
You can now correctly get the actual row that was identified as the one with the lowest score and you can modify the ordering function to use multiple criteria, such as "Show me the earliest game which had the smallest score", etc.
This will work
select * from table
where (id,point) IN (select id,min(point) from table group by id);
As this is tagged with sql only, the following is using ANSI SQL and a window function:
select id, game, point
from (
select id, game, point,
row_number() over (partition by game order by point) as rn
from games
) t
where rn = 1;
Ken Clark's answer didn't work in my case. It might not work in yours either. If not, try this:
SELECT *
from table T
INNER JOIN
(
select id, MIN(point) MinPoint
from table T
group by AccountId
) NewT on T.id = NewT.id and T.point = NewT.MinPoint
ORDER BY game desc
SELECT DISTINCT
FIRST_VALUE(ID) OVER (Partition by Game ORDER BY Point) AS ID,
Game,
FIRST_VALUE(Point) OVER (Partition by Game ORDER BY Point) AS Point
FROM #T
SELECT * from room
INNER JOIN
(
select DISTINCT hotelNo, MIN(price) MinPrice
from room
Group by hotelNo
) NewT
on room.hotelNo = NewT.hotelNo and room.price = NewT.MinPrice;
This alternative approach uses SQL Server's OUTER APPLY clause. This way, it
creates the distinct list of games, and
fetches and outputs the record with the lowest point number for that game.
The OUTER APPLY clause can be imagined as a LEFT JOIN, but with the advantage that you can use values of the main query as parameters in the subquery (here: game).
SELECT colMinPointID
FROM (
SELECT game
FROM table
GROUP BY game
) As rstOuter
OUTER APPLY (
SELECT TOP 1 id As colMinPointID
FROM table As rstInner
WHERE rstInner.game = rstOuter.game
ORDER BY points
) AS rstMinPoints
This is portable - at least between ORACLE and PostgreSQL:
select t.* from table t
where not exists(select 1 from table ti where ti.attr > t.attr);
Most of the answers use an inner query. I am wondering why the following isn't suggested.
select
*
from
table
order by
point
fetch next 1 row only // ... or the appropriate syntax for the particular DB
This query is very simple to write with JPAQueryFactory (a Java Query DSL class).
return new JPAQueryFactory(manager).
selectFrom(QTable.table).
setLockMode(LockModeType.OPTIMISTIC).
orderBy(QTable.table.point.asc()).
fetchFirst();
Try:
select id, game, min(point) from t
group by id

SQL Select highest values from table on two (or more) columns

not sure if there's an elegant way to acheive this:
Data
ID Ver recID (loads more columns of stuff)
1 1 1
2 2 1
3 3 1
4 1 2
5 1 3
6 2 3
So, we have ID as the Primary Key, the Ver as the version and recID as a record ID (an arbitary base ID to tie all the versions together).
So I'd like to select from the following data, rows 3, 4 and 6. i.e. the highest version for a given record ID.
Is there a way to do this with one SQL query? Or would I need to do a SELECT DISTINCT on the record ID, then a seperate query to get the highest value? Or pull the lot into the application and filter from there?
A GROUP BY would be sufficient to get each maximum version for every recID.
SELECT Ver = MAX(Ver), recID
FROM YourTable
GROUP BY
recID
If you also need the corresponding ID, you can wrap this into a subselect
SELECT yt.*
FROM Yourtable yt
INNER JOIN (
SELECT Ver = MAX(Ver), recID
FROM YourTable
GROUP BY
recID
) ytm ON ytm.Ver = yt.Ver AND ytm.recID = yt.RecID
or, depending on the SQL Server version you are using, use ROW_NUMBER
SELECT *
FROM (
SELECT ID, Ver, recID
, rn = ROW_NUMBER() OVER (PARTITION BY recID ORDER BY Ver DESC)
FROM YourTable
) yt
WHERE yt.rn = 1
Getting maximum ver for a given recID is easy. To get the ID, you need to join on a nested query that gets these maximums:
select ID, ver, recID from table x
inner join
(select max(ver) as ver, recID
from table
group by recID) y
on x.ver = y.ver and x.recID = y.recID
You could use a cte with ROW_NUMBER function:
WITH cte AS(
SELECT ID, Ver, recID
, ROW_NUMBER()OVER(PARTITION BY recID ORDER BY Ver DESC)as RowNum
FROM data
)
SELECT ID,Ver,recID FROM cte
WHERE RowNum = 1
straighforward example using a subquery:
SELECT a.*
FROM tab a
WHERE ver = (
SELECT max(ver)
FROM tab b
WHERE b.recId = a.recId
)
(Note: this assumes that the combination of (recId, ver) is unique. Typically there would be a primary key or unique constraint on those columns, in that order, and then that index can be used to optimize this query)
This works on almost all RDBMS-es, although the correlated subquery might not be handled very efficiently (depending on RDBMS). SHould work fine in MS SQL 2008 though.