Select First Row of Every Group in Sql

Select first row in each GROUP BY group?

On databases that support CTE and windowing functions:

WITH summary AS (
    SELECT p.id, 
           p.customer, 
           p.total, 
           ROW_NUMBER() OVER(PARTITION BY p.customer 
                                 ORDER BY p.total DESC) AS rank
      FROM PURCHASES p)
 SELECT *
   FROM summary
 WHERE rank = 1

Supported by any database:

But you need to add logic to break ties:

  SELECT MIN(x.id),  -- change to MAX if you want the highest
         x.customer, 
         x.total
    FROM PURCHASES x
    JOIN (SELECT p.customer,
                 MAX(total) AS max_total
            FROM PURCHASES p
        GROUP BY p.customer) y ON y.customer = x.customer
                              AND y.max_total = x.total
GROUP BY x.customer, x.total

Get top 1 row of each group

;WITH cte AS
(
   SELECT *,
         ROW_NUMBER() OVER (PARTITION BY DocumentID ORDER BY DateCreated DESC) AS rn
   FROM DocumentStatusLogs
)
SELECT *
FROM cte
WHERE rn = 1

If you expect 2 entries per day, then this will arbitrarily pick one. To get both entries for a day, use DENSE_RANK instead

As for normalised or not, it depends if you want to:

• maintain status in 2 places
• preserve status history
• ...

As it stands, you preserve status history. If you want latest status in the parent table too (which is denormalisation) you'd need a trigger to maintain "status" in the parent. or drop this status history table.

Selecting first row per group

SELECT  a, b, c
FROM    (
        SELECT  *, ROW_NUMBER() OVER (PARTITION BY a ORDER BY b, c) rn
        FROM    mytable
        ) q
WHERE   rn = 1
ORDER BY
        a

or

SELECT  mi.*
FROM    (
        SELECT  DISTINCT  a
        FROM    mytable
        ) md
CROSS APPLY
        (
        SELECT  TOP 1 *
        FROM    mytable mi
        WHERE   mi.a = md.a
        ORDER BY
                b, c
        ) mi
ORDER BY
        a

Create a composite index on (a, b, c) for the queries to work faster.

Which one is more efficient depends on your data distribution.

If you have few distinct values of a but lots of records within each a, the second query would be better.

You could improve it even more by creating an indexed view:

CREATE VIEW v_mytable_da
WITH   SCHEMABINDING
AS
       SELECT  a, COUNT_BIG(*) cnt
       FROM    dbo.mytable
       GROUP BY
               a

GO

CREATE UNIQUE CLUSTERED INDEX
       pk_vmytableda_a
ON     v_mytable_da (a)

GO

SELECT  mi.*
FROM    v_mytable_da md
CROSS APPLY
        (
        SELECT  TOP 1 *
        FROM    mytable mi
        WHERE   mi.a = md.a
        ORDER BY
                b, c
        ) mi
ORDER BY
        a

BigQuery/SQL: Select first row of each group

I believe you are looking for the function [FIRST_VALUE][1]?

SELECT 
    landing_page,   
    FIRST_VALUE(URL)  
        OVER ( PARTITION BY landing_page ORDER BY Page_Type DESC) AS first_url
FROM `xxxx.TEST.draft`

How do I select the first row per group in an SQL Query?

declare @sometable table ( foo int, bar int, value int )

insert into @sometable values (47, 1, 100)
insert into @sometable values (47, 0, 10)
insert into @sometable values (47, 2, 10)
insert into @sometable values (46, 0, 100)
insert into @sometable values (46, 1, 10)
insert into @sometable values (46, 2, 10)
insert into @sometable values (44, 0, 2)

WITH cte AS 
(
    SELECT   Foo, Bar, SUM(value) AS SumValue, ROW_NUMBER() OVER(PARTITION BY Foo ORDER BY FOO DESC, SUM(value) DESC) AS RowNumber
    FROM     @SomeTable
    GROUP BY Foo, Bar
)
SELECT * 
FROM cte
WHERE RowNumber = 1

get first row fo each group SQL

If you are runing MySQL 8.0, you can use RANK() in a subquery to rank the records by subject count for each gender, and filter on the top record per group in the outer query (if there are top ties, RANK() preserves them):

SELECT gender, subject, no_of_selections
FROM (
    SELECT  
        se.gender, 
        su.subject, 
        COUNT(*) as no_of_selections,
        RANK() OVER(PARTITION BY se.gender ORDER BY COUNT(*) DESC) rn
    FROM selection se
    JOIN subjects su ON se.subjectID  = su.subjectID
    GROUP BY se.subjectID, se.gender, su.subject
) t
WHERE rn = 1
ORDER BY gender DESC

In earlier versions, where window functions are not availabe, one option is to filter with a having clause that returns to top count per gender:

SELECT  
    se.gender, 
    su.subject, 
    COUNT(*) as no_of_selections
FROM selection se
JOIN subjects su ON se.subjectID  = su.subjectID
GROUP BY se.subjectID, se.gender, su.subject
HAVING COUNT(*) = (
    SELECT COUNT(*) 
    FROM selection se1
    WHERE se1.gender = se.gender
    GROUP BY se1.subjectID, se1.gender
    ORDER BY COUNT(*) DESC
    LIMIT 1
)

Notes:

• I changed the table aliases to make them more meaningful

• You should the subject column to the GROUP BY clause to make your query runnable under sql mode ONLY_FULL_GROUP_MODE, which is by default enabled starting MySQL 5.7

SQL selecting first record per group

GROUP BY u.d (without also listing u1, u2, u3) would only work if u.d was the PRIMARY KEY (which it is not, and also wouldn't make sense in your scenario). See:

• Is it possible to have an SQL query that uses AGG functions in this way?

I suggest DISTINCT ON in a subquery on UTable instead:

SELECT o.d, u.u1, u.u2, u.u3, o.n
FROM  (
   SELECT DISTINCT ON (u.d)
          u.d, u.u1, u.u2, u.u3
   FROM   UTable u
   WHERE  u.gid = 3
   AND    u.gt = 'dog night'
   ORDER  BY u.d, u.timestamp
   ) u
JOIN   OTable o USING (gid, gt, d);

See:

• Select first row in each GROUP BY group?

If UTable is big, at least a multicolumn index on (gid, gt) is advisable. Same for OTable.

Maybe even on (gid, gt, d). Depends on data types, cardinalities, ...

Select first row in each group in sql

You can use distinct on directly with group by:

select distinct on ("Country") Sum("Price"), "In-app Product", "Country"
from cleandatase
group by "Country", "In-app Product"
order by "Country", Sum("Price") desc;

Note: As Thorsten points out, if there are ties and you want all the ties, then distinct on is not the simplest solution.

SQL Server Group By Query Select first row each group

You can GROUP BY StudyID, Year and then in an outer query select the first row from each StudyID, Year group:

SELECT StudyID, Year, minAccess1, minAccess2, minAccess3
FROM (
   SELECT StudyID, Year, min(Access1) minAccess1, min(Access2) minAccess2,
          min(Access3) minAccess3, 
          ROW_NUMBER() OVER (PARTITION BY StudyID ORDER BY Year DESC) AS rn 
   FROM mytable
   GROUP BY StudyID, Year ) t
WHERE t.rn = 1

ROW_NUMBER is used to assign an ordering number to each StudyID group according to Year values. The row with the maximum Year value is assigned a rn = 1.

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