Display only the first row per match - sql

Question

I have a table (AreaPartners), and I want to match only the first "Name" record in each group, order by "ID", grouped by "Area". So for the table below:
Area Name ID
AB ISmith 748
AB AWood 750
AB HArcher 751
AB DMunslow 753
AB DCornelius 754
BH MLee 301
BH NMcClean 307
BH DMiles 309
BH LPayze 325
BH MPinnock 329
I'd want to return the results ISmith for AB and MLee for BH.
How do I go about doing this? I believe it's something to do with the Group By function, but I can't for the life of me get it to work.

Answer 1

Try this:
SELECT yourTable.Area, yourTable.Name
FROM yourTable INNER JOIN (
SELECT MIN(Id) AS MinId
FROM yourTable
GROUP BY Area) M ON yourTable.Id = M.MinId

Answer 2

Update because of comment (There is no table variable and Partition over is not a MS access statement). You can also do it with an IN statement:
SELECT
yourTable.Area,
yourTable.Name
FROM yourTable
WHERE yourTable.Id IN
(
SELECT
MIN(tbl.Id) AS MinId
FROM
yourTable as tbl
GROUP BY
tbl.Area
)
In MSSQL you can write this:
DECLARE #tbl TABLE
(
Area VARCHAR(100),
Name VARCHAR(100),
ID INT
)
INSERT INTO #tbl
SELECT 'AB','ISmith',748
UNION ALL
SELECT 'AB','AWood',750
UNION ALL
SELECT 'AB','HArcher',751
UNION ALL
SELECT 'AB','DMunslow',753
UNION ALL
SELECT 'AB','DCornelius',754
UNION ALL
SELECT 'BH','MLee',301
UNION ALL
SELECT 'BH','NMcClean',307
UNION ALL
SELECT 'BH','DMiles',309
UNION ALL
SELECT 'BH','LPayze',325
UNION ALL
SELECT 'BH','MPinnock',325
;WITH CTE
AS
(
SELECT
RANK() OVER(PARTITION BY tbl.Area ORDER BY ID) AS iRank,
tbl.ID,
tbl.Area,
tbl.Name
FROM
#tbl AS tbl
)
SELECT
*
FROM
CTE
WHERE
CTE.iRank=1