SQL - select distinct only on one column [duplicate]

A very typical approach to this type of problem is to use row_number():

select t.*from (select t.*,             row_number() over (partition by number order by id) as seqnum      from t     ) twhere seqnum = 1;

This is more generalizable than using a comparison to the minimum id. For instance, you can get a random row by using order by newid(). You can select 2 rows by using where seqnum <= 2.

Since you don't care, I chose the max ID for each number.

select tbl.* from tblinner join (select max(id) as maxID, number from tbl group by number) maxIDon maxID.maxID = tbl.id

Query Explanation

 select     tbl.*  -- give me all the data from the base table (tbl)  from     tbl        inner join (  -- only return rows in tbl which match this subquery        select             max(id) as maxID -- MAX (ie distinct) ID per GROUP BY below        from             tbl         group by             NUMBER            -- how to group rows for the MAX aggregation    ) maxID        on maxID.maxID = tbl.id -- join condition ie only return rows in tbl                                 -- whose ID is also a MAX ID for a given NUMBER

You will use the following query:

SELECT * FROM [table] GROUP BY NUMBER;

Where [table] is the name of the table.

This provides a unique listing for the NUMBER column however the other columns may be meaningless depending on the vendor implementation; which is to say they may not together correspond to a specific row or rows.