SQL query to return one single record for each unique value in a column
I have a table in SQL Server 2000 that I am trying to query in a specific way. The best way to show this is with example data.
Name Street City State -------------------------------------------------------- Bob 123 Fake Street Peoria IL Bob 234 Other Street Fargo ND Jim 345 Main Street St Louis MO
This is actually a simplified example of the structure of the actual table. The structure of the table is completely beyond my control. I need a query that will return a single address per name. It doesn't matter which address, just that there is only one. The result could be this:
Name Street City State -------------------------------------------------------- Bob 123 Fake Street Peoria IL Jim 345 Main Street St Louis MO
I found a similar question here, but none of the solutions given work in my case because I do not have access to CROSS APPLY, and calling MIN() on each column will mix different addresses together, and although I don't care which record is returned, it must be one intact row, not a mix of different rows.
Recommendations to change the table structure will not help me. I agree that this table is terrible, (it's worse than shown here) but this is part of a major ERP database that I can not change.
There are about 3000 records in this table. There is no primary key.
Well, this will give you pretty bad performance, but I think it'll work
SELECT t.Name, t.Street, t.City, t.State FROM table t INNER JOIN ( SELECT m.Name, MIN(m.Street + ';' + m.City + ';' + m.State) AS comb FROM table m GROUP BY m.Name ) x ON x.Name = t.Name AND x.comb = t.Street + ';' + t.City + ';' + t.State
Use a temp table or table variable and select a distinct list of names into that. Use that structure then to select the top 1 of each record in the original table for each distinct name.
If you can use a temp table:
select * -- Create and populate temp table into #Addresses from Addresses alter table #Addresses add PK int identity(1, 1) primary key select Name, Street, City, State -- Explicitly name columns here to not return the PK from #Addresses A where not exists (select * from #Addresses B where B.Name = A.Name and A.PK > B.PK)
This solution would not be advisable for much larger tables.
select distinct Name , street,city,state from table t1 where street = (select min(street) from table t2 where t2.name = t1.name)
select Name , street,city,state FROM( select Name , street,city,state, ROW_NUMBER() OVER(PARTITION BY Name ORDER BY Name) AS rn from table) AS t WHERE rn=1
This is ugly as hell, but it sounds like your predicament is ugly, too... so here goes...
select name, (select top 1 street from [Addresses] a1 where a1.name = a0.name) as street, (select top 1 city from [Addresses] a2 where a2.name = a0.name) as city, (select top 1 state from [Addresses] a3 where a3.name = a0.name) as state from (select distinct name from [Addresses]) as a0
A temporary table solution would be as follows
CREATE Table #Addresses ( MyId int IDENTITY(1,1), [Name] NVARCHAR(50), Street NVARCHAR(50), City NVARCHAR(50), State NVARCHAR(50) ) INSERT INTO #Addresses ([Name], Street, City, State) SELECT [Name], Street, City, State FROM Addresses SELECT Addresses1.[Name], Addresses1.Street, Addresses1.City, Addresses1.State FROM #Addresses Addresses1 WHERE Addresses1.MyId = ( SELECT MIN(MyId) FROM #Addresses Addresses2 WHERE Addresses2.[Name] = Addresses1.[Name] ) DROP TABLE #Addresses
I think this is a good candidate for a cursor based solution. It's been so long since I've used a cursor that I won't attempt to write the T-SQL but here's the idea:
- Create temp table with same schema as Addresses
- Select distinct Names into cursor
- Loop through cursor selecting top 1 from Addresses into temp table for each distinct Name
- Return select from temp table
I don't think that you can do that, given your constraints. You can pull out distinct combinations of those fields. But if someone spelled Bob and Bobb with the same address you'd end up with two records. [GIGO] You are correct that any grouping (short of grouping on all of the fields-equivalent to DISTINCT) will mix rows. It's too bad that you don't have a unique identifier for each customer.
You might be able to nest queries together in such as way as to select the top 1 for each name and join all of those together.
A slight modification on the above should work.
SELECT Name, Street, City, State FROM table t INNER JOIN ( SELECT Name, MIN(Street) AS Street FROM table m GROUP BY Name ) x ON x.Name = t.Name AND x.Street = t.Street
Now this won't work if you have the same street but the other pieces of information are different (e.g. with typos).
OR a more complete hash would include all the fields (but you likely have too many for performance):
SELECT Name, Street, City, State FROM table t INNER JOIN ( SELECT Name, MIN(Street + '|' + City + '|' + State) AS key FROM table m GROUP BY Name ) x ON x.Name = t.Name AND x.key = Street + '|' + City + '|' + State
SELECT name, ( SELECT TOP 1 street, city, state FROM addresses b WHERE a.name = b.name ) FROM addresses a GROUP BY name
And still another way:
-- build a sample table DECLARE @T TABLE (Name VARCHAR(50),Street VARCHAR(50),City VARCHAR(50),State VARCHAR(50)) INSERT INTO @T SELECT 'Bob','123 Fake Street','Peoria','IL' UNION SELECT 'Bob','234 Other Street','Fargo','ND' UNION SELECT 'Jim','345 Main Street','St Louis','MO' UNION SELECT 'Fred','234 Other Street','Fargo','ND' -- here is all you do to get the unique record SELECT * FROM @T a WHERE (SELECT COUNT(*) FROM @T b WHERE a.Name = b.name and a.street <= b.street) = 1
select c.*, b.* from companies c left outer join (SELECT *, ROW_NUMBER() OVER(PARTITION BY FKID ORDER BY PKId) AS Seq FROM Contacts) b on b.FKID = c.PKID and b.Seq = 1
SELECT name, street, address, state FROM (SELECT name, street, address, state, DENSE_RANK() OVER (PARTITION BY name ORDER BY street DESC) AS r FROM tbl) AS t WHERE r = 1;