问题
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.
Behold, [Addresses]:
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.
Any ideas?
回答1:
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
回答2:
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.
回答3:
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.
回答4:
select distinct Name , street,city,state
from table t1 where street =
(select min(street) from table t2 where t2.name = t1.name)
回答5:
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
回答6:
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
回答7:
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
回答8:
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
回答9:
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.
回答10:
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
回答11:
SELECT name,
( SELECT TOP 1 street, city, state
FROM addresses b
WHERE a.name = b.name )
FROM addresses a
GROUP BY name
回答12:
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
回答13:
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
回答14:
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;
来源：https://stackoverflow.com/questions/982932/sql-query-to-return-one-single-record-for-each-unique-value-in-a-column