How to implement a ZIP JOIN in T-SQL? - sql-server

Question

Let say I have table #Foo:
Id Color
-- ----
1 Red
2 Green
3 Blue
4 NULL
And table #Bar:
Value
-----
1
2.5
I would like to create table Result using simple statement to get:
Id Color Value
-- ---- -----
1 Red 1
2 Green 2.5
3 Blue NULL
4 NULL NULL
What I have invented so far is:
WITH cte1
AS
(
SELECT [Id], [Color], ROW_NUMBER() OVER (ORDER BY [Id]) AS 'No'
FROM #Foo
),
cte2
AS
(
SELECT [Value], ROW_NUMBER() OVER (ORDER BY [Value]) AS 'No'
FROM #Bar
)
SELECT [Id], [Color], [Value]
FROM cte1 c1
FULL OUTER JOIN cte2 c2 ON c1.[No] = c2.[No]
Do you know faster or more standard way to do ZIP JOIN in T-SQL?

Answer 1

You can simply try this.
;WITH CTE AS
(
SELECT ROW_NUMBER() OVER (ORDER BY (SELECT 1)) AS Id, Value FROM #Bar
)
SELECT F.Id, F.Color, CTE.Value
FROM #Foo F
LEFT JOIN CTE ON CTE.Id = F.Id

Answer 2

You can get rid of the CTE or make your query shorter
with subquery's like this
select Id,Color,Value from
(
SELECT [Id], [Color], ROW_NUMBER() OVER (ORDER BY [Id]) AS 'No'
FROM #Foo
)x full outer join
(
SELECT [Value], ROW_NUMBER() OVER (ORDER BY [Value]) AS 'No'
FROM #Bar
)y
on x.No=y.No

Answer 3

Will this suffice? (Admittedly, I may be miss-interpreting the question)
SELECT
F.ID AS ID,
F.Color AS Color,
B.Value AS Value
FROM #Foo F
LEFT OUTER JOIN #Bar B ON F.ID = FLOOR(B.Value)
--this DOES seem to return the correct output, but I'm not sure that my logic
--is what you are after
SELECT
F.ID AS ID,
F.Color AS Color,
B.Value AS Value
FROM
(
VALUES
(1,'Red'),(2,'Green'),(3,'Blue'),(4, NULL)
) AS F(ID, Color)
LEFT OUTER JOIN
(
VALUES
(1), (2.5)
) AS B(Value)
ON F.ID = FLOOR(B.Value)
Or are you wanting to essentially:
Sort #Foo by ID
Sort #Boo by Value
Match:
"First" row from #Foo with "First" row from #Bar
"Second" row from #Foo with "Second" row from #Bar
etc...
(Sorry, but I am not familiar with what a "ZIP JOIN" is.
I will look at the link provided by #RszardDzegan, though.)

Answer 4

you could try something like this:
DECLARE #Foo TABLE (Id INT, Color VARCHAR(10));
DECLARE #Bar TABLE (Value DECIMAL(2, 1))
INSERT INTO #Foo (Id, Color)
VALUES (1, 'Red'), (2, 'Green'), (3, 'Blue'), (4, NULL)
INSERT INTO #Bar (Value)
VALUES (1), (2.5);
WITH ECROSS
AS (
SELECT F.Id, F.Color, B.Value, DENSE_RANK() OVER (
ORDER BY F.Id
) AS No1, DENSE_RANK() OVER (
ORDER BY B.Value
) AS No2
FROM #Foo F, #Bar B
)
SELECT A.id, A.Color, B.Value
FROM ECROSS A
LEFT JOIN ECROSS B ON A.No1 = B.No2
AND A.No1 = B.No1
GROUP BY A.id, A.Color, B.Value

Answer 5

DECLARE #Foo TABLE (pk_id int identity(1,1), Id INT, Color VARCHAR(10));
DECLARE #Bar TABLE (pk_id int identity(1,1), Value DECIMAL(2, 1))
INSERT INTO #Foo (Id, Color)
VALUES (1, 'Red'), (2, 'Green'), (3, 'Blue'), (4, NULL)
INSERT INTO #Bar (Value)
VALUES (1), (2.5);
SELECT F.id, F.Color, B.Value
FROM #Foo F
LEFT JOIN #Bar B ON F.pk_id = B.pk_id

Answer 6

Try the following code. You just need to provide both data types in the same structure with a row number per group. With that you can use the PIVOT operator to produce the expected result.
WITH
CTE_FOO AS
(
SELECT
[Group]
,[Spread]
,[Aggregate]
FROM
(VALUES
(1, 1, N'Red' )
,(2, 1, N'Green')
,(3, 1, N'Blue' )
,(4, 1, NULL )
) AS FOO([Group], [Spread], [Aggregate])
),
CTE_BAR AS
(
SELECT
[Group]
,[Spread]
,CAST([Aggregate] AS nvarchar(max)) AS [Aggregate]
FROM
(VALUES
(1, 2, 1 )
,(2, 2, 2.5 )
) AS BAR([Group], [Spread], [Aggregate])
),
CTE_FOOBAR AS
(
SELECT [Group], [Spread], [Aggregate] FROM CTE_FOO
UNION ALL
SELECT [Group], [Spread], [Aggregate] FROM CTE_BAR
)
SELECT
[Group] AS [ID]
,[1] AS [Color]
,[2] AS [Value]
FROM
CTE_FOOBAR
PIVOT
(
MAX([Aggregate]) FOR [Spread] IN ([1], [2])
) AS PivotTable

Answer 7

You can skip creating new row numbers for #Foo, since its row numbers in this case are given.
Then the solution will become
SELECT F.Id,F.Color,newBar.Value from #Foo as F
LEFT JOIN
(
SELECT [Value], ROW_NUMBER() OVER (ORDER BY [Value]) AS 'No'
FROM #Bar
) newBar
on F.Id=newBar.No
This solution has been tested and proven. It gives you all values of #Foo and for each a sorted value of #Bar if there is one.