Ordered Value by Column / Row

Question

Its a little difficult to explain. It might be easier to skip to the examples.

A table has an id and four columns that each allow null.

ID, Col1, Co         


        

Answer 1

Use:

SELECT DISTINCT COL1 AS col
  FROM YOUR_TABLE
 WHERE col1 IS NOT NULL
UNION
SELECT DISTINCT COL2 AS col
  FROM YOUR_TABLE
 WHERE col2 IS NOT NULL
UNION 
SELECT DISTINCT COL3 AS col
  FROM YOUR_TABLE
 WHERE col3 IS NOT NULL
UNION
SELECT DISTINCT COL4 AS col
  FROM YOUR_TABLE
 WHERE col4 IS NOT NULL
ORDER BY col

UNION will remove duplicates between the statements; DISTINCT will return a unique list of values per statement. UNION ALL would be faster than UNION, but it doesn't remove duplicates.

Answer 2

Try this one

With MyTables as
(
SELECT  [Col1] as [ColX] FROM [Test].[dbo].[MyTable] 
Union 
SELECT  [Col2] as [ColX]   FROM [Test].[dbo].[MyTable]
Union
SELECT  [Col3] as [ColX]   FROM [Test].[dbo].[MyTable]
Union
SELECT  [Col4] as [ColX]   FROM [Test].[dbo].[MyTable]
)
select ColX from MyTables where ColX is not null

Answer 3

select value from (    
    select col1 as value from table_name where col1 is not null
     union
    select col2 as value from table_name where col2 is not null
     union
    select col3 as value from table_name where col3 is not null
     union
    select col4 as value from table_name where col4 is not null     
) order by value

Answer 4

I may not understand everything that you described you wanted. From reading your problem and comments from others, I am guessing that this is what you are looking for:

Updated version:

with cteOriginal as
(
    select *, RANK() over (partition by [SortOrder] order by id asc) as [NonUniqueSortOrder]
    from
    (
        select id, A as [value], 1 as [SortOrder]
        from #original
        where A is not null
        union all
        select id, B as [value], 2 as [SortOrder]
        from #original
        where B is not null
        union all
        select id, C as [value], 3 as [SortOrder]
        from #original
        where C is not null
        union all
        select id, D as [value], 4 as [SortOrder]
        from #original
        where D is not null
    ) as temp
)
select [value]
from cteOriginal
where id = (select MIN(tmp.id) from cteOriginal tmp where tmp.value = cteOriginal.value)
order by ((([NonUniqueSortOrder] - 1) * 4) + [SortOrder])

I got rid of the duplicate values by picking the one with the smallest id, min(id). You can change it to use max(id).

Initial version:

with cteOriginal as
(
    select *, RANK() over (partition by [column] order by id asc) as [NonUniqueSortOrder]
    from
    (
        select id, A as [value], 'A' as [Column], 1 as [SortOrder]
        from #original
        where A is not null
        union all
        select id, B as [value], 'B' as [Column], 2 as [SortOrder]
        from #original
        where B is not null
        union all
        select id, C as [value], 'C' as [Column], 3 as [SortOrder]
        from #original
        where C is not null
        union all
        select id, D as [value], 'D' as [Column], 4 as [SortOrder]
        from #original
        where D is not null
    ) as temp
)
select [value]
from cteOriginal
order by ((([NonUniqueSortOrder] - 1) * 4) + [SortOrder])

By the way, I am using mssql 2005 for this query. Please comment and we'll refine it.