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order by when there is nothing natural to order by

Posted on 2015-02-19
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Last Modified: 2015-02-20
My sample data looks something like this


1 aaaa
2 zzzz
3 dddd

there is not a natural way of ordering so i can't order by the id or the name.  Theirfore, my query needs to determine the order so it would returns something like

2 zzz
1 aaa
3 ddd

Where i would determine in my SQL what the order should be/
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Question by:vbnetcoder
6 Comments
 
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Expert Comment

by:Jim Horn
ID: 40619819
>it would returns something like
Define for us the logic behind how that sort order is achieved.
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Expert Comment

by:Mike Eghtebas
ID: 40619822
Add a new column (SortOrder Varchar(5)) where you can enter 01, 02, etc. to sort your data using this column.

Mike
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Jim Horn earned 500 total points
ID: 40619825
WIthout any further details, you can use CASE in an ORDER BY clause to pull off a custom sort order.
For example...
CREATE TABLE #tmp (val varchar(10)) 
INSERT INTO #tmp (val) 
VALUES ('1 aaa'), ('2 zzz'), ('3 ddd'), ('goo'), ('foo'), ('boo')

SELECT val
FROM #tmp
ORDER BY CASE LEFT(val, 1) WHEN '2' THEN 1 WHEN '1' THEN 2 WHEN '3' THEN 3 ELSE 4 END

Open in new window


For more reading check out my article on SQL Server CASE Solutions, scroll down about halfway.
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Expert Comment

by:Dave Baldwin
ID: 40619859
'order by' works with some column or field in the result set.  Either method above would work Ok.  The biggest problem is that if you are going to use some order that is not numeric or alpha, then you have to manually define that order and find a way to use it.  On a large table, that can get pretty time consuming if it changes very often.
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Expert Comment

by:PortletPaul
ID: 40620187
well I don't agree there isn't a "natural way to order" that sample data,
  it would either be numeric by id:  ORDER BY ID
  or alphabetic by name then numeric by id: ORDER BY NAME, ID

what you fail to explain at all is the "unnatural" method of order that results in:

2 zzz
1 aaa
3 ddd

that is neither numeric nor alphabetic, and without any explained logic

perhaps: ORDER BY CHECKSUM(NEWID())

(a random order)
using this 4 times produced these results:
| ID | NAME |
|----|------|
|  3 | dddd |
|  1 | aaaa |
|  2 | zzzz |


| ID | NAME |
|----|------|
|  2 | zzzz |
|  1 | aaaa |
|  3 | dddd |


| ID | NAME |
|----|------|
|  1 | aaaa |
|  2 | zzzz |
|  3 | dddd |


| ID | NAME |
|----|------|
|  1 | aaaa |
|  3 | dddd |
|  2 | zzzz |

Open in new window


details
[SQL Fiddle][1]

**MS SQL Server 2008 Schema Setup**:

    
    
    CREATE TABLE Table1
    	([ID] int, [Name] varchar(40))
    ;
    	
    INSERT INTO Table1
    	([ID], [Name])
    VALUES
    	(1, 'aaaa'),
    	(2, 'zzzz'),
    	(3, 'dddd')
    ;

**Query 1**:

    select
    id, name
    from table1
    
    ORDER BY CHECKSUM(NEWID())
    

**[Results][2]**:
    
    | ID | NAME |
    |----|------|
    |  3 | dddd |
    |  1 | aaaa |
    |  2 | zzzz |


**Query 2**:

    select
    id, name
    from table1
    
    ORDER BY CHECKSUM(NEWID())
    

**[Results][3]**:
    
    | ID | NAME |
    |----|------|
    |  2 | zzzz |
    |  1 | aaaa |
    |  3 | dddd |


**Query 3**:

    select
    id, name
    from table1
    
    ORDER BY CHECKSUM(NEWID())
    

**[Results][4]**:
    
    | ID | NAME |
    |----|------|
    |  1 | aaaa |
    |  2 | zzzz |
    |  3 | dddd |


**Query 4**:

    select
    id, name
    from table1
    
    ORDER BY CHECKSUM(NEWID())
    

**[Results][5]**:
    
    | ID | NAME |
    |----|------|
    |  1 | aaaa |
    |  3 | dddd |
    |  2 | zzzz |



  [1]: http://sqlfiddle.com/#!3/c8427/1

  [2]: http://sqlfiddle.com/#!3/c8427/1/0

  [3]: http://sqlfiddle.com/#!3/c8427/1/1

  [4]: http://sqlfiddle.com/#!3/c8427/1/2

  [5]: http://sqlfiddle.com/#!3/c8427/1/3

Open in new window

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Author Closing Comment

by:vbnetcoder
ID: 40620948
ty
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