Order by specific values in a column without using case statement - sql

Question

I would like to get the records in the below format:
if i have a record like
A, B, C, D
and I would like get record in this order -
B, A, C, D, E, F, G, H, so on,
But I need the value B should be at the first row...

Answer 1

try this:
SELECT
*, 1 AS SortBy
FROM YourTable
WHERE YourCol='B'
UNION ALL
SELECT
*, 2 AS SortBy
FROM YourTable
WHERE YourCol!='B'
ORDER BY SortBy, YourCol
You don't give any reason to not want to use CASE. I'd still give it a try and see which is faster, the UNION ALL or the CASE method:
SELECT
*
FROM YourTable
ORDER BY CASE WHEN YourCol='B' then 1 ELSE 2 END, YourCol
EDIT Working example:
DECLARE #YourTable table (YourCol char(1), RowValue varchar(5))
INSERT #YourTable VALUES ('A','aaa')
INSERT #YourTable VALUES ('A','aa')
INSERT #YourTable VALUES ('B','bbb')
INSERT #YourTable VALUES ('B','bb')
INSERT #YourTable VALUES ('C','ccc')
INSERT #YourTable VALUES ('D','ddd')
INSERT #YourTable VALUES ('E','eee')
INSERT #YourTable VALUES ('F','fff')
SELECT
*, 1 AS SortBy
FROM #YourTable
WHERE YourCol='B'
UNION ALL
SELECT
*, 2 AS SortBy
FROM #YourTable
WHERE YourCol!='B'
ORDER BY SortBy, YourCol
OUTPUT:
YourCol RowValue SortBy
------- -------- -----------
B bbb 1
B bb 1
A aaa 2
A aa 2
C ccc 2
D ddd 2
E eee 2
F fff 2
(8 row(s) affected)

Answer 2

SELECT * from mytable where mycolumn = "B";
followed by
SELECT * from mytable where mycolumn != "B" order by mycolumn asc;

Answer 3

Declare and populate table:
DECLARE #t TABLE (col1 CHAR)
INSERT #t
SELECT char(number+ 65)
FROM master..spt_values
WHERE type = 'P' AND number < 6
Query1:
SELECT *, cast(ascii(col1)-66 as bit) * 2 + ascii(col1) [orderby]
FROM #t
ORDER BY cast(ascii(col1)-66 as bit) * 2 + ascii(col1)
Query2:
SELECT *
FROM #t
ORDER BY replace(col1, 'B', ' ')
Result for Query1: (the [orderby] column is included for documentation only)
col1 orderby
---- --------
B 66
A 67
C 69
D 70
E 71
F 72