servidores - sql server data types c# equivalents

Tipo de datos para System.Version en servidor sql

Question

¿Cuál es la mejor manera de almacenar System.Version en SQL Server?

Cuando uso varchar type, el resultado de order by asc es:

1.0.0.0
11.0.0.0
12.0.0.0
2.0.0.0

Answer 1

Para admitir la ordenación entre versiones de longitudes mixtas (por ejemplo, ''1.2'' vs ''1.2.3.4''), se puede realizar una asignación a un decimal (como funciones con valores de tabla en línea).

create function Common.ufn_OrderableVersion(@pVersion nvarchar(100))
returns table
as
/*---------------------------------------------------------------------------------------------------------------------
Purpose: Provide a mapping from Versions of the form ''a.b.c.d'', ''a.b.c, ''a.b'', ''a'', null to
an orderable decimal(25, 0)
Since Parsename() doesn''t apply easily to mixed length comparisions (1.2 vs 1.2.3.4)
Test Cases:
select * from Common.ufn_OrderableVersion(null); -- null
select * from Common.ufn_OrderableVersion(''0''); -- 1000000000000000000000000
select * from Common.ufn_OrderableVersion(''1''); -- 1000001000000000000000000
select * from Common.ufn_OrderableVersion(''1.2.3.4''); -- 1000001000002000003000004
select Version
from
(
select ''1.3.5.3'' as Version
union all
select ''1.2.5.3'' as Version
union all
select ''1.1.5.3'' as Version
union all
select ''1.3.5.2'' as Version
union all
select null as Version
union all
select '''' as Version
union all
select ''2'' as Version
union all
select ''1.2'' as Version
union all
select ''1'' as Version
) v
order by (select Value from Common.ufn_OrderableVersion(Version))
Modified By Description
---------- -------------- ---------------------------------------------------------------------------------------
2015.08.24 crokusek Initial Version
---------------------------------------------------------------------------------------------------------------------*/
return
-- 25 = 1 + VersionPositions * MaxDigitsPerSegment
select convert(decimal(25,0), ''1'' +
stuff((select format(Value, ''000000'')
from
(
select convert(int, Value) as Value, RowNumber
-- Support empty string and partial versions. Null maps to null
from Common.ufn_SplitUsingXml(@pVersion + ''.0.0.0.0'', ''.'') -- pad right
where RowNumber <= 4 -- trim right
) as v
order by RowNumber
for xml path ('''')
), 1, 0, '''')
) as Value
go

Dependencia:

create function Common.ufn_SplitUsingXml
(
@pList nvarchar(max),
@pDelimiter nvarchar(255)
)
returns table
as
/*---------------------------------------------------------------------------------------------------------------------
Purpose: Split an Identifier using XML as an inline table valued function.
Using the SQL Server CLR (C#) capability would be the most efficient way to support this.
Warnings: Will not work if the input contains special XML characters like ''<'', ''>'' or ''&''.
Caller must add "option (maxrecursion 0)" for lists greater than 100 (it can''t be added within the ufn)
Modified By Description
---------- -------------- ---------------------------------------------------------------------------------------
2015.08.24 inet http://sqlperformance.com/2012/07/t-sql-queries/split-strings
---------------------------------------------------------------------------------------------------------------------*/
return
(
select Value = y.i.value(''(./text())[1]'', ''nvarchar(4000)''),
row_number() over (order by (select null)) as RowNumber
from
(
select x = convert(XML, ''<i>''
+ replace(@pList, @pDelimiter, ''</i><i>'')
+ ''</i>'').query(''.'')
) AS a cross apply x.nodes(''i'') AS y(i)
-- option (maxrecursion 0) must be added by caller for lists greater than 100
);
go

Comparación:

alter function Common.ufn_CompareVersions
(
@pVersionA nvarchar(100),
@pVersionB nvarchar(100)
)
returns table
as
/*---------------------------------------------------------------------------------------------------------------------
Purpose: Compare Version of the form ''A.B.C.D''.
Comparing versions of different lengths is also supported ''A.B''.
Test Cases:
select Result from Common.ufn_CompareVersions(''1'', null) -- 1
select Result from Common.ufn_CompareVersions(null, ''1'') -- -1
select Result from Common.ufn_CompareVersions(''1'', ''1'') -- 0
select Result from Common.ufn_CompareVersions(''1'', ''2'') -- -1
select Result from Common.ufn_CompareVersions(''2'', ''1'') -- 1
select Result from Common.ufn_CompareVersions(''1'', ''1.2'') -- -1
select Result from Common.ufn_CompareVersions(''1.2'', ''1'') -- 1
select Result from Common.ufn_CompareVersions(''1.2.3.4'', ''1.2.3.4'') -- 0
select Result from Common.ufn_CompareVersions(''1.2.3'', ''1.2.3.4'') -- -1
select Result from Common.ufn_CompareVersions(''1.2.3.4'', ''1.2.3'') -- 1
select Result from Common.ufn_CompareVersions(''1.9.3.4'', ''1.2.3.4'') -- 1
select Result from Common.ufn_CompareVersions(''1.2.3.4'', ''1.9.3.4'') -- -1
select Result from Common.ufn_CompareVersions(''1.002'', ''1.2'') -- 0
select Result from Common.ufn_CompareVersions(''1.2'', ''1.2.0'') -- 0
Modified By Description
---------- ----------- ------------------------------------------------------------------------------------------
2015.08.24 crokusek Initial Version
---------------------------------------------------------------------------------------------------------------------*/
return
with Compares as
(
select (select IsNull(Value, 0) from Common.ufn_OrderableVersion(@pVersionA)) as A,
(select IsNull(Value, 0) from Common.ufn_OrderableVersion(@pVersionB)) as B
)
select case when A > B then 1
when A < B then -1
else 0
end as Result
from Compares
go

Answer 2

Solo guárdelo como una varchar normal, lo cual es bueno para versiones de hasta 4 partes usando PARSENAME para dividir la cadena y ordenar en 4 columnas separadas.

es decir

ORDER BY PARSENAME(version,4),
PARSENAME(version,3),
PARSENAME(version,2),
PARSENAME(version,1)

Answer 3

puedes usar una columna varchar

podrías ordenar así

SELECT *
FROM t_version
ORDER BY CAST(''/'' + vid + ''/'' AS HIERARCHYID)

El violín de SQL no está funcionando hoy, de lo contrario podría haber mostrado una demostración

Por favor, ejecuta esto para la prueba

 SELECT * FROM
( VALUES
( ''1.0.0.0'' ),
( ''11.0.0.0'' ),
(''12.0.0.0''),
(''2.0.0.0'') ) AS vid ( vid )
ORDER BY CAST(''/'' + vid + ''/'' AS HIERARCHYID)