En fait, c'est VRAIMENT SIMPLE en utilisant du vieux SQL. Utilisez simplement les ET au niveau du bit. J'étais un peu étonné qu'il n'y ait pas de solution simple publiée en ligne (qui n'impliquait pas les UDF). Dans mon cas, je voulais vraiment vérifier si les bits étaient activés ou désactivés (les données proviennent de dotnet eNums).
En conséquence, voici un exemple qui vous donnera séparément et ensemble - des valeurs de bit et une chaîne binaire (la grande union est juste une façon hacky de produire des nombres qui fonctionneront sur les bases de données :
select t.Number
, cast(t.Number & 64 as bit) as bit7
, cast(t.Number & 32 as bit) as bit6
, cast(t.Number & 16 as bit) as bit5
, cast(t.Number & 8 as bit) as bit4
, cast(t.Number & 4 as bit) as bit3
, cast(t.Number & 2 as bit) as bit2
,cast(t.Number & 1 as bit) as bit1
, cast(cast(t.Number & 64 as bit) as CHAR(1))
+cast( cast(t.Number & 32 as bit) as CHAR(1))
+cast( cast(t.Number & 16 as bit) as CHAR(1))
+cast( cast(t.Number & 8 as bit) as CHAR(1))
+cast( cast(t.Number & 4 as bit) as CHAR(1))
+cast( cast(t.Number & 2 as bit) as CHAR(1))
+cast(cast(t.Number & 1 as bit) as CHAR(1)) as binary_string
--to explicitly answer the question, on MSSQL without using REGEXP (which would make it simple)
,SUBSTRING(cast(cast(t.Number & 64 as bit) as CHAR(1))
+cast( cast(t.Number & 32 as bit) as CHAR(1))
+cast( cast(t.Number & 16 as bit) as CHAR(1))
+cast( cast(t.Number & 8 as bit) as CHAR(1))
+cast( cast(t.Number & 4 as bit) as CHAR(1))
+cast( cast(t.Number & 2 as bit) as CHAR(1))
+cast(cast(t.Number & 1 as bit) as CHAR(1))
,
PATINDEX('%1%', cast(cast(t.Number & 64 as bit) as CHAR(1))
+cast( cast(t.Number & 32 as bit) as CHAR(1))
+cast( cast(t.Number & 16 as bit) as CHAR(1))
+cast( cast(t.Number & 8 as bit) as CHAR(1))
+cast( cast(t.Number & 4 as bit) as CHAR(1))
+cast( cast(t.Number & 2 as bit) as CHAR(1))
+cast(cast(t.Number & 1 as bit) as CHAR(1) )
)
,99)
from (select 1 as Number union all select 2 union all select 3 union all select 4 union all select 5 union all select 6
union all select 7 union all select 8 union all select 9 union all select 10) as t
Produit ce résultat :
num bit7 bit6 bit5 bit4 bit3 bit2 bit1 binary_string binary_string_trimmed
1 0 0 0 0 0 0 1 0000001 1
2 0 0 0 0 0 1 0 0000010 10
3 0 0 0 0 0 1 1 0000011 11
4 0 0 0 1 0 0 0 0000100 100
5 0 0 0 0 1 0 1 0000101 101
6 0 0 0 0 1 1 0 0000110 110
7 0 0 0 0 1 1 1 0000111 111
8 0 0 0 1 0 0 0 0001000 1000
9 0 0 0 1 0 0 1 0001001 1001
10 0 0 0 1 0 1 0 0001010 1010
