Всъщност това е НАИСТИНА ПРОСТО с помощта на обикновен стар SQL. Просто използвайте побитови И. Бях малко изумен, че няма просто решение, публикувано онлайн (което не включва UDF). В моя случай наистина исках да проверя дали битовете са включени или изключени (данните идват от dotnet eNums).
Съответно ето един пример, който ще ви даде поотделно и заедно - битови стойности и двоичен низ (големият съюз е просто хакерски начин за създаване на числа, които ще работят в DBs:
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
Получава следния резултат:
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