Diferencia del servidor SQL (opuesto a la intersección)
Si busca la forma más fácil / escalable de hacer una "diferencia" establecida en SQL Server, consulte a continuación.
Si no puede distinguir de la imagen, estoy buscando todo lo que no está en la intersección.
He visto una forma de hacerlo:
select * from (
(select 'test1' as a, 1 as b)
union all
(select 'test2' as a , 2 as b union all select 'test1' as a , 1 as b )
)un group by a,b having count(1)=1
Pero me temo lo que sucedería si usara dos conjuntos grandes (no consultaré desde declaraciones constantes select '', mis consultas se extraerán de tablas reales).
EDITAR:
Solución posible...
drop table #temp_a;
drop table #temp_b;
go
select * into #temp_a from (
select 1 as num, 'String' as two, 'int'as three, 'purple' as four union all
select 2 as num, 'dog' as two, 'int'as three, 'purple' as four union all
select 3 as num, 'dog' as two, 'int'as three, 'cat' as four ) a
select * into #temp_b from (
select 1 as num, 'String' as two, 'decimal'as three, 'purple' as four union all
select 2 as num, 'dog' as two, 'int'as three, 'purple' as four union all
select 3 as num, 'dog' as two, 'int'as three, 'dog' as four ) b
SELECT IsNull(a.num, b.num) A,IsNull(a.two, b.two) B, IsNull(a.three, b.three) C,
IsNull(a.four, b.four) D
FROM #temp_a a
FULL OUTER JOIN #temp_b b ON (a.num=b.num AND a.two=b.two and a.three=b.three and a.four=b.four)
WHERE (a.num is null or b.num is null )
RESULTADOS
1 cuerda int morado
3 perros int cat
1 cuerda dec púrpura
3 perro int perro