MySQL - select count, de dos columnas diferentes misma tabla.

pregunta 1 ---pregunta2---pregunta 3

      1         0           1

      0         1           1

select COUNT(pregunta1) as p1, + COUNT(pregunta2) as p2, + COUNT(pregunta3) as p3, + COUNT(pregunta4)       as p4

   from rs_calificacion_ticket

   where pregunta1 =1 or pregunta2=1 or pregunta3=1 or pregunta4=0

   group by pregunta1, pregunta2

select Id_calificacionticket, id_tiket, pregunta1, comentario1, pregunta2, comentario2, pregunta3, comentario3, pregunta4, comentario4, suma = pregunta1+ pregunta2 + pregunta3+ pregunta4

from rs_calificacion_ticket

;

with a as (select id = cast(null as char(1)) , pregunta1 = cast(null as bit)

, pregunta2 = cast(null as bit)

, pregunta3 = cast(null as bit)

, pregunta4 = cast(null as bit)

union

select 'a',1,1,0,1 union

select 'b',1,1,0,0

),

man as (select id, pregunta = 'pregunta1', valor = pregunta1 from a union

select id, pregunta = 'pregunta2', pregunta2 from a union

select id, pregunta = 'pregunta3', pregunta3 from a union

select id, pregunta = 'pregunta4', pregunta4 from a),

unpiv as ( -- Aqui bajo de normalización un poquito

SELECT id, pregunta, valor

FROM

   (SELECT id, pregunta1, pregunta2, pregunta3, pregunta4

   FROM a) p

UNPIVOT

   (valor fOR pregunta IN

      (pregunta1, pregunta2, pregunta3, pregunta4)

)AS unpvt

),

con_un_crossjoin AS ( -- solucion con un crossjoin

select * from (select si = count(valor)  from unpiv where valor = 1) [si]

cross join (select [no]= count(valor)  from unpiv where valor = 0) [no]

),

con_un_pivot as ( -- solución con la función PIVOT

select  [1] as [Si] , [0] as [No]

from  (

		select pregunta , valor from unpiv

		) p

pivot (

		count(pregunta) for valor in ([1],[0])

		) as pvt

)

-- resultados por ambos métodos...

select 'Usando un cross join ' as [opcion] , * from con_un_crossjoin

union

select 'Usando un UnPivot ', * from con_un_pivot