QUIMFV ... mira lo probe .. pero me tira error .. te dejo codigo terminado :
Código:
SELECT if(cal_calendario.tarifa_normal<"100","de $0 a $100",
* * * * * * * * * if(cal_calendario.tarifa_normal<"500","de $100 a $500",
* * * * * * * * * * * * * *if(cal_calendario.tarifa_normal<"1000","de $500 a $1000",
* * * * * * * * * * * * * * * * * *if(cal_calendario.tarifa_normal>"1000","mas de $1000")))) as rango, count(cal_calendario.tarifa_normal) FROM alojamientos , habitaciones, cal_calendario, ciudades WHERE alojamientos.tipo NOT LIKE "ca%" AND habitaciones.alojar_id = alojamientos.alojar_id AND cal_calendario.id_habitacion = habitaciones.hab_id AND habitaciones.pax_max >= "$pax1" AND cal_calendario.tarifa_normal > "0" AND cal_calendario.fecha ="$fechabus11" AND cal_calendario.estancia_minima <= "$dias_diferencia1" AND ciudades.Country = alojamientos.pais AND ciudades.Subdivision = alojamientos.provincia AND ciudades.Location = alojamientos.ciudad
GROUP BY *if(cal_calendario.tarifa_normal<100,"de $0 a $100",
* * * * * * * * * * if(cal_calendario.tarifa_normal<500,"de $100 a $500",
* * * * * * * * * * * * * *if(cal_calendario.tarifa_normal<1000,"de $500 a $1000",
* * * * * * * * * * * * * * * * * if(cal_calendario.tarifa_normal>1000,"mas de $1000"))))
y el error es :
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')))) as rango, count(cal_calendario.tarifa_normal) FROM alojamientos , habitacio' at line 4
espero tu ayuda