@Jordan, Luke M
Hi!
I did this:
- formatted cell A1 as text
- entered Jordan's formula in cell B1
- entered Luke M's formula in cell C1
- typed "12/345" (unquoted) in cell A1
- retrieved both FALSE in B1:C1
Am I doing something wrong?
Regards!
PS: Just for the records... who in the hell was supposing you were using cell D6 instead of A1???
EDITED:
PS2: I didn't like something from the beginning and remembered that once got into same error, which now I don't realize what it was... But now yes!
The above formulas don't work for this case:
"1 /34 " (unquoted)
as Excel retrieves as number a string with trailing and leading spaces from a VALUE function.
So I propose to change the condition to:
=Y(LARGO(A1)=6;EXTRAE(A1;3;1)="/";NO(ESTEXTO(IZQUIERDA(A1;2)));NO(ESTEXTO((DERECHA(A1;3))))) -----> in english: =AND(LEN(A1)=6,MID(A1,3,1)="/",NOT(ISTEXT(LEFT(A1,2))),NOT(ISTEXT((RIGHT(A1,3)))))
Regards!
PS3: Yeah, I know I used A1
EDITED: It doesn't work for correct values, I think I should include LEN somewhere... be back in a while.
Back...
=Y(LARGO(A1)=6;EXTRAE(A1;3;1)="/";ESNUMERO(VALOR(IZQUIERDA(A1;2)));LARGO(ESPACIOS(IZQUIERDA(A1;2)))=2;ESNUMERO(VALOR(DERECHA(A1;3)));LARGO(ESPACIOS(DERECHA(A1;3)))=3) -----> in english: =AND(LEN(A1)=6,MID(A1,3,1)="/",ISNUMBER(VALUE(LEFT(A1,2))),LEN(TRIM(LEFT(A1,2)))=2,ISNUMBER(VALUE(RIGHT(A1,3))),LEN(TRIM(RIGHT(A1,3)))=3)
I think it works now. It isn't the way I did it that time but I don't remember how... but if works now, who cares? Does it work?
Regards!