Natural Statements - MASK or IS

, ,
1

Hi,

I am trying to find a variable length by using MASK parameter, but its not working.

eg:
Local
01 #INPUT (a8)
end-define

IF #INPUT EQ MASK(AAAAAAAA)
write stmts
else
write stmts
END-IF

The #INPUT value may have 5 or 8 byte of alpha numeric data. If its 5 byte of data, then it should fall on else part. But its not happening. Kindly advice

2

#INPUT (a 8)

3

#INPUT (a 8 )

4

To get the length of the contents of a variable use EXAMINE.

DEFINE DATA LOCAL
1 #A (A8)   INIT <'12345'>
1 #L (I4)
END-DEFINE
EXAMINE #A FOR PATTERN '.' GIVING LENGTH #L
DISPLAY #A #L
END
Page     1                                                   02/04/19  09:25:11
 
   #A        #L
-------- -----------
 
12345              5
5

Thanks… Is it not possible do via MASK stmt

6

If valid values were restricted to alphabetic characters or numeric digits, then MASK could be used, as long as there were no leading or embedded spaces.

DEFINE DATA LOCAL
1 #A (A8) INIT <'AbCdEf'>
1 #L (I4)
END-DEFINE
IF  #A = MASK (AAAAAAAA)
  THEN
    WRITE 'length is 8'
  ELSE
    IF  #A = MASK (AAAAA'   ')
      THEN
        WRITE 'length is 5'
      ELSE
        WRITE 'length is bad'
    END-IF
END-IF
END
DEFINE DATA LOCAL
1 #A (A8) INIT <'1234567'>
1 #L (I4)
END-DEFINE
IF  #A = MASK (nnnnnnnn)
  THEN
    WRITE 'length is 8'
  ELSE
    IF  #A = MASK (nnnnn'   ')
      THEN
        WRITE 'length is 5'
      ELSE
        WRITE 'length is bad'
    END-IF
END-IF
END

But you can’t mix alpha and numeric because the mask for alphanumeric, C, includes ‘space’ as a valid character.

7

If there is a value of five characters are the last three characters always blanks?

If so, the most efficient solution would be:

1 #INPUT (a8)
1 redefine #input
2 filler 5x
2 #lastthree (a3)

:::
IF #lastthree = ’ ’ /* three blanks in the apostrophes
** guaranteed a five character value
else
guaranteed an eight character value
end-if

8

Thanks Ralph & Steve

9

Hi Steve,

Nope, the value on the field sometimes carries 3 or 5 or 8 byte of data and cant guarantee last characters with blank.

May be I can move the value to left justified and try it out

10

Hi Prakash,

Originally you posed your problem with just 5 or 8 characters. You just changed that to 3, 5, or 8.

So which is it , 5-8 or 3-5-8 ?

If the last three characters are not blank for a 5 character value, how can you differentiate 5 from 8 characters? Can the first three characters be blank and the next 5 characters the value?

If it is really a 3-5-8 problem and not a 5-8 problem, what are the possibilities?

Would a three character value always have 5 leading blanks or 5 trailing blanks? Or, could it be, for example, bbXXXbbb, where X is a character and b a blank.

Please clarify the problem.