Hi, VB Experts:
I have a simple question.  Here is a string:  "1   2 3   4" and
I want to read it into four variables.  How do I do that?  The spaces vary from
string to string, but they only have four numbers.  Your help is
greatly appreciated.

Sincerely,

Frank

Frank,

I don't know if this "1 2 3 4 "thing is just an example or
if you want to do something more with expressions.
If you fi. also want 10 from "2*5" etc then you should
look at ABC\Algorithms. There are a few polish expression-
parsers(one by William Yiu himself)which basically does the
thing. Also on my homepage you can find several sources
adressing expressionparsing from a different point of view
(recursive).

Otherwise, if you only want to read stringintegers to integers
then there are a few easy measures to take:

1) Read in a string. When you reach a separator( in your case a space
or cr), you quit inputting. The inputstring can then be translated
using: a%=val(ltrim$(rtrim$(valstring$)))

For the way you have to read one key at the time you could use
Don's source in ABC\FAQs about INKEY.

Since probably you are not allowing for long integers/doubles, a good
program must also test the stringvalue if its in the bounds( you can
see some source for that in the opcode generator on my homepage).

This way it doesn't really matter how much spaces there are between
the integers.  If you want you can also ignore cr's when the last
integer is not reached with something like this:

nrints=10
dim intarray[nrints-1]
for i=0 to nrints-1 'say you need 10 integers.
     do: i$=getkey$: loop while iswhite(i$)
         instring$="":                
         do  
        intstring$=intstring$+i$
        i$=getkey$ 'see Don's sources..
         loop until i$=' ' or i$=chr$(13) or i$=chr$(10)
intarray[i]=val(ltrim$(rtrim$(intstring$)))
next                    

'As an exercise I left the  iswhite function.

Rick
[Now you tell me where in Visual Basic you put your
basic code( I only saw windows,buttons and the like ;))))

  If it's the end of the String,
     Exit the loop
 Otherwise
     Go back and reassign the String you just built
      and loop through again

Done!

DEFINT A-Z
DIM num(1 TO 4) AS INTEGER

CLS
test$ = "     123   35 67    5678       "
PRINT "The test char string is '"; test$; "'"

s$ = LTRIM$(RTRIM$(test$))   'strip leading and trailing spaces
s$ = s$ + " "                'add 1 trailing space for convenience

FOR n = 1 TO 4
    p = INSTR(s$, " ")  'space location behind present first "number"
    number$ = LEFT$(s$, p - 1)
    num(n) = VAL(number$)   'nth integer from string to array variable
    s$ = LTRIM$(RIGHT$(s$, LEN(s$) - p))  'deletes first "number"
NEXT n

PRINT : PRINT "The numbers entered are:": PRINT
FOR n = 1 TO 4
    PRINT "num("; n; ") =";
    PRINT USING "#####     "; num(n);
    PRINT "num("; n; ") MOD 10 = "; num(n) MOD 10
NEXT n

PRINT
PRINT "The MOD values show these are integers NOT char strings."

END

END

Try this Frank,

CLS : DIM v$(5): s$ = "1    23   45   6"

FOR i = 1 TO LEN(s$)

  m$ = MID$(s$, i, 1)

  IF m$ = " " THEN z = z + 1

  IF z = 1 THEN n = n + 1

  IF m$ <> " " THEN
     z = 0
     v$(n) = v$(n) + m$
  END IF

NEXT

FOR i = 0 TO n

  PRINT v$(i)        ' Listing the variables

NEXT

--

                                   is not at the mercy of
               *                   the one with the argument.  

Remove MAPSON if replying via email

Sub Parse$(Text$, Separator$, Array$())
NoOfWords=0
Do Until Instr(Text$, Separator$) = 0
        Place = Instr(Text$, Separator$)
        NoOfWords = NoOfWords + 1
        Array$(NoOfWords) = Left$(Place - 1)
        Text$=Mid$(Text$,Place+1)
Loop

NoOfWords = NoOfWords+1
Array$(NoOfWords)=Text$

End Sub

Rob Davis MSc MIAP
Anstey, Leicester UK. 0976 379489