Counting digits, extracting numbers (resolved)

I can get the number of digits in an integer in two functions:

Digits = Len(Str(intX)) - 1

But is there a function that does this in one?



also . . .



I have various strings like:

1 day
2 days
14 days
365 days

And I want to extract just the numbers into another string. Is this the slickest way:

strNumber = Replace(strDays, " day", vbNullString)
strNumber = Replace(strDays, "s", vbNullString)



I know I'm quibbling when it could be said, "if it ain't broke... " but I'm just as concerned with creating streamlined code that's easy to read.

check out the "val" function

Val("21 days") won't do the trick.

Oh my God! It certainly does! :eek: (I honestly just assumed that was going to be a 0 returned there, thanks.) ;)

Val will search until it hits a space. Try both of these

VB Code:

MsgBox Val("21 Days")
MsgBox Val("day 21")

Aha. Interesting.

I'm embarrassed for not having tried it myself but I just kind of assumed you had to strip all the text out before using Val or it'd just be = 0.

There is no single function in VB that tells you the number of digits in an integer that i know of... But IMO you should be using Cstr and tehre is no need for the -1

VB Code:

Dim X As Integer
X = 1111
MsgBox Len(CStr(X))

Thanks, Arc. I read one time long ago to stay away from Variants if at all possible and it always stuck with me. Perhaps I should read up a bit more on them.

Why do you say that? were you using a variant?.. If so then you could just use Len(MyInt) and it would return the result.

Well, this function may work better if there is no space between the number and the rest of the string:

VB Code:

Public Function NumbersOnly(Expression As Variant) As Long
 Dim x As String
 x = Expression
 For i = 1 To 47
  x = Replace(x, Chr(i), "")
 Next
 For i = 58 To 255
  x = Replace(x, Chr(i), "")
 Next
 NumbersOnly = CLng(x)
End Function

It will remove EVERYTHING but the numbers themselves from the
string.

That's like the nuclear option. Very thorough indeed! Put that one in the code bank.

More in the same vein:

Here's what I want to do:

a$="9x99"

If a$ = val(a$) then print "all numbers"

Of course, what I'm getting is a type-mismatch.

So, what's the best way to see if a string is all numbers (can I avoid a loop?)

Hey Mikey, try IsNumeric.

Hey, thanks Mikey!