Counting digits, extracting numbers (resolved)

[mikeycorn]

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.

[phinds]

check out the "val" function

[mikeycorn]

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

[mikeycorn]

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

[MarkT]

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

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

[mikeycorn]

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.

[Arc]

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

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

[mikeycorn]

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.

[Arc]

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.

[macai]

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

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.

[mikeycorn]

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

[mikeycorn]

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?)

[mikeycorn]

Hey Mikey, try IsNumeric.

Hey, thanks Mikey!