How to split a string

the way I would do it is:

----------------------------

dim strarray() as string
dim realarray() as string

for x=0 to len(stringvar)
   strarray(x)=mid(stringvar,x,1)
next

redim realarray(x)

for y=0 to x
   realarray(y) = strarray(y)
next

--------------------

of course if your string had a delimiter like:

A B C D

you could use the split() function like so:

------------------------

dim strarray() as string
dim realarray() as string

strarray() = split(stringvar," ")

redim realarray(ubound(strarray()))

-------------------------

hope that helps

This doesn't work from some reason

Sub a()


Dim strarray() As String

stringvar = ABCD
For x = 0 To Len(stringvar)
   strarray(x) = Mid(stringvar, x, 1)

MsgBox strarray(0)
Next

End Sub

Dim strarray() As String

stringvar = ABCD
For x = 0 To Len(stringvar)
   strarray(x) = Mid(stringvar, x, 1)

MsgBox strarray(x) '<-- Changed this from 0 To x as your variable is x in your for next loop
Next

End Sub

Try that and let us know how it goes :)

I hope that helps you out :)

Kind regards

Gecko

You basically need what dr binks suggested :)

Well, it will not work because the MsgBox will display whateveris requested. It seem the actual problem is somewhere here.

stringvar = ABCD
For x = 0 To Len(stringvar)
   strarray(x) = Mid(stringvar, x, 1)

It says Invalid call procedure or argument in the

strarray(x) = Mid(stringvar, x, 1)

Sub a()


Dim strarray() As String

stringvar = "ABCD"
For x = 0 To Len(stringvar)
   strarray(x) = Mid(stringvar, x, 1)

MsgBox strarray(0)
Next

End Sub

Did you try this?

hehe, I just tried the code, I found this works:

Dim strarray(1000) As String

stringvar = "ABCD"
For x = 1 To Len(stringvar)
   strarray(x) = Mid(stringvar, x, 1)

MsgBox strarray(x)
Next

(forgot x has to start at 1 because there is no position 0 in a string, my bad :/ )

anyways, I have work tommorow and its getting quite late, im sure gecko_au2003 and Shauli  can help you out from here zzwin.
I shall check in the morning to see if there was a solution and if not I shall help you out :)

nite

Just a question- what means
 Dim strarray(1000)

That means that you fix the size of the array to 1000 elements. There are two ways to define an array: Either you define the amount of elemnts before you load the array, as in  Dim strarray(1000), or you extend the amount of elemnt while you add elemnts, as in:

Dim myArray() as string, Counter as integer

Redim Preserve myArray(Counter)
myArray(Counter) = anyvalue
Counter = Counter + 1

That way, you add an elemnt to the array "on the fly" instead fixing the size of the array.

S

Thanx a lot guys.

To implement the "on the fly" option to the suggested code:

Dim strarray() As String, Counter As Integer

stringvar = "ABCD"
For x = 1 To Len(stringvar)
   ReDim Preserve strarray(x)
   strarray(x) = Mid(stringvar, x, 1)
   Counter  = Counter  + 1
Next

S

>>That means that you fix the size of the array to 1000 elements.<<
Actually no, it is 1001.  Since an array defined as:
Dim strarray(1000) is zero based than that would be 1001 elements.

It is a common mistake, and is especially true in .NET

although it is a common mistake, in most other languages it would infact be only 1000 elements.

In general, acperkins would be correct. But not at this time, as if the asker chooses to go with Dim strarray(1000), and as the loop to add the elements starts from 1 (For x = 1 To Len(stringvar)), then in this case it is 1000 and not 1001.

S

>>But not at this time<<
Really? Than I am afraid to say, you have no clue.  Please take the time to read up on arrays and in particular LBound and UBound.  Here is a little exercise for you:

Dim strarray(1000)
Debug.Print "Elements in the array:"; UBound(strarray) - LBound(strarray) + 1

Of course if you had done this:
Dim strarray(1 to 1000)

Or even this:
Option Base 1
Dim strarray(1000)

Then you would of course be correct (both of those arrays have only 1000 eleements), but then I would not be posting here then, would I?