# How to split a string

There is an ABCD I would like to split and put into an array with elemets A,B,C,D. How to do this universally so next time if I have an ABCDE just to redim the array and get an array whose elements are A,B,C,D,E
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Commented:
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

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

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

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Author Commented:
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
0
2nd Line Desktop SupportCommented:
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
0
2nd Line Desktop SupportCommented:
You basically need what dr binks suggested :)
0
Author Commented:
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)
0
Author Commented:
It says Invalid call procedure or argument in the

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

I dont know if that was a lypo... but it should be:

stringvar = "ABCD"   << you need the " " other wise it takes the value out of the variable called 'ABCD'
For x = 0 To Len(stringvar)
strarray(x) = Mid(stringvar, x, 1)
0

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Author Commented:
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?
0
Commented:
dr_binks :
It will still say Invalid call procedure or argument and that is because you begin the loop from 0 (zero) instead of 1 (one).
Mid cannot accept 0 (zero) as argument. So your loop should be:

For x = 1 To Len(stringvar)

S
0
Commented:
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 :/ )
0
Commented:
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
0
Author Commented:
Just a question- what means
Dim strarray(1000)
0
Commented:
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
0
Author Commented:
Thanx a lot guys.
0
Commented:
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
0
Commented:
>>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
0
Commented:
although it is a common mistake, in most other languages it would infact be only 1000 elements.
0
Commented:
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
0
Commented:
>>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
0
Commented:
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?
0
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