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From: Bill Betournay on 19 Mar 2010 18:14
I've spent half the day today on try to figure out how to parse this string.

Here it is

I have a string like this: "1 1.1 1.2 1.3 1.4 1.5 1.6" and sometimes that
string may have more that one carriage return like I have in my example
between 1 & 1.1

What I need is to separate out each number like

1
1.1
1.2
1.3
1.4
1.5
1.6

I've been googling and sifting through help files looking for a way to parse
it out but I can't seem to find any. This has to be an easy task and I sure
would appreciate any help at all.

Thank you

Bill



From: Mike Williams on 19 Mar 2010 18:41
"Bill Betournay" <bill(a)REMOVEdatapacks.com> wrote in message
news:eGFx9F7xKHA.2012(a)TK2MSFTNGP04.phx.gbl...

> I've spent half the day today on try to figure out how
> to parse this string. Here it is. I have a string like this:
> "1 1.1 1.2 1.3 1.4 1.5 1.6" and sometimes that string
> may have more that one carriage return like I have in
> my example between 1 & 1.1

You mean more than one space?

> What I need is to separate out each number like 1
> 1.1
> 1.2 . . etc

There are loads of different ways of doing it. Here's one way that will
parse it no matter how many spaces there are between the individual values:

Private Sub Command1_Click()
Dim s1 As String, s2() As String, n As Long
s1 = "1 1.1 1.2 1.3 1.4 1.5 1.6"
Do
n = Len(s1)
s1 = Replace(s1, " ", " ")
Loop Until Len(s1) = n
s2 = Split(s1, " ")
' display the result:
For n = LBound(s2) To UBound(s2)
Print n, s2(n)
Next n
End Sub

Mike


From: dpb on 19 Mar 2010 18:44
Bill Betournay wrote:
....

> I have a string like this: "1 1.1 1.2 1.3 1.4 1.5 1.6" and sometimes that
> string may have more that one carriage return like I have in my example
> between 1 & 1.1

You don't show cr/lf pairs, only a couple spaces in your example.

You _really_ have a string w/ embedded crlf characters separating entries?

Even if do,

idx = instr(DataString, vbCrLf)

will return position of next match if exists, 0 if not. Put in a loop
until 0 or reach end of string, converting each subsection of nonzero
length on each pass.

I'm stuck w/ VB5 so don't recall if the VB6 Split() function would
handle this or not otomh...

--
From: Mike Williams on 19 Mar 2010 18:58
"Bill Betournay" <bill(a)REMOVEdatapacks.com> wrote in message
news:eGFx9F7xKHA.2012(a)TK2MSFTNGP04.phx.gbl...

> I've spent half the day today on try to figure out
> how to parse this string . . .

By the way, in the code I just posted there are the lines:

s1 = Replace(s1, " ", " ")
Loop Until Len(s1) = n
s2 = Split(s1, " ")

In the first of those three lines there are two spaces in the first set of
quotes and one space in the second set, and in the third of those three
lines there is one space in the set of quotes. So if you are typing it into
your code rather than pasting it in then make sure you type it correctly.
Otherwise, if you wish, you can change those three lines to the following so
that the code is more easily readable:

s1 = Replace(s1, Space$(2), Space$(1))
Loop Until Len(s1) = n
s2 = Split(s1, Space$(1))

Note that in the array in which the results are placed the individual items
are Strings, so in the rest of your code you might like to use Csng or some
other suitable function to convert them into Singles or whatever you are
using.

Mike



From: Nobody on 19 Mar 2010 20:15
What version of VB are you using?


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