I am hoping that someone can offer help with this matter.

I have the following text that I need to break into separate columns to
get the sentences separated:
"1.2.3 This is an example. That I need to break down"

What I get when I do a texttocolumn using the "." as the separator:
"1" "2" "3" "This is an example" "That I need to
break down"

What I am trying to get is:
"1.2.3" "This is an example" "That I need to break down"

Is there someway of breaking the "1.2.3" out first?

Thanks for any help.
Troy Vincent

Dim s As String
Dim FirstPart As String
Dim OtherParts() As String

s = "1.2.3 This is an example. That I need to break down"
FirstPart = Split(s, " ", 2)(0)
OtherParts = Split(Split(s, " ", 2)(1), ".")

FirstPart returns a single string
OtherParts returns a (zero-based) array


"Viking" wrote:

I am hoping that someone can offer help with this matter.

I have the following text that I need to break into separate columns to
get the sentences separated:
"1.2.3 This is an example. That I need to break down"

What I get when I do a texttocolumn using the "." as the separator:
"1" "2" "3" "This is an example" "That I need to
break down"

What I am trying to get is:
"1.2.3" "This is an example" "That I need to break down"

Is there someway of breaking the "1.2.3" out first?

Thanks for any help.
Troy Vincent

Charlie

That worked great for the most part. When I ran a test to the text
that I am working with I found that the following happened.

when the numbers changed to something like "1.4.0 This is an example.
That I need to break down"

I get "1.4" "This is an example" "That I need to break down"

It is droping the .0
Any Ideas on how I would be able to keep it.
Thanks in advance
Troy Vincent

Charlie wrote:
Dim s As String
Dim FirstPart As String
Dim OtherParts() As String

s = "1.2.3 This is an example. That I need to break down"
FirstPart = Split(s, " ", 2)(0)
OtherParts = Split(Split(s, " ", 2)(1), ".")

FirstPart returns a single string
OtherParts returns a (zero-based) array


"Viking" wrote:

I am hoping that someone can offer help with this matter.

I have the following text that I need to break into separate columns to
get the sentences separated:
"1.2.3 This is an example. That I need to break down"

What I get when I do a texttocolumn using the "." as the separator:
"1" "2" "3" "This is an example" "That I need to
break down"

What I am trying to get is:
"1.2.3" "This is an example" "That I need to break down"

Is there someway of breaking the "1.2.3" out first?

Thanks for any help.
Troy Vincent

That's unusual. Could you post the exact code and data you are using? I
will be hitting the road shortly but someone else may be able to figure it
out.

"Viking" wrote:

Charlie

That worked great for the most part. When I ran a test to the text
that I am working with I found that the following happened.

when the numbers changed to something like "1.4.0 This is an example.
That I need to break down"

I get "1.4" "This is an example" "That I need to break down"

It is droping the .0
Any Ideas on how I would be able to keep it.
Thanks in advance
Troy Vincent

Charlie wrote:
Dim s As String
Dim FirstPart As String
Dim OtherParts() As String

s = "1.2.3 This is an example. That I need to break down"
FirstPart = Split(s, " ", 2)(0)
OtherParts = Split(Split(s, " ", 2)(1), ".")

FirstPart returns a single string
OtherParts returns a (zero-based) array


"Viking" wrote:

I am hoping that someone can offer help with this matter.

I have the following text that I need to break into separate columns to
get the sentences separated:
"1.2.3 This is an example. That I need to break down"

What I get when I do a texttocolumn using the "." as the separator:
"1" "2" "3" "This is an example" "That I need to
break down"

What I am trying to get is:
"1.2.3" "This is an example" "That I need to break down"

Is there someway of breaking the "1.2.3" out first?

Thanks for any help.
Troy Vincent