need to break up a string by character count

[Joe Reynolds]

say i have a string that is "hello this is my string and i like it very
much" and i want to insert a newline at every 20th character, how could i go
about doing this?

[Bob Barrows [MVP]]

Joe Reynolds wrote:

say i have a string that is "hello this is my string and i like it
very much" and i want to insert a newline at every 20th character,
how could i go about doing this?

I'm sure someone will post a regex solution for this, but until then:

dim buffer, i, newstring,s
s="hello this is my string and i like it very much"
buffer=20
for i = 1 to len(s) step 20
newstring=newst ring & mid(s,i,20) & vbcrlf
next
msgbox newstring
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[Joe Reynolds]

that just might work

:)
"Bob Barrows [MVP]" <re******@NOyah oo.SPAMcomwrote in message
news:%2******** *******@TK2MSFT NGP06.phx.gbl.. .

Joe Reynolds wrote:

>say i have a string that is "hello this is my string and i like it
very much" and i want to insert a newline at every 20th character,
how could i go about doing this?

I'm sure someone will post a regex solution for this, but until then:

dim buffer, i, newstring,s
s="hello this is my string and i like it very much"
buffer=20
for i = 1 to len(s) step 20
newstring=newst ring & mid(s,i,20) & vbcrlf
next
msgbox newstring
--
Microsoft MVP -- ASP/ASP.NET
Please reply to the newsgroup. The email account listed in my From
header is my spam trap, so I don't check it very often. You will get a
quicker response by posting to the newsgroup.

[Dave Anderson]

Bob Barrows [MVP] wrote:

dim buffer, i, newstring,s
s="hello this is my string and i like it very much"
buffer=20
for i = 1 to len(s) step 20
newstring=newst ring & mid(s,i,20) & vbcrlf
next
msgbox newstring

Won't that put an extra line break at the end, Bob?

--
Dave Anderson

Unsolicited commercial email will be read at a cost of $500 per message. Use
of this email address implies consent to these terms.

that doesnt hurt me

:)

"Dave Anderson" <NY**********@s pammotel.comwro te in message
news:%2******** ********@TK2MSF TNGP05.phx.gbl. ..

Bob Barrows [MVP] wrote:

>dim buffer, i, newstring,s
s="hello this is my string and i like it very much"
buffer=20
for i = 1 to len(s) step 20
newstring=newst ring & mid(s,i,20) & vbcrlf
next
msgbox newstring

Won't that put an extra line break at the end, Bob?

--
Dave Anderson

Unsolicited commercial email will be read at a cost of $500 per message.
Use of this email address implies consent to these terms.

[Bob Barrows [MVP]]

Dave Anderson wrote:

Bob Barrows [MVP] wrote:

>dim buffer, i, newstring,s
s="hello this is my string and i like it very much"
buffer=20
for i = 1 to len(s) step 20
newstring=newst ring & mid(s,i,20) & vbcrlf
next
msgbox newstring

Won't that put an extra line break at the end, Bob?

Ummmm ... errr. .. I left that as an exercise for the reader ... yeah,
that's the ticket!

--
Microsoft MVP - ASP/ASP.NET
Please reply to the newsgroup. This email account is my spam trap so I
don't check it very often. If you must reply off-line, then remove the
"NO SPAM"

[Anthony Jones]

"Bob Barrows [MVP]" <re******@NOyah oo.SPAMcomwrote in message
news:%2******** *******@TK2MSFT NGP06.phx.gbl.. .

Joe Reynolds wrote:

say i have a string that is "hello this is my string and i like it
very much" and i want to insert a newline at every 20th character,
how could i go about doing this?

I'm sure someone will post a regex solution for this, but until then:

Here's then:-

s = InsertStringAtI nterval(s, vbCrLf, 20))

Function InsertStringAtI nterval(rsSourc e, rsInsert, rlInterval)

Dim rgx
Set rgx = new RegExp
rgx.Pattern = "([\s\S]{" & rlInterval & "})"
rgx.Global = true

InsertStringAtI nterval = rgx.Replace(rsS ource, "$1" & rsInsert)

End Function
BTW Joe, I still think you should persue a different solution than this.
If you use this in the TD you will actually need:-

s = InsertStringAtI nterval(Server. HTMLEncode(s), "<br />", 20))

dim buffer, i, newstring,s
s="hello this is my string and i like it very much"
buffer=20
for i = 1 to len(s) step 20
newstring=newst ring & mid(s,i,20) & vbcrlf
next
msgbox newstring
--
Microsoft MVP -- ASP/ASP.NET
Please reply to the newsgroup. The email account listed in my From
header is my spam trap, so I don't check it very often. You will get a
quicker response by posting to the newsgroup.

[Joe Reynolds]

thank you. that works great, but i had another thought (posted a new
question also)
what i want to do is first check the users input. if i find any strings of
characters that are longer than 40 characters that dont contain a space THEN
i would pass it to your function. as long as there is at least 1 space for
every 40 chars, im ok

"Anthony Jones" <An*@yadayadaya da.comwrote in message
news:%2******** ********@TK2MSF TNGP03.phx.gbl. ..

>
"Bob Barrows [MVP]" <re******@NOyah oo.SPAMcomwrote in message
news:%2******** *******@TK2MSFT NGP06.phx.gbl.. .

>Joe Reynolds wrote:

say i have a string that is "hello this is my string and i like it
very much" and i want to insert a newline at every 20th character,
how could i go about doing this?

I'm sure someone will post a regex solution for this, but until then:

Here's then:-

s = InsertStringAtI nterval(s, vbCrLf, 20))

Function InsertStringAtI nterval(rsSourc e, rsInsert, rlInterval)

Dim rgx
Set rgx = new RegExp
rgx.Pattern = "([\s\S]{" & rlInterval & "})"
rgx.Global = true

InsertStringAtI nterval = rgx.Replace(rsS ource, "$1" & rsInsert)

End Function
BTW Joe, I still think you should persue a different solution than this.
If you use this in the TD you will actually need:-

s = InsertStringAtI nterval(Server. HTMLEncode(s), "<br />", 20))

>dim buffer, i, newstring,s
s="hello this is my string and i like it very much"
buffer=20
for i = 1 to len(s) step 20
newstring=newst ring & mid(s,i,20) & vbcrlf
next
msgbox newstring
--
Microsoft MVP -- ASP/ASP.NET
Please reply to the newsgroup. The email account listed in my From
header is my spam trap, so I don't check it very often. You will get a
quicker response by posting to the newsgroup.

[Bob Lehmann]

every 40 chars, im ok

How did you determine that 40 chars is OK?

Is that when I have my font size set to any of smallest, smaller, medium,
larger or largest in IE, or use anthing but normal in FireFox?

Bob Lehmann
"Joe Reynolds" <[e-mail] te in message
news:%2******** ********@TK2MSF TNGP02.phx.gbl. ..

thank you. that works great, but i had another thought (posted a new
question also)
what i want to do is first check the users input. if i find any strings of
characters that are longer than 40 characters that dont contain a space

THEN

i would pass it to your function. as long as there is at least 1 space for
every 40 chars, im ok

"Anthony Jones" <An*@yadayadaya da.comwrote in message
news:%2******** ********@TK2MSF TNGP03.phx.gbl. ..

"Bob Barrows [MVP]" <re******@NOyah oo.SPAMcomwrote in message
news:%2******** *******@TK2MSFT NGP06.phx.gbl.. .

Joe Reynolds wrote:
say i have a string that is "hello this is my string and i like it
very much" and i want to insert a newline at every 20th character,
how could i go about doing this?

I'm sure someone will post a regex solution for this, but until then:

Here's then:-

s = InsertStringAtI nterval(s, vbCrLf, 20))

Function InsertStringAtI nterval(rsSourc e, rsInsert, rlInterval)

Dim rgx
Set rgx = new RegExp
rgx.Pattern = "([\s\S]{" & rlInterval & "})"
rgx.Global = true

InsertStringAtI nterval = rgx.Replace(rsS ource, "$1" & rsInsert)

End Function
BTW Joe, I still think you should persue a different solution than

this.

If you use this in the TD you will actually need:-

s = InsertStringAtI nterval(Server. HTMLEncode(s), "<br />", 20))

dim buffer, i, newstring,s
s="hello this is my string and i like it very much"
buffer=20
for i = 1 to len(s) step 20
newstring=newst ring & mid(s,i,20) & vbcrlf
next
msgbox newstring
--
Microsoft MVP -- ASP/ASP.NET
Please reply to the newsgroup. The email account listed in my From
header is my spam trap, so I don't check it very often. You will get a
quicker response by posting to the newsgroup.