Cut String & Insert in String

[vunet.us]

Hello,
In ASP, vbscript:
How can I display the first 20 characters of a long string?
How can I insert some character inside of the string on every 5th
character?
Thank you for help.

[Bob Barrows [MVP]]

vu******@gmail. com wrote:

Hello,
In ASP, vbscript:
How can I display the first 20 characters of a long string?

first20 = left(longstring ,20)

How can I insert some character inside of the string on every 5th
character?

There may be a regular expression solution for this, but for
simplicity's sake:

dim i, oldstring,newst ring, stringseg,subst _char
for i = 1 to len(oldstring) step 5
stringseg=mid(o ldstring,i,5)
if len(newstring) 0 then
newstring = newstring & subst_char & stringseg
else
newstring=strin gseg
end if
next

Thank you for help.

The vbscript documentation can be downloaded from:
http://www.microsoft.com/downloads/d...DisplayLang=en

or it can be read online at
http://msdn2.microsoft.com/en-us/library/t0aew7h6.aspx

--
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.

[vunet.us]

On May 14, 12:06 pm, "Bob Barrows [MVP]" <reb01...@NOyah oo.SPAMcom>
wrote:

vunet...@gmail. com wrote:

Hello,
In ASP, vbscript:
How can I display the first 20 characters of a long string?

first20 = left(longstring ,20)

How can I insert some character inside of the string on every 5th
character?

There may be a regular expression solution for this, but for
simplicity's sake:

dim i, oldstring,newst ring, stringseg,subst _char
for i = 1 to len(oldstring) step 5
stringseg=mid(o ldstring,i,5)
if len(newstring) 0 then
newstring = newstring & subst_char & stringseg
else
newstring=strin gseg
end if
next

Thank you for help.

The vbscript documentation can be downloaded from:http://www.microsoft.com/downloads/d...D=01592c48-207...

or it can be read online athttp://msdn2.microsoft .com/en-us/library/t0aew7h6.aspx

--
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.

thank you veeeery much

[Evertjan.]

Bob Barrows [MVP] wrote on 14 mei 2007 in
microsoft.publi c.inetserver.as p.general:

>How can I insert some character inside of the string on every 5th
character?

There may be a regular expression solution for this, but for
simplicity's sake:

dim i, oldstring,newst ring, stringseg,subst _char
for i = 1 to len(oldstring) step 5
stringseg=mid(o ldstring,i,5)
if len(newstring) 0 then
newstring = newstring & subst_char & stringseg
else
newstring=strin gseg
end if
next

There may be a regular expression solution for this

"may", "simplicity "?

<%
reponse.write insertFifth('12 345678901234567 890123','x')
%>

<script language='javas cript' runat='server'>
function insertFifth(s,i nser){
return s.replace(/(.{5})/g,"$1"+inser) ;
};
</script>

--
Evertjan.
The Netherlands.
(Please change the x'es to dots in my emailaddress)

[Bob Barrows [MVP]]

Evertjan. wrote:

>There may be a regular expression solution for this

"may", "simplicity "?

<%
reponse.write insertFifth('12 345678901234567 890123','x')
%>

<script language='javas cript' runat='server'>
function insertFifth(s,i nser){
return s.replace(/(.{5})/g,"$1"+inser) ;
};
</script>

Yes! Simplicity!
Go ahead. Tell the user how to interpret that regexp, as well as how to
create it, the next time he needs something like this.

Simplicity isn't always about the number of lines of code involved.

--
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.

[Evertjan.]

Bob Barrows [MVP] wrote on 14 mei 2007 in
microsoft.publi c.inetserver.as p.general:

Evertjan. wrote:

>>There may be a regular expression solution for this

"may", "simplicity "?

<%
reponse.writ e insertFifth('12 345678901234567 890123','x')
%>

<script language='javas cript' runat='server'>
function insertFifth(s,i nser){
return s.replace(/(.{5})/g,"$1"+inser) ;
};
</script>

Yes! Simplicity!
Go ahead. Tell the user how to interpret that regexp, as well as how to
create it, the next time he needs something like this.

Sorry Bob, I fail to see that.
The earlier vbscirpt example was not explained.

Usenet is an interactive medium, so anyone is free to ask.

Simplicity isn't always about the number of lines of code involved.

Simplicity is in the hands of the beholder.

--
Evertjan.
The Netherlands.
(Please change the x'es to dots in my emailaddress)