Mischa what I am doing with what you sent me, I am asking because I cant think 
any more it seems.

Still getting back the same error even with x = 5
  ----- Original Message ----- 
  From: Mischa Kroon 
  To: [email protected] 
  Sent: Wednesday, May 11, 2005 3:13 PM
  Subject: Re: [AspNetAnyQuestionIsOk] VB.net String maniplulation problem


  ok try 5 :)

  ----- Original Message ----- 
  From: "Tony Trapp" <[EMAIL PROTECTED]>
  To: <[email protected]>
  Sent: Thursday, May 12, 2005 12:27 AM
  Subject: Re: [AspNetAnyQuestionIsOk] VB.net String maniplulation problem


  > Ok I dim x as integer
  >
  > now I get back an error:
  >
  >
  > Server Error in '/' Application.
  > 
--------------------------------------------------------------------------------
  >
  > Input string was not in a correct format.
  > Description: An unhandled exception occurred during the execution of the 
  > current web request. Please review the stack trace for more information 
  > about the error and where it originated in the code.
  >
  > Exception Details: System.FormatException: Input string was not in a 
  > correct format.
  >
  > Source Error:
  >
  > Line 330:        newstring_withnum_tostr = words1.substring(location + x, 
  > 1)
  > Line 331:
  > Line 332: words1.Replace(newstring_withnum_tostr, 
  > Cint(newstring_withnum_tostr) + 1)
  > Line 333:
  > Line 334:    ' newstr = words1.Insert(newstring_withnum_tostr + 4, "_" & 
  > Counter + 1)
  >
  >  ----- Original Message ----- 
  >  From: Mischa Kroon
  >  To: [email protected]
  >  Sent: Wednesday, May 11, 2005 3:02 PM
  >  Subject: Re: [AspNetAnyQuestionIsOk] VB.net String maniplulation problem
  >
  >
  >  x stands for I'm not certain at this point :)
  >  I think it's 5
  >  but could be 4
  >
  >  Pizza sounds good btw :)
  >
  >
  >
  >  ----- Original Message ----- 
  >  From: "Tony Trapp" <[EMAIL PROTECTED]>
  >  To: <[email protected]>
  >  Sent: Thursday, May 12, 2005 12:11 AM
  >  Subject: Re: [AspNetAnyQuestionIsOk] VB.net String maniplulation problem
  >
  >
  >  > Micha fallowing along I know x in math represents the unknown and of
  >  > course if its not declared I will get back an error so what is x
  >  > representing in this equation?
  >  >
  >  > Thanks for your help by the way, can I buy you a pizza?   LOL
  >  >
  >  > Tony...
  >  >
  >  >  ----- Original Message ----- 
  >  >  From: Mischa Kroon
  >  >  To: [email protected]
  >  >  Sent: Wednesday, May 11, 2005 2:48 PM
  >  >  Subject: Re: [AspNetAnyQuestionIsOk] VB.net String maniplulation 
  > problem
  >  >
  >  >
  >  >  whoops didn't read the full message.
  >  >
  >  >  here is where you go wrong i think :)
  >  >
  >  >
  >  >  >   location = words1.IndexOf("_new")
  >  >  >
  >  >  >   If location > 0 or not -1 then
  >  >  >
  >  >
  >  >  location will give the location of the start of "_new"
  >  >  you want the next char not the position of the "_"
  >  >
  >  >  then:        newstring_withnum_tostr = words1.substr (location + x, 1)
  >  >  words1.replace (newstring_withnum_tostr, cint 
  > (newstring_withnum_tostr) +
  >  >  1 )
  >  >
  >  >  check it:
  >  >
  >  >  response.write (words1)
  >  >
  >  >
  >  >
  >  >
  >  >
  >  >  ----- Original Message ----- 
  >  >  From: "Tony Trapp" <[EMAIL PROTECTED]>
  >  >  To: <[email protected]>
  >  >  Sent: Wednesday, May 11, 2005 11:13 PM
  >  >  Subject: [AspNetAnyQuestionIsOk] VB.net String maniplulation problem
  >  >
  >  >
  >  >  > Hey guys here is what I want to do.
  >  >  >
  >  >  > Lets say I have a file name is my database my_pic_new1.jpg first 
  > off.
  >  >  >
  >  >  > Now I want to change this file name to make it a uniqe file name but
  >  >  > I need to do a few things to the string first.
  >  >  >
  >  >  > Is in the string _new is there then I want to see if a number is
  >  >  > there.
  >  >  >
  >  >  > If there is a number there I want to take that number and what ever
  >  >  > it is increment that number up by one make the file now
  >  >  > my_pic_new2.jpg
  >  >  >
  >  >  > and so on.
  >  >  >
  >  >  > I am not asking for someone to write the code for me but point me to
  >  >  > where I can gain viable info on how to make this hapeen or a small
  >  >  > code example.
  >  >  >
  >  >  > Here is my code at present,
  >  >  >
  >  >  >
  >  >  >  Dim words1 as String = tmnescort_currentpic_small
  >  >  >
  >  >  >   Dim location as Integer
  >  >  >   Dim Counter as Integer
  >  >  >   Dim newstr as Integer
  >  >  >   Dim newstring_withnum as Integer
  >  >  >   Dim countvar as Integer
  >  >  >   Dim newstring_withnum_tostr
  >  >  >   dim addnum as String
  >  >  >
  >  >  >   Counter = 0
  >  >  >   location = words1.IndexOf("_new")
  >  >  >
  >  >  >   If location > 0 or not -1 then
  >  >  >
  >  >  >        newstring_withnum_tostr = instr(1, words1, "1")
  >  >  >
  >  >  >
  >  >  > newstr = System.Convert.ToInt16
  >  >  > (newstring_withnum_tostr + 1)
  >  >  > newstring_withnum = newstr & Counter + 1
  >  >  > addnum = System.Convert.ToString(newstring_withnum)
  >  >  >
  >  >  >     str_msg.Visible = True
  >  >  >     str_msg.Text = "This is the pic name in DB&nbsp;" &
  >  >  > tmnescort_currentpic_small & "<br><br>new string was found<br><br>" 
  > &
  >  >  > addnum & "<br><br>"
  >  >  >
  >  >  >        else
  >  >  >
  >  >  > str_msg.Visible = True
  >  >  >     str_msg.Text = "This is the pic name in DB&nbsp;" &
  >  >  > tmnescort_currentpic_small & "<br>new string was not found<br><br>" 
  > &
  >  >  > addnum & "<br><br>"
  >  >  >
  >  >  >    end if
  >  >  >
  >  >  >
  >  >  > what I get back printed out is
  >  >  >
  >  >  > This is the pic name in DB my_pic_new1.jpg
  >  >  >
  >  >  > new string was found
  >  >  >
  >  >  > 161 << this is what happens after my string manipualtion I dont want
  >  >  > this.
  >  >  >
  >  >  > Can any please help me?
  >  >  >
  >  >  > Thanks,
  >  >  >
  >  >  > Tony...
  >  >  >
  >  >  >
  >  >  >
  >  >  >
  >  >  >
  >  >  >
  >  >  > Yahoo! Groups Links
  >  >  >
  >  >  >
  >  >  >
  >  >  >
  >  >  >
  >  >  >
  >  >  >
  >  >  > -- 
  >  >  > No virus found in this incoming message.
  >  >  > Checked by AVG Anti-Virus.
  >  >  > Version: 7.0.308 / Virus Database: 266.11.8 - Release Date: 
  > 5/10/2005
  >  >  >
  >  >  >
  >  >
  >  >
  >  >
  >  > 
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