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Normal Topic Replace Problem! (Read 776 times)
Amor
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Replace Problem!
Jun 24th, 2010 at 9:05pm
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Hello!
This Code  does not work for me. Pease help me.

     If @lt(vergebnis,1) = @Ch(60)
{
vErgebnis = @replfir(vErgebnis,@Ch(60),"")
vwertungErgebnis = @Ch(60)
}

or

     If @lt(vergebnis,1) = "<" then
{
vErgebnis = @replfir(vErgebnis,"<","")
vwertungErgebnis = "<"
}
  
Dr. med. Amor Belhareth&&Medizin Labor &&Germany
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Carl Underwood
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Re: Replace Problem!
Reply #1 - Jun 24th, 2010 at 11:51pm
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Are you working with the correct variable? Because it works as I would expect it to. Both of your approaches are producing the same results.

The results of the following code (which is mostly the same as your example, except that I added the variable declarations, and some WriteLn commands to view the results) produces the output at the end of this message.

Code
Select All
var vergebnis as string
var vwertungErgebnis as string

vergebnis = "<TestThis"


   If @lt(vergebnis,1) = @Ch(60)  
 {
 vErgebnis = @replfir(vErgebnis,@Ch(60),"")
 vwertungErgebnis = @Ch(60)  
}

WriteLn(vErgebnis)
WriteLn(vwertungErgebnis)
WriteLn("")

//or

     If @lt(vergebnis,1) = "<" then  
{
 vErgebnis = @replfir(vErgebnis,"<","")
 vwertungErgebnis = "<"  
}

WriteLn(vErgebnis)
WriteLn(vwertungErgebnis)

WriteLn("")


Quote:
TestThis
<

TestThis
<
  

Carl Underwood
CDU Computer Consulting LLC
Epsom, New Hampshire
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Amor
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Re: Replace Problem!
Reply #2 - Jun 26th, 2010 at 10:00am
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Thank you Carl for your code. I have to say your method works better. 

  
Dr. med. Amor Belhareth&&Medizin Labor &&Germany
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Carl Underwood
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Re: Replace Problem!
Reply #3 - Jun 26th, 2010 at 6:28pm
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Quote:
Thank you Carl for your code.

It's really your code, with a couple lines I added to it.  Wink
  

Carl Underwood
CDU Computer Consulting LLC
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