Help with C# to VB.Net ... Please

[Rob]

I am trying to convert the following C# routine (below) to Vb.Net. I
am having troubles with the following line of code

{ result.Append(chars[b % (chars.Length - 1)]); }

My C# to Vb.Net routine converted the line above to be:

result.Append(chars(b % (chars.Length - 1)))

But I am getting error "Character is not Valid" on the "%".

Any help would be appreciated.

Cheers,
Rob Panosh
Advanced Software Designs


private string GetUniqueKey()
{
int maxSize = 8 ;
int minSize = 5 ;
char[] chars = new char[62];
string a;
a = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890";
chars = a.ToCharArray();
int size = maxSize ;
byte[] data = new byte[1];
RNGCryptoServiceProvider crypto = new RNGCryptoServiceProvider();
crypto.GetNonZeroBytes(data) ;
size = maxSize ;
data = new byte[size];
crypto.GetNonZeroBytes(data);
StringBuilder result = new StringBuilder(size) ;
foreach(byte b in data )
{ result.Append(chars[b % (chars.Length - 1)]); }
return result.ToString();
}

 

[Larry Lard]

I am trying to convert the following C# routine (below) to Vb.Net. I
am having troubles with the following line of code

{ result.Append(chars[b % (chars.Length - 1)]); }

My C# to Vb.Net routine converted the line above to be:

result.Append(chars(b % (chars.Length - 1)))

But I am getting error "Character is not Valid" on the "%".

Any help would be appreciated.

% is the modulus operator in C#, as google easily shows.

 

[IdleBrain]

Try result.Append(chars(b Mod (chars.Length - 1)))

 

[Herfried K. Wagner [MVP]]

I am trying to convert the following C# routine (below) to Vb.Net.

Converting code between .NET programming languages
<URL:http://dotnet.mvps.org/dotnet/faqs/?id=languageconverters&lang=en>

 

[Guest]

result.Append(chars(b Mod (chars.Length - 1)))
--
David Anton
www.tangiblesoftwaresolutions.com
Instant C#: VB to C# converter
Instant VB: C# to VB converter
Instant C++: C# to C++ converter
Instant C++: VB to C++ converter

 

[Rob]

Everybody,

Thanks for your help.

Cheers,
Rob