Generate random strings in vb.net - vb.net

Question

I need to generate random strings in vb.net, which must consist of (randomly chosen) letters A-Z (must be capitalized) and with random numbers interspersed. It needs to be able to generate them with a set length as well.
Thanks for the help, this is driving me crazy!

Answer 1

If you can convert this to VB.NET (which is trivial) I'd say you're good to go (if you can't, use this or any other tool for what's worth):
/// <summary>
/// The Typing monkey generates random strings - can't be static 'cause it's a monkey.
/// </summary>
/// <remarks>
/// If you try hard enough it will eventually type some Shakespeare.
/// </remarks>
class TypingMonkey
{
private const string legalCharacters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890";
static Random random = new Random();
/// <summary>
/// The Typing Monkey Generates a random string with the given length.
/// </summary>
/// <param name="size">Size of the string</param>
/// <returns>Random string</returns>
public string TypeAway(int size)
{
StringBuilder builder = new StringBuilder();
char ch;
for (int i = 0; i < size; i++)
{
ch = legalCharacters[random.Next(0, legalCharacters.Length)];
builder.Append(ch);
}
return builder.ToString();
}
}
Then all you've got to do is:
TypingMonkey myMonkey = new TypingMonkey();
string randomStr = myMonkey.TypeAway(size);

Answer 2

Why don't you randomize a number 1 to 26 and get the relative letter.
Something like that:
Dim output As String = ""
Dim random As New Random()
For i As Integer = 0 To 9
output += ChrW(64 + random.[Next](1, 26))
Next
Edit: ChrW added.
Edit2: To have numbers as well
Dim output As String = ""
Dim random As New Random()
Dim val As Integer
For i As Integer = 0 To 9
val = random.[Next](1, 36)
output += ChrW(IIf(val <= 26, 64 + val, (val - 27) + 48))
Next

Answer 3

C#
public string RandomString(int length)
{
Random random = new Random();
char[] charOutput = new char[length];
for (int i = 0; i < length; i++)
{
int selector = random.Next(65, 101);
if (selector > 90)
{
selector -= 43;
}
charOutput[i] = Convert.ToChar(selector);
}
return new string(charOutput);
}
VB.Net
Public Function RandomString(ByVal length As Integer) As String
Dim random As New Random()
Dim charOutput As Char() = New Char(length - 1) {}
For i As Integer = 0 To length - 1
Dim selector As Integer = random.[Next](65, 101)
If selector > 90 Then
selector -= 43
End If
charOutput(i) = Convert.ToChar(selector)
Next
Return New String(charOutput)
End Function

Answer 4

How about:
Private Function GenerateString(len as integer) as String
Dim stringToReturn as String=""
While stringToReturn.Length<len
stringToReturn&= Guid.NewGuid.ToString().replace("-","")
End While
Return left(Guid.NewGuid.ToString(),len)
End Sub

Answer 5

Here's a utility class I've got to generate random passwords. It's similar to JohnIdol's Typing Monkey, but has a little more flexibility in case you want generated strings to contain uppercase, lowercase, numeric or special characters.
public static class RandomStringGenerator
{
private static bool m_UseSpecialChars = false;
#region Private Variables
private const int m_MinimumLength = 8;
private const string m_LowercaseChars = "abcdefghijklmnopqrstuvqxyz";
private const string m_UppercaseChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
private const string m_NumericChars = "123456890";
private const string m_SpecialChars = "~?/##!£$%^&*+-_.=|";
#endregion
#region Public Methods
/// <summary>
/// Generates string of the minimum length
/// </summary>
public static string Generate()
{
return Generate(m_MinimumLength);
}
/// <summary>
/// Generates a string of the specified length
/// </summary>
/// <param name="length">The number of characters to generate</param>
public static string Generate(int length)
{
return Generate(length, Environment.TickCount);
}
#endregion
#region Private Methods
/// <summary>
/// Generates a string of the specified length using the specified seed
/// </summary>
private static string Generate(int length, int seed)
{
// Generated strings must contain at least 3 of the following character groups: uppercase letters, lowercase letters
// numerals, and special characters (!, #, $, £, etc)
// The generated string must be at least 4 characters so that we can add a single character from each group.
if (length < 4) throw new ArgumentException("String length must be at least 4 characters");
StringBuilder SB = new StringBuilder();
Random rand = new Random(seed);
// Ensure that we add all of the required groups first
SB.Append(GetRandomCharacter(m_LowercaseChars, rand));
SB.Append(GetRandomCharacter(m_UppercaseChars, rand));
SB.Append(GetRandomCharacter(m_NumericChars, rand));
if (m_UseSpecialChars)
SB.Append(GetRandomCharacter(m_SpecialChars, rand));
// Now add random characters up to the end of the string
while (SB.Length < length)
{
SB.Append(GetRandomCharacter(GetRandomString(rand), rand));
}
return SB.ToString();
}
private static string GetRandomString(Random rand)
{
int a = rand.Next(3);
switch (a)
{
case 1:
return m_UppercaseChars;
case 2:
return m_NumericChars;
case 3:
return (m_UseSpecialChars) ? m_SpecialChars : m_LowercaseChars;
default:
return m_LowercaseChars;
}
}
private static char GetRandomCharacter(string s, Random rand)
{
int x = rand.Next(s.Length);
string a = s.Substring(x, 1);
char b = Convert.ToChar(a);
return (b);
}
#endregion
}
To use it:
string a = RandomStringGenerator.Generate(); // Generate 8 character random string
string b = RandomStringGenerator.Generate(10); // Generate 10 character random string
This code is in C# but should be fairly easy to convert to VB.NET using a code converter.

Answer 6

Just be simple. for vb just do:
Public Function RandomString(size As Integer, Optional validchars As String = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789abcdefghijklmnopqrstuvwxyz") As String
If size < 1 Or validchars.Length = 0 Then Return ""
RandomString = ""
Randomize()
For i = 1 To size
RandomString &= Mid(validchars, Int(Rnd() * validchars.Length) + 1, 1)
Next
End Function
This function allows for a base subset of chars to use or user can choose anything. For example you could send ABCDEF0123456789 and get random Hex. or "01" for binary.

Answer 7

Try this, it's the top answer already converted to VB!
Private Function randomStringGenerator(size As Integer)
Dim random As Random = New Random()
Dim builder As Text.StringBuilder = New Text.StringBuilder()
Dim ch As Char
Dim legalCharacters As String = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz1234567890"
For cntr As Integer = 0 To size
ch = legalCharacters.Substring(random.Next(0, legalCharacters.Length), 1)
builder.Append(ch)
Next
Return builder.ToString()
End Function