I am trying to use a formula to get a letter of the alphabet.

Formula:

```
=Keytable(RANDOM,ROW())
```

Function:

```
Function KeyTable(seed As Long, position As Long) As String
Dim i As Long
Stop
Dim calpha(1 To 26) As String
Dim alpha(1 To 26) As String
For i = 1 To 26
alpha(i) = Chr(i + UPPER_CASE - 1)
Next i
For i = 1 To 26
calpha(i) = alpha(seed Mod 27 - i)
Next i
Stop
KeyTable = calpha(position)
End Function
```

Result:

```
#Value!
```

When I step through the function, it never gets to the second stop. What is wrong?

`RANDOM`

is not a function in Excel. `RAND()`

is and it returns a `float`

between 0 and 1. You need an `integer`

to do modulus calculations.

To get a random integer, use:

```
INT ((upperbound - lowerbound + 1) * RAND() + lowerbound)
```

Then, once `seed Mod 27 - i`

becomes 0 or less, the function dies because arrays can’t be indexed with 0 or less in VBA (or most languages).

But really all you need to do for a random letter is this:

```
=CHAR(RANDBETWEEN(65,90))
```

### Answer：

This code will return random letter of alphabet:

```
Function GetLetter()
Dim letters As String
Dim randomIndex As Byte
letters = "abcdefghijklmnopqrstuvwxyz"
randomIndex = Int((26 - 1 + 1) * Rnd() + 1) //Get random number between 1 and 26
GetLetter = VBA.Mid$(letters, randomIndex, 1)
End Function
```

Tags: excel-vbaexcel, function