Weird results when splitting strings in VB.NET

String.Split does not have an overload that takes only a String. The argument is a Char array or String array. Your string is probably being converted to a char array. Explicitly pass a string array like so:

testString.Split(New String() { "10" }, StringSplitOptions.None)

Vb treats the delimiter argument only as a single character.

This is a tricky scenario that I have seen trip people up before, so I think it is worth a little more explanation than the other answers give. In your original format Split(testString,"10")(1).Split("20")(0), you are unknowingly using two DIFFERENT Split functions.

The first Split(testString,"10") is using the Microsoft.VisualBasic.Strings.Split function, which takes String type parameters. http://msdn.microsoft.com/en-us/library/microsoft.visualbasic.strings.split(v=vs.110).aspx

The second .Split("20")(0) is using System.String.Split method, which does not have an overload that takes a String parameter. http://msdn.microsoft.com/en-us/library/System.String.Split(v=vs.110).aspx

So what was happening is:

1. Split(testString,"10") uses Microsoft.VisualBasic.Strings.Split, which returns new String() {"123456789", "11121314151617181920"}
2. (1) means get 1st position of the returned array, which is "11121314151617181920"
3. "11121314151617181920".Split("20")(0) uses System.String.Split, and attempts to split on string separator "20"
4. NOTE: The string "20" param gets implicitly converted to a char "2" because the only single parameter overload of String.Split has a signature of Public Function Split (ParamArray separator As Char()) As String(). The ParamArray parameter option allows you to pass a comma delimited list of values into the function, similar to how String.Format works with a dynamic # of replacement values. http://msdn.microsoft.com/en-us/library/538f81ec.aspx
5. Step 3 code becomes "11121314151617181920".Split(new Char() {CChar("20")})(0), which using literal values is "11121314151617181920".Split(new Char() {"2"c})(0). The result is {"111", "13141516171819", "0"}. Get the 0th position, returns "111".

So to avoid confusion, you should convert your code to use the same version of Split on both sides.

Either of the 2 examples below should work:

Example 1: Using Microsoft.VisualBasic.Strings.Split:

Split( Split(testString,"10")(1), "20" )(0)

Example 2: Using System.String.Split:

testString _ 
    .Split(New String() {"10"}, StringSplitOptions.None)(1) _ 
    .Split(New String() {"20"}, StringSplitOptions.None)(0)