Question

Background:

I needed to split a string of numerous words into an array which separates the words for me for further use later on in my code. However, I needed to get rid of any numbers which might be present in the string so I declared a string contaning characters which I wanted to use as separators/delimiters as follows:

dim Separators As String = ' 1234567890' 

so my code became more or less as follows:

''//USING SPLIT-FUNCTION dim MyTestString as String = 'This is 9 a 567 test'  dim Separators As String = ' 1234567890' dim ResultWithNoNumbers() as String  ResultWithNoNumbers = Split(MyTestString,Separators.ToCharArray) Messagebox.Show(ResultWithNoNumbers.Length.ToString) ''//Result is 1 and no split was performed 

This did not work since the Split-functions did not consider my Separators as individual characters

however this worked..

''//USING SPLIT-METHOD dim MyTestString as String = 'This is 9 a 567 test'  dim Separators As String = ' 1234567890' dim ResultWithNoNumbers() as String  ResultWithNoNumbers = MyTestString.Split(Separators.ToCharArray,                                     _StringSplitOptions.RemoveEmptyEntries) Messagebox.Show(ResultWithNoNumbers.Length.ToString) ''//Result is 4 and split was performed 

So far, so good – since one has more options with the Split-method compared to the Split-function, I managed to resolve my problem.

Now to my question, let’s say I only needed the standard space-character (‘ ‘) as a delimiter (and no need to get rid of numbers like in above example), then both ways would work. So which one would you use? Apart from various options available, are there any advantages using a particular one? Perhaps one is faster, less memory-hungry?

EDIT: I’m developing for Windows Mobile and that’s why speed- and memory-issues get important.

Thank you.

Answer 1

I would use regular expressions

Dim MyTestString As String = 'This is 9 a 567 test 23424234 this is  23 another test 23 and 3 again and one 34234 more' Dim reg_exp As New Text.RegularExpressions.Regex('\d') Dim reg_exp2 As New Text.RegularExpressions.Regex('\s{2,}')  MyTestString = reg_exp.Replace(MyTestString, String.Empty) MyTestString = reg_exp2.Replace(MyTestString, ' ')  Dim strArr() As String strArr = MyTestString.ToString.Split(' ')  Console.WriteLine(MyTestString) Console.ReadLine() 

OUTPUT:

This is a test this is another test and again and one more