Deleting certain character from right and left of

问题:

I want to delete some certain characters that used wrongly in a string.

"........I.wanna.delete.only.the.dots.outside.of.this.text..............."

as you can see I can't use replace for this. I must find a way to delete only the characters at left and right of a string and those dots only an example of characters that I want to delete. I have an array of those unwanted characters. So after the process string should look like

"I.wanna.delete.only.the.dots.outside.of.this.text"

but I couldn't find a way to get this work.

回答1:

VB6 has a Trim() function but that only removes spaces.

To remove characters from both ends, you'll need to check each end in turn removing the character until you get something else:

Function TrimChar(ByVal Text As String, ByVal Characters As String) As String
  'Trim the right
  Do While Right(Text, 1) Like "[" & Characters & "]"
    Text = Left(Text, Len(Text) - 1)
  Loop

  'Trim the left
  Do While Left(Text, 1) Like "[" & Characters & "]"
    Text = Mid(Text, 2)
  Loop

  'Return the result
  TrimChar = Text
End Function

Result:

?TrimChar("........I.wanna.delete.only.the.dots.outside.of.this.text...............", ".")
I.wanna.delete.only.the.dots.outside.of.this.text

This is far from optimised but you could expand on it to just work out the end positions, then do a single Mid() call.

回答2:

    Public Function Remove(text As String) As String

      text = Replace(text, "..", "")

      If Left(text, 1) = "." Then
         text = Right(text, Len(text) - 1)
      End If

      If Right(text, 1) = "." Then
         text = Left(text, Len(text) - 1)
      End If

      Remove = text

    End Function

If you remove any instances of ".." then you may or may not be left with a single leading dot and a single trailing dot to deal with. If you are lucky enough to be able to guarantee that the dots will always be an even number then

 text = Replace(text, "..", "")

is all you need.