parse this...

[Scorpionz]

how would i be able to parse a string like this:

373 3 3 113 "hello there" "sup all" 4 4

into this:

373
3
3
113
hello there
sup all
4
4

[dsy5]

Well if you have VB6 you can use the Split function with a " " as delimiter. If you have VB5, no such command.

But, tada, I have a VB5 Split function for you!

Public Function WordSplit(sWords As String, sDelim As String) As Variant

	Dim sTemp As String
	Dim vWords() As Variant
	Dim n, nCount As Integer

	nCount = 0
	sTemp = sWords

	Do
		n = InStr(1, sTemp, sDelim)
		ReDim Preserve vWords(nCount)
		If n = 0 Then
			vWords(nCount) = sTemp
		Else
			vWords(nCount) = Left$(sTemp, n)
			If Right(vWords(nCount), 1) = sDelim Then
				vWords(nCount) = Mid(vWords(nCount), 1, Len(vWords(nCount)) - 1)
			End If
			sTemp = Mid$(sTemp, n + 1)
		End If

		nCount = nCount + 1

	Loop Until n = 0

	WordSplit = vWords

End Function

The code above was not written by me Alas, I do not know who the author was.

[gwdash]

Try this:
VB6 ONLY!!! (otherwise, maybe someone will post the split code for vb5)

Sub SplitIt(MyStr As String)
Dim Splited

Splited = Split(MyStr, " ")
For i = LBound(Splited) To UBound(Splited)
    Debug.Print TrimQuotes(Splited(i))
End If

End Sub

Function TrimQuotes(String1 As String) As String
Dim FirstChr As String
Dim LastChr As String
FirstChr = Left(String1, 1)
LastChr = Right(String1, 1)

If FirstChr = Chr(34) And LastChr = Chr(34) Then
    TrimQuotes = Mid(String1, 2, Len(String1) - 1)
Else
    TrimQutoes = String1
End If
End Sub

[gwdash]

I See someone did post the VB5 Split Function!!! Before Me!!!
(I haven't tested my trim quotes function, but it should work)

[Scorpionz]

naa, thats not gonna work...

i need to parse out the first 4 strings then parse out the words in ""s, but at the same time keeping there spacing, and then parse out the last 2 strings so that it comes out as:

373
3
3
113
hello there
sup all
4
4