Однажды я столкнулся с проблемой, и это еще один метод, который может помочь, однако он разбивает сами строки, т. Е. Перед использованием этого метода необходимо сначала разбить строку на строки
Также предполагается, что он содержится в модуле с именем Module1
''Perfoms a smart split that takes care of the ""
Public Function SmartSplit(Str As String) As Variant
''New collection
Dim Quote As String
Dim Delimiter As String
Dim MyString As String
Dim Sample As String
Dim StrCollection As New Collection
Dim Array_1() As String
Dim HasSeenQuote As Boolean
Dim index As Long
Quote = "" & CStr(Chr(34))
Delimiter = "" & CStr(Chr(44))
HasSeenQuote = False
Array_1 = Split(Str, Delimiter)
For index = LBound(Array_1) To UBound(Array_1)
Sample = Array_1(index)
If Module1.StartsWith(Sample, Quote, False) Then
HasSeenQuote = True
End If
''We append the string
If HasSeenQuote Then
MyString = MyString & "," & Sample
End If
''We add the term
If Module1.EndsWith(Sample, Quote, False) Then
HasSeenQuote = False
MyString = Replace(MyString, Quote, "")
MyString = Module1.TrimStartEndCharacters(MyString, ",", True)
MyString = Module1.TrimStartEndCharacters(MyString, Quote, True)
StrCollection.Add (MyString)
MyString = ""
GoTo LoopNext
End If
''We did not see a quote before
If HasSeenQuote = False Then
Sample = Module1.TrimStartEndCharacters(Sample, ",", True)
Sample = Module1.TrimStartEndCharacters(Sample, Quote, True)
StrCollection.Add (Sample)
End If
LoopNext:
Next index
''Copy the contents of the collection
Dim MyCount As Integer
MyCount = StrCollection.Count
Dim RetArr() As String
ReDim RetArr(0 To MyCount - 1) As String
Dim X As Integer
For X = 0 To StrCollection.Count - 1 ''VB Collections start with 1 always
RetArr(X) = StrCollection(X + 1)
Next X
SmartSplit = RetArr
End Function
''Returns true of false if the string starts with a string
Public Function EndsWith(ByVal Str As String, Search As String, IgnoreCase As Boolean) As Boolean
EndsWith = False
Dim X As Integer
X = Len(Search)
If IgnoreCase Then
Str = UCase(Str)
Search = UCase(Search)
End If
If Len(Search) <= Len(Str) Then
EndsWith = StrComp(Right(Str, X), Search, vbBinaryCompare) = 0
End If
End Function
''Trims start and end characters
Public Function TrimStartEndCharacters(ByVal Str As String, ByVal Search As String, ByVal IgnoreCase As Boolean) As String
If Module1.StartsWith(Str, Search, IgnoreCase) Then
Str = Right(Str, (Len(Str) - Len(Search)))
End If
If Module1.EndsWith(Str, Search, IgnoreCase) Then
Str = Left(Str, (Len(Str) - Len(Search)))
End If
TrimStartEndCharacters = Str
End Function
''Returns true of false if the string starts with a string
Public Function StartsWith(ByVal Str As String, Search As String, IgnoreCase As Boolean) As Boolean
StartsWith = False
Dim X As Integer
X = Len(Search)
If IgnoreCase Then
Str = UCase(Str)
Search = UCase(Search)
End If
If Len(Search) <= Len(Str) Then
StartsWith = StrComp(Left(Str, X), Search, vbBinaryCompare) = 0
End If
End Function