用逗号分隔多个逗号分隔的csv文件

Question

我试图在访问中导入逗号分隔的csv文件。我遇到的问题是其中一列“金额”在数据本身中有逗号，例如“1,433.36”。这些数据中总会有逗号。

我怎样才能导入成功？

样本数据：

sjonn,one,"1,855.9" 
ptele,two,344.0 
jrudd,one,334.8 

在此先感谢

Answer 1

我会分隔符更改为一个不同的角色，就像一个管道 “|”。

Answer 2

将该文件另存为制表符分隔的文本文件并导入。

Answer 3

使用输入读取文件处理的报价为您

Dim f1 As String 
Dim f2 As String 
Dim f3 As String 

Open "d:\test.txt" For Input As #1 

Input #1, f1, f2, f3 
Debug.Print f1, f2, f3 
Input #1, f1, f2, f3 
Debug.Print f1, f2, f3 

Close #1 ' 

给

sjonn   one   1,855.9 
ptele   two   344.0 

Answer 4

如果DoCmd.TransferText不为你工作，那么你可以定义一个方法，这样做“手动” ：

Set fs = Server.CreateObject("Scripting.FileSystemObject") 
Set objFile = fs.GetFile("import.txt") 
Set objFileTextStream = objFile.OpenAsTextStream(1, 2) 

objFileTextStream.skipLine 'if the file contains the header 

Do While objFileTextStream.AtEndOfStream <> True 
    strLine = objFileTextStream.ReadLine 'read a line 
    strLinePart = split(strLine,",") 'Split the line using the , delimiter 
    firstField = strLinePart(0) 
    secondField = strLinePart(1) 
    thirdField = strLinePart(2) 
    strSQL = "INSERT INTO myTable Values('"& firstField &"','"& secondField &"','"& thirdField &"')" 
    conn.Execute strSQL 
Loop 

objFileTextStream.Close: Set objFileTextStream = Nothing 
Set fs = Nothing 
conn.Close: Set conn = Nothing 

Answer 5

我曾经遇到过这个问题，这是另一种方法，也许会有帮助，但是它的分割线本身，即你必须使用这种方法 其还假定它包含在一个命名为模块

''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