将2D变体数组的一个二维抽取到一个string

类似于： 将Variant数组转换为string – 但接受的答案不适用于我的情况。

我的代码：

SQLStr = "SELECT lineID FROM alldata where asdf = '" & asdf & "'" Set rs = New ADODB.Recordset rs.Open SQLStr, conn, adOpenStatic Dim arr Dim arrString rs.MoveFirst arr = rs.GetRows arrString = Join(arr, ", ") 

在语句arr = rs.GetRows执行后，它在我的控制台中具有以下值：

我试图join这两个值，用comma space分隔一个string，所以最终的结果是：

arrString = "25616, 99607"

但是，当我运行Join行时，我得到一个Invalid procedure call or argument错误。 我究竟做错了什么？

Answer 1

虽然我确定有一个更优雅的方式来完成您的任务，但我确实写了一些看起来正在成功运行的代码。

 Option Explicit Private Sub TestArray() Dim arr(0, 1) As Variant 'Mimic the structure of your array Dim arrString As String Dim sDelimiter As String Dim j As Long 'Mimic the structure of your array arr(0, 0) = "25616" arr(0, 1) = "99607" sDelimiter = ", " 'Set your delimiter 'Loop through the inner array For j = 0 To UBound(arr, 2) 'Concatenate the elements arrString = arrString & sDelimiter & arr(0, j) Next j 'Remove the leading delimiter / cleanup arrString = Replace(arrString, sDelimiter, "", 1, 1, vbTextCompare) MsgBox arrString End Sub 

有趣的代码是从sDelimiter = ", "和向下。

我无法弄清楚如何将数组转换成一维数组（从一个二维数组数组），所以我决定循环内部数组并连接值。

串联会在列表的前面产生一个额外的分隔符，所以有一行代替这个分隔符。

Answer 2

join只接受一维数组。 这是一个非循环的方式来获取logging到一个逗号分隔的string。

 Sub TestArray() Dim cn As ADODB.Connection Dim rs As ADODB.Recordset Dim sArr As String Set cn = New ADODB.Connection cn.Open gsCON Set rs = cn.Execute("SELECT * FROM InvoiceLine") rs.MoveFirst sArr = Join(Split(rs.GetString, vbTab), ", ") Debug.Print sArr rs.Close cn.Close Set cn = Nothing End Sub 

GetRows将数据放在错误的行/列方向。 你可以使用Application.Transpose来正确的，但是你仍然不能将它传递给Join，因为它是2D的。

GetString返回一个tab / lf分隔string。 我把它分成了标签，然后在逗号空间重新join，这是很好的去。

Answer 3

就在两天前，有一个问题促使我编写一个函数，我称之为MultiSplit() ，它使用了一个string和2个分隔符，并将其分成一个2维数组。 这个现在的问题需要一些反面的东西：

 Function MultiJoin(A As Variant, Optional ItemDelimiter As String = ",", Optional RowDelimiter As String = ";") As String 'assumes that A is a 2-dimensional array of strings Dim s As String, i As Long, j As Long Dim lb1 As Long, lb2 As Long, ub1 As Long, ub2 As Long lb1 = LBound(A, 1) lb2 = LBound(A, 2) ub1 = UBound(A, 1) ub2 = UBound(A, 2) For i = lb1 To ub1 For j = lb2 To ub2 s = s & A(i, j) & IIf(j < ub2, ItemDelimiter, "") Next j s = s & IIf(i < ub1, RowDelimiter, "") Next i MultiJoin = s End Function 

testing像：

 Sub test() Dim A As Variant ReDim A(0 To 0, 0 To 1) A(0, 0) = "25616" A(0, 1) = "99607" Debug.Print MultiJoin(A, ", ") ReDim A(0 To 1, 0 To 1) A(0, 0) = 1 A(0, 1) = 2 A(1, 0) = 3 A(1, 1) = 4 Debug.Print MultiJoin(A, " ", vbCrLf) End Sub 

输出：

 25616, 99607 1 2 3 4 

Answer 4

正如你在Excel中，你可以使用Index切片并返回一维数组

  Dim arr(0, 2) As Variant arr(0, 0) = "25616" arr(0, 1) = "99607" arr(0, 2) = "12345" MsgBox Join(Application.Index(arr, 0), ", ")