无法粘贴 - Excel VBA

Question

新的VBA试图弄清楚。使用代码尝试从各种文件自动复制到具有公式的文件，计算并再返回。 当我尝试粘贴时，遇到了'运行时错误'1004'，范围类的pastespecial方法失败'消息。该消息仅在使用变量声明第一个单元格时出现，以便复制一系列值。 当即时通讯使用直接单元格描述一切工作正常。 我还使用自定义函数来获取给定字段名称的列字母。

Function ActiveColumnName(fieldname As String, fieldnames_line As Integer) As String 

Range("A" & fieldnames_line & ":AB" & fieldnames_line).NumberFormat = "@" 

Cells.find(What:=fieldname, After:=ActiveCell, LookIn:=xlFormulas, LookAt _ 
    :=xlPart, SearchOrder:=xlByRows, SearchDirection:=xlNext, MatchCase:= _ 
    False, SearchFormat:=False).Activate 

ActiveColumnNumber = ActiveCell.Column 


Dim m As Integer 
Dim ActiveColumnName As String 

ActiveColumnName = "" 

Do While (ActiveColumnNumber > 0) 
    m = (ActiveColumnNumber - 1) Mod 26 
    ActiveColumnName = Chr(65 + m) + ActiveColumnName 
    ActiveColumnNumber = Int((ActiveColumnNumber - m)/26) 
Loop 

End Function 


sub main() 

Dim firstrow_data_main As Integer 
Dim firstrow_fieldnames_main As Integer 

firstrow_data_main = 16 
firstrow_fieldnames_main = 15 


Range(ActiveColumnName("<FIELDNAME>", firstrow_fieldnames_main) & firstrow_data_main, Range(ActiveColumnName("ÄÅÔÅ", firstrow_fieldnames_main) & Rows.Count).End(xlUp).Offset(-1)).Select 
Application.CutCopyMode = False 
Selection.Copy 

Workbooks.Open help_file '"help_file" is any given .xls path with formulas 


Dim firstrow_data_help As Integer 
Dim firstrow_fieldnames_help As Integer 

firstrow_data_help = 7 
firstrow_fieldnames_help = 4 

'NOW WHEN I USE THIS, DOESNT WORK: 

-> Range(ActiveColumnName("<FIELDNAME>", firstrow_fieldnames_help) & firstrow_data_help).Select 

'WHEN I USE THIS, WORKS FINE: 

-> Range("L7").Select 

Selection.PasteSpecial Paste:=xlPasteValues, Operation:=xlNone, SkipBlanks:=False, Transpose:=False 
    Application.CutCopyMode = False 

End Sub 

当它不工作时，它打开.xls和它的确选择了理想的细胞，但没有脑袋。我知道这与剪贴板有关，但无法弄清楚。有什么建议么？

Answer 1

1. 通过直接引用单元格删除所有的选择和激活，见HERE获取更多信息。
2. 查看Cells（）而不是Range，并避免将列号转换为字母的整个需求，因为Cells（）使用数字。
3. 当值是唯一需要的值时避免剪贴板，并将值简单地分配给新单元格（这将要求两个范围的大小相同，因此请使用Resize（））
4. 总是表示范围，它会减少错误。

重构的代码

Sub main() 

Dim firstrow_data_main As Integer 
Dim firstrow_fieldnames_main As Integer 
Dim rng As Range 

Dim tWb As Workbook 
Dim ws As Worksheet 
Dim tWs As Worksheet 
Dim firstrow_data_help As Integer 
Dim firstrow_fieldnames_help As Integer 



Set ws = ThisWorkbook.ActiveSheet 
Set tWb = Workbooks.Open(help_file) 
Set tWs = tWb.ActiveSheet 

firstrow_data_main = 16 
firstrow_fieldnames_main = 15 
firstrow_data_help = 7 
firstrow_fieldnames_help = 4 

With ws 
    Set rng = .Range(.Cells(firstrow_data_main, firstrow_fieldnames_main), .Cells(.Rows.Count, firstrow_fieldnames_main).End(xlUp).Offset(-1)) 
    tWs.Cells(firstrow_data_help, firstrow_fieldnames_help).Resize(rng.Rows.Count, rng.Columns.Count).Value = rng.Value 
End With 

End Sub 

Answer 2

我认为这个问题可能是在这里：

ActiveColumnName("<FIELDNAME>", firstrow_fieldnames_main) 

这意味着ActiveColumnName是n x (1 to 2)维度的矩阵。如果你想连接一个名字给一个变量，你必须使用（例如）：

"YourStringHere" & YourVariableHere & "AnotherString" 

而你的情况应该是：

ActiveColumnName("<FIELDNAME>" & firstrow_fieldnames_main) 

所以，如果我理解正确的（<FIELDNAME>有点晦涩），整个命令应该是：

Range(ActiveColumnName("<FIELDNAME>" & firstrow_fieldnames_help) & "," & firstrow_data_help).Select 

Answer 3

首先，运行之前，你应该编译VBA。在VBA编译器捕获这种蝙蝠：
Dim ActiveColumnName As String
是不必要的，因为你分配ActiveColumnName作为字符串，当你在第1行

定义的函数可以使用很多活性细胞和细胞选择参考。这被称为导致运行时错误。看到这篇文章："How to avoid using Select in Excel Vba Macros"。

我怀疑字段名不在你认为它应该在你的help_file中的位置，即它不在第4行。这将意味着代码不知道在哪里粘贴数据。通常，调试的最佳方式是将代码分成尽可能最小的动作，以查看导致错误的原因（请参阅SpreadSheet Guru Strategies）。你可以运行下面的代码来看看输出是什么？

Function ActiveColumnName(fieldname As String, fieldnames_line As Integer) As String 

Range("A" & fieldnames_line & ":AB" & fieldnames_line).NumberFormat = "@" 
Cells.Find(What:=fieldname, After:=ActiveCell, LookIn:=xlFormulas, LookAt _ 
    :=xlWhole, SearchOrder:=xlByRows, SearchDirection:=xlNext, MatchCase:= _ 
    False, SearchFormat:=False).Activate 
ActiveColumnNumber = ActiveCell.Column 
Dim m As Integer 
ActiveColumnName = "" 

Do While (ActiveColumnNumber > 0) 
    m = (ActiveColumnNumber - 1) Mod 26 
    ActiveColumnName = Chr(65 + m) + ActiveColumnName 
    ActiveColumnNumber = Int((ActiveColumnNumber - m)/26) 
Loop 

End Function 
Sub main() 

Workbooks.Open "help_file" '"help_file" is any given .xls path with formulas 
Dim firstrow_data_help As Integer 
Dim firstrow_fieldnames_help As Integer 
firstrow_data_help = 7 
firstrow_fieldnames_help = 4 

MessageBox = ActiveColumnName("FIELDNAME", firstrow_fieldnames_help) & firstrow_data_help 

End Sub 

Answer 4

谢谢大家的回复！我尝试了所有的建议，但一路上遇到了各种问题。不过，您的所有建议都有助于我对这个问题发展出不同的看法。我最终得到的解决方案源于你的建议，放弃“.select”作为参考方式，并使用“rng”变量，当然也可以去掉双引用“ActiveColumnName”。我知道我有一个很长的路要走，但目前这个东西的作品!!谢谢！

子主（）

Dim firstrow_data_main As Integer 
Dim firstrow_fieldnames_main As Integer 
Dim firstrow_data_help As Integer 
Dim firstrow_fieldnames_help As Integer 



firstrow_data_main = 16 
firstrow_fieldnames_main = 15 
firstrow_data_help = 7 
firstrow_fieldnames_help = 4 


Dim rng1 As Range 
Dim rng2 As Range 

Set rng1 = Range(ActiveColumnName("<FIELDNAME>", firstrow_fieldnames_main) & firstrow_data_main, Range(ActiveColumnName("<FIELDNAME>", firstrow_fieldnames_main) & Rows.Count).End(xlUp).Offset(-1)) 

cells_selected = rng1.Rows.Count 

Workbooks.Open <help_file> 


Set rng2 = Range(ActiveColumnName("<FIELDNAME>", firstrow_fieldnames_help) & firstrow_data_help, Range(ActiveColumnName("<FIELDNAME>", firstrow_fieldnames_help) & cells_selected + firstrow_data_help - 1)) 

rng1.Copy rng2 

结束子