由于括号所在的位置,我很难让我的函数识别该过程。
以下代码不起作用。
功能
Public Function KonKatenate(rIN As range) As String
Dim r As range
For Each r In rIN
KonKatenate = Replace(KonKatenate & r.Text, ".", "")
Next r
End Function
程序
Sub LoopThroughUntilBlanks()
Dim xrg As range
Cells(3, 951).Select
' Set Do loop to stop when two consecutive empty cells are reached.
Application.ScreenUpdating = False
i = 3
Do Until IsEmpty(ActiveCell) And IsEmpty(ActiveCell.Offset(0, -2).Value)
Cells(i, 951).Value = KonKatenate(range("AJE" & i & ":AJG" & i & ")"))
ActiveCell.Offset(1, 0).Select
i = i + 1
Loop
Application.ScreenUpdating = False
End Sub
当我完全删除括号并使用例如静态数字时,这有效:
Cells(i, 951).Value = KonKatenate(range("AJE3:AJG3"))
但是我需要 3 作为变量 i 以便循环向下超越行
非常需要建议
最佳答案
您的 KonKatenate 函数在循环范围时不断覆盖其自身的结果。您需要不断地将新的字符串连接到结果上。您的原始文件中没有分隔符,但我添加了一种简单的方法来包含分隔符。
Public Function KonKatenate(rIN As range) As String
Dim r As range, d as string
d = ""
For Each r In rIN
KonKatenate = KonKatenate & d & Replace(r.Text, ".", "")
Next r
KonKatenate = mid(KonKatenate, len(d)+1)
End Function
您的 LoopThroughUntilBlanks 子过程应该使用它声明的变量并声明它使用的变量。 For ... Next 循环可能更合适。
Sub LoopThroughUntilBlanks()
dim lr as long, i as long
Application.ScreenUpdating = False
with activesheet '<~~ would be better as a defined worksheet
lr = application.max(.cells(.rows.coun, "AJO").end(xlup).row, _
.cells(.rows.coun, "AJO").Offset(0, -2).end(xlup).row)
for i=3 to lr
.Cells(i, "AJO").Value = KonKatenate(.range(.cells(i, "AJE"), .cells(i, "AJG")))
next i
end with
Application.ScreenUpdating = False
End Sub
关于vba - 括号阻止函数运行,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50884095/