它的作用:我有一段代码在单元格中找到子串TL,并通过在&#34之后立即添加或删除0来强制其后面的数字为长度为6 ; TL - &#34 ;. (即TL-00072 -> TL-000072
,TL-034 -> TL-000034
,TL-000000789 -> TL-000789
)
我想要它做什么:但是,有时一个单元格中有多个TL值。我需要找出TL是否有第二次出现,如果是,则删除第二次出现及其后的所有内容。
示例:
Start: Output:
TL-000789 TL-000187 TL-000773 -> TL-000789
TL-000689 TL -000787 -> TL-000689
TL-000982 TL - 980819 -> TL-000982
这是尝试使用split进行的代码(不正确且无法正常工作)(也许修剪也会起作用?),它会找到TL的第二次出现并删除它之后的所有内容。完整的工作代码。
当前的代码尝试
Dim splitValues As Variant
If Str(str, "TL" + 1) 'do not know how to get SECOND occurrence
splitValues = Split(theValue, "TL")
theValue = splitValues(0)
End If
工作代码
[将新代码添加到开头](问题顶部的解释)
注意:StartSht是值和代码一样的工作簿。 所有被更改的值都在列" C"
中'force length of TL/CT to be 6/4 numbers long, eliminate spaces
Dim str As String, ret As String, tmp As String, j As Integer, k As Integer
For k = 2 To StartSht.Range("C2").End(xlDown).Row
ret = ""
str = StartSht.Range("C" & k).Value
'for TL numbers
''''''''''''''new code goes here''''''''''''''''''
If InStr(str, "TL") > 0 Then
For j = 1 To Len(str)
tmp = Mid(str, j, 1)
If IsNumeric(tmp) Then ret = ret + tmp
Next j
'force to 6 numbers if too short; add 0s immediately after "TL-"
For j = Len(ret) + 1 To 6
ret = "0" & ret
Next j
'force to 6 numbers if too long; eliminate 0s immediately after "TL-"
If Len(ret) > 6 Then
Debug.Print Len(ret)
For j = Len(ret) To 7 Step -1
If Mid(ret, 1, 1) = "0" Then
ret = Right(ret, j - 1)
End If
Next j
End If
'eliminate superfluous spaces around "TL-"
ret = "TL-" & ret
StartSht.Range("C" & k).Value = ret
'for CT numbers
ElseIf InStr(str, "CT") > 0 Then
For j = 1 To Len(str)
tmp = Mid(str, j, 1)
If IsNumeric(tmp) Then ret = ret + tmp
Next j
'force to 4 numbers if too short; add 0s immediately after "CT-"
For j = Len(ret) + 1 To 4
ret = "0" & ret
Next j
'force to 4 numbers if too long; eliminate 0s immediately after "CT-"
If Len(ret) > 4 Then
Debug.Print Len(ret)
For j = Len(ret) To 5 Step -1
If Mid(ret, 1, 1) = "0" Then
ret = Right(ret, j - 1)
End If
Next j
End If
'eliminate superfluous spaces around "CT-"
ret = "CT-" & ret
StartSht.Range("C" & k).Value = ret
End If
Next k
答案 0 :(得分:1)
如果找到第二个“TL”,看起来你可以重新定义str。行后:
If InStr(str, "TL") > 0 Then
添加另一个IF声明:
If InStr(3, str, "TL") > 0 Then str = Mid(str, 1, Instr(3, str, "TL") - 2)
然后使用新的str
继续使用其余代码。