VLookup字母数字值 – 删除行

现在我真的很担心VLookup的情况。 我正在努力做一些真正在我头上的事情，但是我认为只需要一点帮助，我就可以做到。

1. 对于初学者，我创build了一个单独的表和一个表，所以我可以给一些字母数字，所以我可以量化他们的价值。

2. 有了这些信息，我想详细说明一个VBA公式，该公式可以检查列中的那些字母（带有一个数字）的表格，并删除每一行，表示一个低于input字母给定数字的字母值。 （忽略列I中的N）

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3. 我到目前为止是这样的：

   Sub DeletarIndices() indice = InputBox("Digite o IC/IV Desejado", "GBG Pneus") Set planilhaV = Sheets("IV") Dim vValores As String sResult = Application.VLookup("Y", planilhaV.Range("A2:B11"), 2) Application.ScreenUpdating = False Application.Calculation = xlCalculationManual Dim i As Long For i = Range("A" & Rows.Count).End(xlUp).Row To 1 Step -1 If Not (Range("A" & i).Value > sResult) Then Range("A" & i).EntireRow.Delete End If Next i Application.Calculation = xlCalculationAutomatic Application.ScreenUpdating = True End Sub 

发生什么事是，当我运行代码时，一个input框打开，并要求在代码中使用的值。 该字母将被转换为数字，并将被用作比较以删除较低的值。 但是…在这里：

 If Not (Range("A" & i).Value > sResult) Then 

这个字母还是一个数字，所以当它检查表格时，只有低于该值的数字才会被删除，而不是低的字母。 我需要的就是这个。 这些字母被分析，行被删除，而不是数字。

感谢任何可能帮助我的人！

Answer 1

假设你的信件的优先顺序总是按字母顺序排列的，你不需要将字母转换成数字，你可以简单地比较字母和字母。 我也会注意到你的方法来确定最后一行（ Range("A" & Rows.Count).End(xlUp).Row ）并不是很好的做法，我build议你找一个更可靠的方法。

假设您的数据如您所示 – H列中的数据为3个字符的代码，其中字母是最后一个字符（例如“91B”，“89D”等） – 让我们一步一步来：注意没有查找工作表planilhaV了）

1）声明我们的variables

 Dim indice As String ' To hold our user input letter Dim rowLetter As String ' To hold the current row letter value Dim firstRow As Long ' First row of your data Dim lastRow As Long ' Last row of your data Dim currentRow As Long ' Current row for your loop counter Dim sht As Worksheet ' To hold the worksheet you're working on 

2）分配一些值

 Set sht = ThisWorkbook.Worksheets("*WORKSHEET NAME*") ' Use the name of your worksheet firstRow = 1 lastRow = sht.Range("A" & Rows.Count).End(xlUp).Row 

3）获取用户input

 indice = UCase(InputBox("Digite o IC/IV Desejado", "GBG Pneus")) ' Convert user input to upper case and store 

4）做你的处理

 For currentRow = lastRow To firstRow Step -1 rowLetter = UCase(Right(sht.Range("H" & currentRow).Value, 1)) ' Get letter from code in column H, converted to upper case If rowLetter < indice Then ' Compare the letters, and if smaller than user input... sht.Rows(currentRow).EntireRow.Delete ' Delete the row End If Next currentRow 

而这应该关于做到这一点。 声明你打算使用的所有variables是一个好主意，并且使用Option Explicit将确保你不能在没有声明的情况下使用任何东西。

编辑：

@LeonVFX：如果字母的值不总是按字母顺序增加，那么你可以使用查找表，或者你可以在代码中进行比较。

如果您select使用类似于您的原始示例中的查找表，那么您可以调整上面的代码，如下所示：

 Dim indice As String ' To hold our user input letter Dim indiceValue As Long ' To hold the numeric value of our user input letter Dim rowLetter As String ' To hold the current row letter Dim rowLetterValue As Long ' To hold the numeric value of the current row letter Dim firstRow As Long ' First row of your data Dim lastRow As Long ' Last row of your data Dim currentRow As Long ' Current row for your loop counter Dim sht As Worksheet ' To hold the worksheet you're working on Dim planilhaV As Worksheet ' To hold your lookup worksheet Set sht = ThisWorkbook.Worksheets("*WORKSHEET NAME*") ' Use the name of your worksheet Set planilhaV = ThisWorkbook.Worksheets("IV") ' As in your original example firstRow = 1 lastRow = sht.Range("A" & Rows.Count).End(xlUp).Row indice = UCase(InputBox("Digite o IC/IV Desejado", "GBG Pneus")) ' Convert user input to upper case and store indiceValue = CLng(Application.VLookup(indice, planilhaV.Range("A2:B11"), 2, False)) ' Creating numeric indice value with lookup table For currentRow = lastRow To firstRow Step -1 rowLetter = UCase(Right(sht.Range("H" & currentRow).Value, 1)) ' Get letter from code in column H, converted to upper case rowLetterValue = CLng(Application.VLookup(rowLetter, planilhaV.Range("A2:B11"), 2, False)) ' Creating numeric value for current row letter with lookup table If rowLetterValue < indiceValue Then ' Compare the numeric letter values, and if smaller than user input... sht.Rows(currentRow).EntireRow.Delete ' Delete the row End If Next currentRow 

如果你发现它一次只删除一行，我的猜测是使用sht.Range("A" & Rows.Count).End(xlUp).Rowfind最后一行是原因，因为你不A列没有任何数据？ 如果你的数据在你的原始示例中的H中，那么在代码中把A改为H ，你应该没问题，或者试着找一个更可靠的方法来设置你的lastRowvariables值。

Answer 2

在论坛上得到一些帮助后，我得到了这个答案，如果有人遇到这样的情况，我会把它留在这里。

 ' Verifies if the digit is a number. If it is, returns True, False if it isn't. Function ehNumero(ByVal digito As String) As Boolean a = Asc(digito) If a >= 48 And a <= 57 Then ehNumero = True Else enNumero = False End If End Function ' Separates the code in a numeric and a alphabetic part. Function separaCodigo(ByVal codigo As String, ByRef numero As Integer, ByRef letras As String) As Boolean p = 0 For i = 1 To Len(codigo) digito = Mid(codigo, i, 1) If Not ehNumero(digito) Then ' Found the point break when it finds the first non-numeric digit. p = i Exit For End If Next i If p = 0 Or p = 1 Then numero = 0 letras = "" separaCodigo = False Else codigo = UCase(codigo) numero = Int(Mid(codigo, 1, p - 1)) letras = Mid(codigo, p) separaCodigo = True End If End Function ' Gets the values from the configuration table. Function valorDasLetras(ByVal letras As String) As Integer On Error GoTo trataErro valorDasLetras = Application.VLookup(letras, Worksheets("Configuração").Range("A2:B11"), 2, False) Exit Function trataErro: valorDasLetras = 0 ' End Function 'Deletes the lines in the table in the interval. Function deletar(ByVal numero As Integer, letras As String) As Integer valor = valorDasLetras(letras) If valor = 0 Then deletar = -1 Exit Function End If limInf = numero limSup = valor Dim dados As Worksheet Set dados = ActiveWorkbook.ActiveSheet Dim linhasPraDeletar As Range totalLinhas = 0 linha = 1 Do While True curCodigo = dados.Cells(linha, 7) ' Using Column G If curCodigo = "" Then Exit Do End If Dim curNumero As Integer Dim curLetras As String If Not separaCodigo(curCodigo, curNumero, curLetras) Then deletar = -1 Exit Function End If ' Gets the values for the letters from the table curValor = valorDasLetras(curLetras) If curValor = 0 Then deletar = -1 Exit Function End If If curNumero < limInf Or curValor < limSup Then If linhasPraDeletar Is Nothing Then Set linhasPraDeletar = dados.Rows(linha) Else Set linhasPraDeletar = Union(linhasPraDeletar, dados.Rows(linha)) End If totalLinhas = totalLinhas + 1 End If linha = linha + 1 Loop linhasPraDeletar.Select linhasPraDeletar.EntireRow.Delete ' <===== Comment to select, but do not delete, the line deletar = totalLinhas End Function ' Run the code Sub LimparValores() 'On Error GoTo trataErro ' Reads the user code msg = "Input your code" codigo = InputBox(msg, "Código") If codigo = "" Then ' Cancelado Exit Sub End If ' Separates the user code number from letters Dim numero As Integer Dim letras As String If Not separaCodigo(codigo, numero, letras) Then MsgBox ("Invalid code: " & codigo) Exit Sub End If ' Calls the delete function linhas = deletar(numero, letras) If linhas = -1 Then MsgBox ("There was an error with the code (the letter does not exist in configuration)") Else If linhas = 0 Then MsgBox ("There are no rows in the interval - no row was deleted") Else MsgBox (linhas & " rows deleted!") End If End If Exit Sub trataErro: MsgBox ("The code is not in the expected format.") End Sub 

希望它可以帮助任何人在未来！