VBA：检查每个头是否在范围内罚款

你好，我开始在VBA，并要检查我的范围内的每个标题是否处于良好的位置。 所以第一的位置是：“Dana wartosc”第二：“a”等等。但问题是，在每一个循环中，都把我带到别的地方（这意味着头是坏的），我不明白为什么， m检查RangeHolder(1)通过Debug.Print RangeHolder(1)在代码中它显示我适当的价值。

你能告诉我我错过了什么吗？

我想要的是当它在检查开关第一个标题。 消息是说，第一个头是好的，它是在循环检查每个头

 Option Explicit Sub Szukanka() Dim UpRow As Integer, DownRow As Integer, RangeHolder As Range Dim x x = 1 UpRow = 1 DownRow = 5 Set RangeHolder = Range(Cells(UpRow, 1), Cells(DownRow, 4)) RangeHolder.Select For x = 1 To 4 Select Case RangeHolder(x) Case RangeHolder(1) = "Dana wartosc" MsgBox ("Its good") Case RangeHolder(2) = "a" MsgBox ("Its good") Case RangeHolder(3) = "b" MsgBox ("Its good") Case RangeHolder(4) = "c" MsgBox ("Its good") Case Else MsgBox ("Its bad" + RangeHolder(x)) End Select Next x End Sub 

Answer 1

您的Select Case语句被错误地写入了，您应该将RangeHolder当作一个二维（行x列）对象。 （将其视为一维不会造成错误，但不太可能成为您想要做的事情。请参阅此问题以解决混淆其他用户的情况。）

 Option Explicit Sub Szukanka() Dim UpRow As Integer, DownRow As Integer, RangeHolder As Range Dim x x = 1 UpRow = 1 DownRow = 5 Set RangeHolder = Range(Cells(UpRow, 1), Cells(DownRow, 4)) Dim Good As Boolean For x = 1 To 4 Good = False Select Case RangeHolder(1, x).Value Case "Dana wartosc" Good = x = 1 Case "a" Good = x = 2 Case "b" Good = x = 3 Case "c" Good = x = 4 End Select If Good Then MsgBox "It's good" Else MsgBox "It's bad " & RangeHolder(1, x) End If Next x End Sub 

我试图写上面的Select Case来做我认为你正在做的事情（即检查第一行，以确保它具有正确的值作为标题？）。 但是，这实际上并不适用于Select Case结构，因此最好将其写为If语句

 Option Explicit Sub Szukanka() Dim UpRow As Integer, DownRow As Integer, RangeHolder As Range Dim x x = 1 UpRow = 1 DownRow = 5 Set RangeHolder = Range(Cells(UpRow, 1), Cells(DownRow, 4)) Dim CorrectValues As Variant CorrectValues = Array("Dana wartosc", "a", "b", "c") For x = 1 To 4 If RangeHolder(1, x).Value = CorrectValues(x - 1) Then MsgBox "It's good" Else MsgBox "It's bad " & RangeHolder(1, x) End If Next x End Sub 

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你的Select Case不工作的原因：

如果您的原始Select Case语句被重写为等效块If语句，则看起来像这样：

 If RangeHolder(x) = (RangeHolder(1) = "Dana wartosc") Then MsgBox ("Its good") ElseIf RangeHolder(x) = (RangeHolder(2) = "a") Then MsgBox ("Its good") ElseIf RangeHolder(x) = (RangeHolder(3) = "b") Then MsgBox ("Its good") ElseIf RangeHolder(x) = (RangeHolder(4) = "c") Then MsgBox ("Its good") Else MsgBox ("Its bad" + RangeHolder(x)) End If 

如果RangeHolder(x)的值是"a" （ RangeHolder(1)是"Dana wartosc" ， RangeHolder(2)是"a" ， RangeHolder(3)是"b" ， RangeHolder(4)是"c" ），然后变成：

 If "a" = True Then MsgBox ("Its good") ElseIf "a" = True Then MsgBox ("Its good") ElseIf "a" = True Then MsgBox ("Its good") ElseIf "a" = True Then MsgBox ("Its good") Else MsgBox ("Its bad" + "a") End If 

因为"a"不是= True （如果这样写，实际上会产生types不匹配，但不会在Select Case语法中），则会调用Elseexpression式。

Answer 2

select语句被错误地编写，而且overral方法也不是一个好方法，因为它会导致代码的可读性差。 删除RangeHolder.Select所有行，并将其replace为这样，以便您的正确值以可读方式出现在代码中：

  Dim correctValues: correctValues = Array("", _ "Dana wartosc", "a", "b", "c") '<-- write correct sequence here For x = 1 To UBound(correctValues) If RangeHolder(x) <> correctValues(x) Then msgBox (RangeHolder(x) & " is not at the correct position :(") Exit Sub End If Next x msgBox "All is good :) " End Sub 

Answer 3

你的语法有缺陷。 请试试这个：

 Sub Szukanka() Dim UpRow As Long, DownRow As Long ' rows and columns are generally Longs Dim RangeHolder As Range Dim x As Long ' it's a column, right? x = 1 UpRow = 1 DownRow = 5 Set RangeHolder = Range(Cells(UpRow, 1), Cells(UpRow, 4)) ' RangeHolder.Select ' don't need to select anything For x = 1 To 4 Select Case RangeHolder(x).Value Case "Dana wartosc" MsgBox ("Its good") Case "a" MsgBox ("Its good") Case "b" MsgBox ("Its good") Case "c" MsgBox ("Its good") Case Else MsgBox ("Its bad" + RangeHolder(x)) End Select Next x End Sub