Complie错误 - 否则没有如果VBA

Question

我是vba的新手，所以请原谅我的无知。我创建了一个将通过一个下拉列表中运行，并打印在下拉列表中每个名字的宏 - 和它的作品：

Sub PrintAll() 

    Dim Cell As Range 
    Dim Rng As Range 
    Dim Wks As Worksheet 

    Set Wks = Worksheets("PRINT PAGE") 
    Set Rng = ThisWorkbook.Names("Brokers").RefersToRange 


    For Each Cell In Rng 
    If Cell <> "" Then 
     Wks.Range("$B$5").Value = Cell.Text 
     Wks.PrintOut 
    End If 
    Next Cell 

End Sub 

但是，该工作簿具有多个工作表进行读取，因此我需要的VBA从读多个范围，所以我想这

Sub PrintAll() 

    Dim Cell As Range 
    Dim Cell2 As Range 
    Dim Rng As Range 
    Dim Wks As Worksheet 

Set Wks = Worksheets("PRINT PAGE") 

If "$A$5" = "Company1" Then Rng = ThisWorkbook.Names("1Brokers").RefersToRange 
ElseIf "$A$5" = "Company2" Then Rng = ThisWorkbook.Names("2Brokers").RefersToRange 
Else: Set Rng = ThisWorkbook.Names("3Brokers").RefersToRange 
End If 

    For Each Cell In Rng 
    If Cell <> "" Then 
     Wks.Range("$B$5").Value = Cell.Text 
     Wks.PrintOut 
    End If 
    Next Cell 

End Sub 

问题是我不断收到“编译错误否则没有如果” IF语句的。我是如何设置If语句或者如何在代码中使用它的？

Answer 1

该块为我编译。请测试它。我始终在then之后开始设置if。

Sub PrintAll() 

    Dim Cell As Range 
    Dim Cell2 As Range 
    Dim Rng As Range 
    Dim Wks As Worksheet 

'Set Wks = Worksheets("PRINT PAGE") 

If "$A$5" = "Company1" Then 
    Rng = ThisWorkbook.Names("1Brokers").RefersToRange 
ElseIf "$A$5" = "Company2" Then 
    Rng = ThisWorkbook.Names("2Brokers").RefersToRange 
Else 
    Set Rng = ThisWorkbook.Names("3Brokers").RefersToRange 
End If 

    For Each Cell In Rng 
    If Cell <> "" Then 
     Wks.Range("$B$5").Value = Cell.Text 
     Wks.PrintOut 
    End If 
    Next Cell 

End Sub 

Answer 2

不是：

If "$A$5" = "Company1" Then Rng = ThisWorkbook.Names("1Brokers").RefersToRange 

使用：

If "$A$5" = "Company1" Then Set Rng = ThisWorkbook.Names("1Brokers").RefersToRange 

有可能是在你的代码的其他问题。

Answer 3

使用

If Range("$A$5").Value = "Company1".. 

同样为 “$ A $ 5”。

“$ A $ 5”只是一个字符串，您将字符串与字符串进行比较。你想要的是一个范围对象

编辑

关于你得到，你必须使用Then

的语法（Hiten004后让我意识到这一点）

If <Cond> Then 
ElseIF <Cond> Then 
End If 

错误