当我运行我的程序时,我得到错误。我想将图像更改为另一个图像,但图像不会更改。请帮我找出来。
这是我的代码:
Public Image1Color As String
Public Image1 As Image
Public Sub CheckIn()
If Image1Color = "Green" Then
Image1.Picture = LoadPicture("Color\red1.jpg")
Image1Color = "Red"
Else
MsgBox ("This table is already occupied!")
End If
End Sub发布于 2012-07-17 01:33:03
试一试:
Set Image1.Picture = LoadPicture("Color\red1.jpg")(请注意Set)
发布于 2012-07-17 02:02:40
您可以使用stdpicture
Dim pic As New StdPicture
Set pic = LoadPicture("")
Set Me.Picture = pic在您的代码中,如:
Public Image1Color As String
Public Image1 As New StdPicture
Public Sub CheckIn()
If Image1Color = "Green" Then
Set Image1 = LoadPicture("Color\red1.jpg")
Image1Color = "Red"
Else
MsgBox ("This table is already occupied!")
End If
End Sub发布于 2014-12-17 18:14:02
您必须在代码中的某个位置初始化Image1Color = "Green"。
如果不初始化Image1Color,那么它将是一个空字符串,并且总是转到语句的Else部分
编辑
我刚刚读了你对错误消息的评论。
将Image控件添加到窗体并使用该Image控件比在代码中声明它更容易。
https://stackoverflow.com/questions/11508936
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