我只是新手,我试图在Visual Basic 6中创建一个简单的程序.代码几乎相当于教科书中的代码.它本来是一种油漆程序.令人惊讶的是,它无法使用此问题标题中给出的错误进行编译.
这是代码：
这是代码：
Option Explicit
Dim Col As Long
Private Sub Form_Load()
AutoRedraw = True
BackColor = vbWhite
Col = vbBlack
DrawWidth = 3
End Sub
Private Sub Command1_Click()
CommonDialog1.ShowOpen
Form1.Picture = LoadPicture(CommonDialog1.FileName)
End Sub
Private Sub Command2_Click()
CommonDialog1.ShowSave
SavePicture Image,CommonDialog1.FileName
End Sub
Private Sub Command3_Click()
CommonDialog1.ShowColor
Col = CommonDialog1.Color
End Sub
Private Sub Form_MouseMove(Button As Integer,Shift As Integer,X As Single,Y As Single)
PSet (X,Y),Col
End Sub
Private Sub Toolbar1_ButtonClick(ByVal Button As MSComctlLib.Button)
Select Case Button.Key
Case "Line1"
DrawWidth = 3
Case "Line2"
DrawWidth = 20
End Select
End Sub
应用程序崩溃在以下行：
Private Sub Toolbar1_ButtonClick(ByVal Button As MSComctlLib.Button)
有错误：
procedure declaration does not match description of event or procedure
having the same name
问题出在这里：
原文链接：/vb/255439.htmlPrivate Sub Toolbar1_ButtonClick(ByVal Button As MSComctlLib.Button)
好吧,因为你在VB6编码,你可以学习VB6剧本中的一些技巧.暂时将方法重命名为qqToolbar_ButtonClick之类的其他内容,然后转到设计器并单击工具栏中的按钮以在代码中重新生成事件.
另一项检查是查看ToolBar1是否已添加到控件数组中？在这种情况下,方法签名需要如下所示：
Private Sub Toolbar1_ButtonClick(ByVal Index as Integer,ByVal Button As MSComctlLib.Button)
我希望其中一个有助于解决问题.