[RESOLVED] VB6 - Mouse move on Shape1

I want to execute an instruction each time the mouse moves over a shape (a rectangle).
Picture1 control, for example, has that property. But when I try to find the same option in shape it doesn't exists.

How can I solve this situation?

I've thought about creating a matrix of coordenates (x,y) and if the mouse position is at some point of those coordinates then, execute the instruction. In any case, explain me how can I accomplish that.

Need to check coordinates on Form_Move event (or relative container of the shape).
Code:

Code:

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Private Sub Form_MouseMove(Button As Integer, Shift As Integer, X As Single, Y As Single)
    If X >= Shape1.Left And X <= (Shape1.Left + Shape1.Width) And Y >= Shape1.Top And Y <= (Shape1.Top + Shape1.Height) Then
        Label1.Caption = "IN"
    Else
        Label1.Caption = "OUT"
    End If
End Sub

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Quote:

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Originally Posted by Jose_VB

I've thought about creating a matrix of coordenates (x,y) and if the mouse position is at some point of those coordinates then, execute the instruction. In any case, explain me how can I accomplish that.

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Try this

Code:

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Private Sub Form_MouseMove(Button As Integer, Shift As Integer, X As Single, Y As Single)
    Const x1 As Single = 1000
    Const x2 As Single = 4000
    Const y1 As Single = 1000
    Const y2 As Single = 4000
   
    If X > x1 And X < x2 And Y > y1 And Y < y2 Then
        Me.Caption = "moving inside region"
    Else
        Me.Caption = "moving outside region"
    End If
End Sub

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from 4x2y's example

Code:

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Private Sub Form_MouseMove(Button As Integer, Shift As Integer, X As Single, Y As Single)
    Dim x1 As Single, x2 As Single
    Dim y1 As Single, y2 As Single
   
    x1 = Shape1.Left
    x2 = Shape1.Left + Shape1.Width
    y1 = Shape1.Top
    y2 = Shape1.Top + Shape1.Height
   
    If X > x1 And X < x2 And Y > y1 And Y < y2 Then
        Me.Caption = "moving inside region"
    Else
        Me.Caption = "moving outside region"
    End If
End Sub

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Thank you very much to all.
It works perfectly.