VB6迁移至VB.NET：添加ShockwaveFlash控件报错咨询

The error you’re hitting boils down to a critical difference between VB6 and VB.NET’s Controls.Add method: VB6 let you pass a ProgID string to dynamically spawn a control, but VB.NET requires an actual Control object instead. That’s why your old VB6 syntax breaks here.

Here’s how to fix this properly:

Step 1: Confirm Required References #

First, make sure your VB.NET project has the correct interop assemblies for the ShockwaveFlash control:

• AxShockwaveFlashObjects (the Windows Forms wrapper for the control)
• ShockwaveFlashObjects (the underlying COM interop assembly)

You can add these via the Toolbox (right-click → Choose Items → COM Components → select "Shockwave Flash Object") or through your project’s References dialog.

Step 2: Rewrite the Control Creation Code #

Replace your existing code with this VB.NET-compatible version:

Public WithEvents ShockwaveFlash1 As AxShockwaveFlashObjects.AxShockwaveFlash

' Put this in your form's initialization method (like Form_Load or the form's constructor):
ShockwaveFlash1 = New AxShockwaveFlashObjects.AxShockwaveFlash()
With ShockwaveFlash1
    .Name = "ShockwaveFlash1"
    .Parent = Me
    ' Set essential display properties (adjust values as needed):
    .Location = New System.Drawing.Point(10, 10)
    .Size = New System.Drawing.Size(400, 300)
    .Visible = True
End With
Me.Controls.Add(ShockwaveFlash1)

Why This Works #

• Instead of passing a ProgID string to Controls.Add, we directly instantiate the AxShockwaveFlash control with New.
• We explicitly set key properties (name, parent, position, size) before adding it to the form’s controls collection.
• The WithEvents declaration still functions exactly like it did in VB6, so you can handle the control’s events without extra changes.

Quick Troubleshooting Note #

If the control fails to render, try calling ShockwaveFlash1.CreateControl() right after adding it to the controls collection. In most cases, the Parent assignment and Controls.Add will handle this automatically, but it’s a handy fallback.

内容的提问来源于stack exchange，提问作者Himanshu