如何在VB6中检测显示器的分辨率？

Private Sub Form_Load()
    ResWidth = Screen.Width \ Screen.TwipsPerPixelX
    ResHeight = Screen.Height \ Screen.TwipsPerPixelY
    ScreenRes = ResWidth & "x" & ResHeight
    MsgBox (ScreenRes)
End Sub

还有其他几个类似的代码我用Google搜索过.问题是,我总是得到一个消息框说我的分辨率是1200×1200,虽然我的实际分辨率是1920×1200.为什么我的成绩不好？

Private Declare Function GetSystemMetrics Lib "user32" _
    (ByVal nIndex As Long) As Long

然后,当您需要屏幕宽度和高度时,请定义以下常量：

Private Const SM_CXSCREEN = 0
Private Const SM_CYSCREEN = 1

然后,您可以在任何需要的地方使用GetSystemMetrics.如果将声明和常量添加到模块(.BAS)更有意义,那么只需声明声明和常量.

Dim width as Long, height as Long
width = GetSystemMetrics(SM_CXSCREEN)
height = GetSystemMetrics(SM_CYSCREEN)

GetSystemMetrics on Microsoft Support