vb.net 數(shù)組顯示矩陣的一道題

'這是在vb6中的代碼，在vb.net中基本差不多，你可以參考一下

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Private Sub cmdCommand1_Click()

Me.AutoRedraw = True

Dim Grp

Grp = Array(1, 2, 3, 4, 5)

Dim i, j As Long

Dim StrPrt As String

For i = 0 To UBound(Grp)

'i為位移量

StrPrt = ""

For j = i To UBound(Grp)

StrPrt = StrPrt Grp(j)

Next j

For j = 0 To i - 1

StrPrt = StrPrt Grp(j)

Next j

Me.Print StrPrt

Next i

End Sub

vb.net矩陣的問題

要實現(xiàn)什么樣的功能呢?矩陣就是二維表吧,在.Net中有許多方法可以實現(xiàn)二維表,根據(jù)不同的需求選擇適合的方法,你應(yīng)該詳細一點說明

VB.NET：繪圖后，如何才能取得所繪圖形的顏色值？

加入一個TextBox控件,一個Command控件

代碼:

Private Declare Function GetPixel Lib "gdi32" (ByVal hDC As Long, ByVal X As Long, ByVal Y As Long) As Long

Private Declare Function GetWindowDC Lib "user32" (ByVal hWnd As Long) As Long

Private Sub Command1_Click()

Dim Color As Long

WindowDC = GetWindowDC(0) '獲取屏幕的設(shè)備場景

Color = GetPixel(WindowDC, 500, 100) '獲指定點的顏色

'分解RGB顏色值

R = (Color Mod 256) '紅色

b = (Int(Color \ 65536)) '藍色

G = ((Color - (b * 65536) - R) \ 256) '綠色

Text1.BackColor = RGB(R, G, b)

End Sub

VBNET怎么獲取控件顏色的RGB值，又怎么用RGB將顏色賦給控件顏色。

R/G/B值最小是0最大是255屬Byte值類型

Dim cr As Color = 控件.BackColor '獲取控件背景色

Dim alpha As Byte = cr.A '透明度

Dim R As Byte = cr.R 'R值

Dim G As Byte = cr.G 'G值

Dim B As Byte = cr.B 'B值

Dim outAcr As Color = Color.FromArgb(alpha, R, G, B) '創(chuàng)建帶有透明通道的ARGB顏色

Dim outcr As Color = Color.FromArgb(R, G, B) '創(chuàng)建不透明的RGB顏色

vb.net拾色器設(shè)計，要求：能獲取圖片任意位置的顏色

VB可使用Point方法來獲取圖片指定點的顏色。

Point 方法

按照長整數(shù)，返回在 Form 或 PictureBox 上所指定磅的紅-綠-藍 (RGB) 顏色。

語法

object.Point(x, y)

'窗體判色代碼:

Private Sub Form1_MouseMove(Button As Integer, Shift As Integer, X As Single, Y As Single)

Text1 = X

Text2 = Y

Text3 = Point(X, Y)

Text4 = (Val(Text3) Mod 65536) Mod 256 'Red

Text5 = (Val(Text3) Mod 65536) \ 256 'Green

Text6 = Val(Text3) \ 65536 'Blue

Shape1.FillColor = RGB(Val(Text4), Val(Text5), Val(Text6))

End Sub

'PictureBox判色代碼:

Private Sub Picture1_MouseMove(Button As Integer, Shift As Integer, X As Single, Y As Single)

Text1 = X

Text2 = Y

Text3 = Picture1.Point(X, Y)

Text4 = (Val(Text3) Mod 65536) Mod 256 'Red

Text5 = (Val(Text3) Mod 65536) \ 256 'Green

Text6 = Val(Text3) \ 65536 'Blue

Shape1.FillColor = RGB(Val(Text4), Val(Text5), Val(Text6))

End Sub