如何在asp.net中实现“验证码”的功能

Question

string ValidateString=(string)Session["ValidateString"];
            int height=20;
int length=ValidateString.Length*12;
System.Drawing.Bitmap image=new Bitmap(length,height);
            System.Drawing.Graphics graphic=Graphics.FromImage(image);
graphic.DrawString(ValidateString,(new Font("Lucida Calligraphy",12)),(new SolidBrush(Color.Blue)),3,0);
MemoryStream stream=new MemoryStream();
image.Save(stream,ImageFormat.Png);
Response.ClearContent();
Response.BinaryWrite(stream.ToArray());
            graphic.Dispose();
image.Dispose();
Response.End();

Answer 1

--showcode.aspx<asp:Image id="Image1" runat="server" ImageUrl="Code.aspx"></asp:Image>---code.aspx<%@ import namespace="System"%>
<%@ import namespace="System.IO"%>
<%@ import namespace="System.Drawing"%>
<%@ import namespace="System.Drawing.Imaging"%>
<%@ import namespace="System.Drawing.Drawing2D"%>
<script language="C#" runat="server">
private Bitmap validateimage;
private Graphics g;
 public void  Page_Load(object Sender   ,EventArgs e  )
 {
    if (!IsPostBack){ 
    string  VNum  =MakeValidateCode( );
Session["VNum"]=VNum;
    ValidateCode(VNum);
}
}public void  ValidateCode(string VNum)
{
validateimage = new Bitmap(70, 20, PixelFormat.Format24bppRgb);
g = Graphics.FromImage(validateimage); g.DrawString(VNum, new Font("宋体",15),new SolidBrush(Color.White),new PointF(8,0));
g.FillRectangle(new LinearGradientBrush(new Point(0,0), new Point(110,20), Color.FromArgb(0,0,0,0),Color.FromArgb(255,255,60,40)),0,0,120,30); 
g.Save();
MemoryStream ms=new MemoryStream();
validateimage.Save(ms,System.Drawing.Imaging.ImageFormat.Gif);
Response.ClearContent();
Response.ContentType="image/gif";
Response.BinaryWrite(ms.ToArray());
Response.End();
}string MakeValidateCode()
{
char[] s = new char[]{'0','1', '2','3','4','5','6','7','8','9','a'
,'b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q'
,'r','s','t','u','v','w','x','y','z','A','B','C','D','E','F','G'
,'H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W'
,'X','Y','Z'};
string num = "";
Random r = new Random();
for(int i = 0; i < 5; i++)
{
num += s[r.Next(0, s.Length)].ToString(); 
}
 return num;
}
</script>