彩色验证码图片可以防御别人的攻击?
因为当别人用轮询技术模拟登录的时候,他并不知道你的验证码是什么,也获取不到,因为这是一张图片,电脑并不能识别里面的数字是什么(除非破解验证码里面的干扰,再利用相关的图片识别技术有可能读出验证码,这里先不扯这个)。读不出验证码就没有机会轮询访问了,当然我们后台判断的时候一定要先判断验证码是否正确,以防止占用服务器资源。
3、随机数 code
①数字随机数
1 ///
2 /// 数字随机数
3 ///
4 ///
5 private string getrndnum()
6 {
7 string code = string.empty;
8 random random = new random();
9 for (int i = 0; i < 4; i++)
10 {
11 code = code + random.next(9).tostring();
12 }
13 return code;
14 }
②字符串随机数
1 ///
2 /// 字符串验证码
3 ///
4 ///
5 private string getrndstr()
6 {
7 string vchar = "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";
8 string[] vcarray = vchar.split(',');
9 string checkcode = string.empty;
10 random rand = new random();
11 for (int i = 0; i < 4; i++)
12 {
13 rand = new random(unchecked((int)datetime.now.ticks));//为了得到不同的随机序列
14 int t = rand.next(vcarray.length);// the exclusive upper bound of the random number to be generated. maxvalue must be greater than or equal to zero,下标从0开始
15 checkcode += vcarray[t];
16 }
17 return checkcode;
18 }
1 ///
2 /// 随机中文码
3 ///
4 ///
5 private string getrndch()
6 {
7 system.text.encoding gb = system.text.encoding.default;//获取gb2312编码页(表)
8 object[] bytes = createregioncode(4);//调用函数产生4个随机中文汉字编码
9 string[] str = new string[4];
10 system.text.stringbuilder sb = new system.text.stringbuilder();
11 for (int i = 0; i < 4; i++)
12 {
13 //根据汉字编码的字节数组解码出中文汉字
14 str[i] = gb.getstring((byte[])convert.changetype(bytes[i], typeof(byte[])));
15 sb.append( str[i].tostring());
16 }
17 return sb.tostring ();
18 }
19
20
21 ///
22 /// 产生随机中文汉字编码
23 ///
24 ///
25 ///
26 private static object[] createregioncode(int strlength)
27 {
28 //定义一个字符串数组储存汉字编码的组成元素
29 string[] rbase = new string[16] { "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "a", "b", "c", "d", "e", "f" };
30 random rnd = new random();
31 object[] bytes = new object[strlength];
32
33 for (int i = 0; i < strlength; i++)
34 {
35 //区位码第1位
36 int r1 = rnd.next(11, 14);
37 string str_r1 = rbase[r1].trim();
38
39 //区位码第2位
40 rnd = new random(r1 * unchecked((int)datetime.now.ticks) + i);
41 int r2;
42 if (r1 == 13)
43 {
44 r2 = rnd.next(0, 7);
45 }
46 else
47 {
48 r2 = rnd.next(0, 16);
49 }
50 string str_r2 = rbase[r2].trim();
51
52 //区位码第3位
53 rnd = new random(r2 * unchecked((int)datetime.now.ticks) + i);//更换随机种子
54 int r3 = rnd.next(10, 16);
55 string str_r3 = rbase[r3].trim();
56
57 //区位码第4位
58 rnd = new random(r3 * unchecked((int)datetime.now.ticks) + i);
59 int r4;
60 if (r3 == 10)
61 {
62 r4 = rnd.next(1, 16);
63 }
64 else if (r3 == 15)
65 {
66 r4 = rnd.next(0, 15);
67 }
68 else
69 {
70 r4 = rnd.next(0, 16);
71 }
72 string str_r4 = rbase[r4].trim();
73
74 //定义两个字节变量存储产生的随机汉字区位码
75 byte byte1 = convert.tobyte(str_r1 + str_r2, 16);
76 byte byte2 = convert.tobyte(str_r3 + str_r4, 16);
77
78 //将两个字节变量存储在字节数组中
79 byte[] str_r = new byte[] { byte1, byte2 };
80
81 //将产生的一个汉字的字节数组放入object数组中
82 bytes.setvalue(str_r, i);
83 }
84 return bytes;
85 }
1 ///
2 /// 画图片的背景图,干扰
3 ///
4 ///
5 ///
6 private bitmap createimages(string checkcode,string type)
7 {
8 int step=0;
9 if(type=="ch")
10 {
11 step=5;//中文字符比较大,所以字距要比较大
12 }
13 int iwidth = (int)(checkcode.length * (13 + stepw));
14 system.drawing.bitmap image = new system.drawing.bitmap(iwidth, 22);
15 graphics g = graphics.fromimage(image);
16
17 g.clear(color.white);//清除背景色
18
19 color[] c = { color.black, color.red, color.darkblue, color.green, color.orange, color.brown, color.darkcyan, color.purple };//定义随机颜色
20
21 string[] font = { "verdana", "microsoft sans serif", "comic sans ms", "arial", "宋体" };
22 random rand = new random();
23
24 for (int i = 0; i < 50; i++)
25 {
26 int x1 = rand.next(image.width);
27 int x2 = rand.next(image.width);
28 int y1 = rand.next(image.height);
29 int y2 = rand.next(image.height);
30 g.drawline(new pen(color.lightgray,1), x1,y1,x2,y2);//根据坐标画线
31 }
32
33 for (int i = 0; i < checkcode.length; i++)
34 {
35 int cindex = rand.next(7);
36 int findex = rand.next(5);
37
38 font f = new system.drawing.font(font[findex], 10, system.drawing.fontstyle.bold);
39 brush b = new system.drawing.solidbrush(c[cindex]);
40 int ii = 4;
41 if ((i + 1) % 2 == 0)
42 {
43 ii = 2;
44 }
45 g.drawstring(checkcode.substring(i, 1), f, b, 3 + (i * (12 + stepw)), ii);
46 }
47
48 g.drawrectangle(new pen(color.black, 0), 0, 0, image.width - 1, image.height - 1);
49
50 system.io.memorystream ms = new system.io.memorystream();
51 return image;
52 }
5、总结
根据你要的随机数和背景就可以返回bitmap数组,然后把bitmap数组以图片形式存到内存流,就可以返回了