一个for循环

Question

我想取下面数据 192.168.0.1 -192.168.1.100得到的结果应该是192.168.0.1-192.168.0.254； 192.168.1.1-192.168.1.100这个for循环该如果实现呢。

Answer 1

不太清楚你的意思
拆成两个循环就可以实现呀
            for (int i = 1; i <= 254; i++)
            {
                Console.WriteLine("192.168.0." + i.ToString());
            }            for (int i = 1; i <= 100; i++)
            {
                Console.WriteLine("192.168.1." + i.ToString());
            }

Answer 2

for (int i = 0; i < 254; i++)
            {
                Console.WriteLine("192.168.0." + i.ToString());
                Console.WriteLine("192.168.1." + i.ToString());
            }
//或者
            for (int i = 0; i < 2; i++)
            {
                for (int j = 0; j < 254; j++)
                {
                    Console.WriteLine("192.168." + i.ToString() + "." + j.ToString());
                }
            }

Answer 3

你自己说只要0段个1段的啊..
要100段？
for (int i = 0; i < 100; i++)
            {
                ThreadPool.QueueUserWorkItem(new WaitCallback(addIP),i); 
            }public void addIP(Object state)
        {
            int i = Convert.ToInt32(state);
            for (int j = 0; j < 255; j++)
            {
                Console.WriteLine("192.168."+j.ToString()+"." + i.ToString());
                Console.WriteLine("192.168." + j.ToString() + "." + i.ToString());
            }
        }
没试过,也不想试.
输出顺序我不保证,但是运行结束 应该能把100段的全弄出来

Answer 4

public void addIP(Object state)
        {
            int i = Convert.ToInt32(state);
            for (int j = 0; j < 255; j++)
            {
                Console.WriteLine("192.168."+j.ToString()+"." + i.ToString());
                            }
        }
- -戳,多写了一句,,,,看样子真困了...

Answer 5

 (int i = 0; i < 254; i++)
            {
                Console.WriteLine("192.168.0." + i.ToString());
                Console.WriteLine("192.168.1." + i.ToString());
            }            for (int i = 0; i < 2; i++)
            {
                for (int j = 0; j < 254; j++)
                {
                    Console.WriteLine("192.168." + i.ToString() + "." + j.ToString());
                }
            }

Answer 6

在for 里面加判断条件 

Answer 7

string ip1 = "192.168.0.10";
        string ip2 = "192.168.1.20";
        //ip的倒数第二位和最后一位
        int ip1_last2 = int.Parse(ip1.Split('.')[2]), ip1_last = int.Parse(ip1.Split('.')[3]);
        int ip2_last2 = int.Parse(ip2.Split('.')[2]), ip2_last = int.Parse(ip2.Split('.')[3]);
        while (ip1_last2 < ip2_last2 || (ip1_last2 == ip2_last2 && ip1_last <= ip2_last))
        {
            Console.WriteLine(string.Format("192.168.{0}.{1}", ip1_last2, ip1_last));
            ip1_last++;
            if (ip1_last == 255)
            {
                ip1_last = 1;
                ip1_last2++;
            }
        }

Answer 8

for (int i = 0; i < 254; i++)             {                 Console.WriteLine("192.168.0." + i.ToString());                 Console.WriteLine("192.168.1." + i.ToString());             } //或者             for (int i = 0; i < 2; i++)             {                 for (int j = 0; j < 254; j++)                 {                     Console.WriteLine("192.168." + i.ToString() + "." + j.ToString());                 }             } 

Answer 9

for (int i = 0; i < 254; i++)             {                 Console.WriteLine("192.168.0." + i.ToString());                 Console.WriteLine("192.168.1." + i.ToString());             } //或者             for (int i = 0; i < 2; i++)             {                 for (int j = 0; j < 254; j++)                 {                     Console.WriteLine("192.168." + i.ToString() + "." + j.ToString());                 }             }  

Answer 10

            string str1 = "192.168.0.1";
            string str2 = "192.168.22.100";
            string reg = @"192\.168\.(\d{1,3})\.(\d{1,3})";
            Match m = Regex.Match(str1, reg);
            int istart = int.Parse(Regex.Match(str1, reg).Groups[1].Value);
            int istop = int.Parse(Regex.Match(str2, reg).Groups[1].Value);
            int jstart = int.Parse(Regex.Match(str1, reg).Groups[2].Value);
            int jstop = int.Parse(Regex.Match(str2, reg).Groups[2].Value);
            for (int i = istart; i <= istop; i++)
            {
                if (istart == istop)
                {
                    Console.WriteLine("192.168." + istart + "." + jstart + "-192.168." + istop + "." + jstop);
                }                if (i == istart && i < istop)
                {
                    Console.WriteLine("192.168." + i + "." + jstart + "-192.168." + i + ".254");
                }
                if (i > istart && i < istop)
                {
                    Console.WriteLine("192.168." + i + ".1-192.168." + i + ".254");
                }
                if (i > istart && i == istop)
                {
                    Console.WriteLine("192.168." + i + ".1-192.168." + i+"."+jstop);
                }
            }
            Console.ReadKey();亲测可用，支持192.168.0.1-192.168.254.254之间的任意IP

Answer 11

for (int i = 0; i <= 1; i++)
                {
                    for (int j = 1; j <= 255; j++)
                    {
                        if (i == 0)
                        {
                            Console.WriteLine("192.168.0." + j.ToString());
                        }
                        else if (i == 1)
                        {
                            Console.WriteLine("192.168.1." + j.ToString());
                            if (j == 100)
                            {
                                break;
                            }
                        }
                    }
                }
            //可能比较啰嗦，但是思路挺清晰

Answer 12

 - -晕 上次是在太困了, 理解错意思了
现在知道LZ想表达什么了..
就是遍历出2个合法IP之间的所有IP
如：开始ip：192.168.0.0  结束IP：192.168.1.100
那么就要输出 开始IP段的所有IP 和结束IP段的前100条IP
从192.168.0.0到192.168.1.100结束重新贴下代码,我用的是winform,其实差不多.
string IP_S = "";
        string IP_E = "";
        private void button1_Click(object sender, EventArgs e)
        {
            //开始IP
            IP_S = textBox2.Text;
            //结束IP
            IP_E = textBox3.Text;
            //拆分
            string[] IP_SList = IP_S.Split('.');
            string[] IP_EList = IP_E.Split('.');            //不满足循环
            if ((IP_S == IP_E) || (int.Parse(IP_EList[2]) < int.Parse(IP_SList[2])))
            {
                return;
            }
            else
            {
                //开始循环输出
                getIP(IP_SList, IP_EList);
            }
        }
        /// <summary>
        /// 循环输出
        /// </summary>
        /// <param name="IP_SList">开始IP拆分后的集合</param>
        /// <param name="IP_EList">结束IP拆分后的集合</param>
        private void getIP(string[] IP_SList,string[] IP_EList)
        {
            //如果2个IP段相同
            if (int.Parse(IP_EList[2]) == int.Parse(IP_SList[2]))
            {
                //循环IP差
                for (int i = int.Parse(IP_SList[3]); i < int.Parse(IP_EList[3])+1; i++)
                {
                    //输出IP
                    textBox1.Text += IP_SList[0] + "." + IP_SList[1] + "." + IP_SList[2] + "." + i.ToString()+"\r\n";
                }
            }
            else
            {
                //循环255次
                for (int i = int.Parse(IP_SList[3]); i < 256; i++)
                {
                    //输出IP
                    textBox1.Text += IP_SList[0] + "." + IP_SList[1] + "." + IP_SList[2] + "." + i.ToString() + "\r\n";
                }
                //开始IP段+1
                IP_SList[2] = (int.Parse(IP_SList[2]) + 1).ToString();
                //开始IP归0
                IP_SList[3] = "0";
                //递归循环输出
                getIP(IP_SList, IP_EList);
            }
        }

Answer 13

public List<string> IP_Adress(string begin, string end)
        {
            //下面的z.c.x.y分别对应192.168.1.1
            List<string> list = new List<string>();
            //获取开始和结束(z.c.x.y的x)
            int _temp_begin_one = Convert.ToInt32(begin.Substring(8, 1));
            int _temp_end_one = Convert.ToInt32(end.Substring(8, 1));
            //获取开始和结束(z.c.x.y的y)
            int _temp_begin_two = Convert.ToInt32(begin.Substring(10));
            int _temp_end_two = Convert.ToInt32(end.Substring(10));
            //获取开头部分z.c.x.y的z和c
            int _temp_z = Convert.ToInt32(end.Substring(0,3));
            int _temp_c = Convert.ToInt32(end.Substring(4,3));
            //从x开始循环
            for (int i = _temp_begin_one; i <= _temp_end_one; i++)
            {
                
                int _for_begin;//y的开始;
                int _for_end;  //y的结束                //如果x的开始和结束是否相等
                if (_temp_begin_one != _temp_end_one )
                {
                    _for_begin = _temp_begin_two;
                    _for_end = 254;
                    _temp_begin_one++;
                }
                else
                {
                    _for_begin = _temp_begin_two;
                    _for_end = _temp_end_two;
                }
                for (int j = _for_begin; j <= _for_end; j++)
                {
                    list.Add(_temp_z + "." + _temp_c + "." + i + "." + j);
                }
            }            return list;
        }看成254进制的数就好了..调用时直接输入IP就行了..如string begin="192.168.0.1" string end="192.168.1.12"

Answer 14

public List<string> IP_Adress(string begin, string end)
        {
            //下面的z.c.x.y分别对应192.168.1.1
            List<string> list = new List<string>();
            string reg = @"(\d{1,3})\.(\d{1,3})\.(\d{1,3})\.(\d{1,3})";
            Match m = Regex.Match(begin, reg);
            //获取开头部分z.c.x.y的z和c
            int _temp_z = int.Parse(Regex.Match(begin, reg).Groups[1].Value);
            int _temp_c = int.Parse(Regex.Match(begin, reg).Groups[2].Value);
            //获取开始和结束(z.c.x.y的x)
            int _temp_begin_one = int.Parse(Regex.Match(begin, reg).Groups[3].Value);
            int _temp_end_one = int.Parse(Regex.Match(end, reg).Groups[3].Value);
            //获取开始和结束(z.c.x.y的y)
            int _temp_begin_two = int.Parse(Regex.Match(begin, reg).Groups[4].Value);
            int _temp_end_two = int.Parse(Regex.Match(end, reg).Groups[4].Value);
            //从x开始循环
            for (int i = _temp_begin_one; i <= _temp_end_one; i++)
            {
                
                int _for_begin;//y的开始;
                int _for_end;  //y的结束                //如果x的开始和结束是否相等
                if (_temp_begin_one != _temp_end_one )
                {
                    _for_begin = _temp_begin_two;
                    _for_end = 254;
                    _temp_begin_one++;
                }
                else
                {
                    _for_begin = _temp_begin_two;
                    _for_end = _temp_end_two;
                }
                for (int j = _for_begin; j <= _for_end; j++)
                {
                    list.Add(_temp_z + "." + _temp_c + "." + i + "." + j);
                }
            }            return list;
        }一开始没详细看你的要求..这个改过了..偷了下21楼的正则表达式..1.1.1.1到999.999.999.999都行
开始和结束的开头2位(如192,168)要一样..如果不一样就按结束的开头做内容了

Answer 15

看楼上的都太复杂了，我有个简单的方法，代码贴下
 int num = 0;
            for (int i = 0; i < 2; i++)
            {
                if (i == 0)
                {
                    num = 255;
                }
                else
                {
                    num = 100;
                }
                for (int j = 1; j <= num; j++)
                {
                    listBox1.Items.Add("192.168." + i + "." + j);
                    messagebox.show("192.168." + i + "." + j);
                }
            }

Answer 16

谢谢各位的热心回答，这个问题已经解决了，只要加个判断条件就行。  if(i==1 && j==100)break;