本文介绍了此代码中的唯一字符密码生成的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
您好,我使用此代码生成密码,但是我想要唯一的字符密码.密码中的字符必须唯一.请帮助我
私有 静态 字符串 MakePassword( int pl) { 字符 [] comb = 新 字符 [ pl]; 字符串可能= " ZYXWVUTSRQPONMLKJIHGFEDCAB" ; 字符 []密码= 新 字符 [ pl]; 随机rd = 新 Random(); 尝试 { 字符 ss =可能性[rd.Next( 0 ,可能性.长度)]; char sp =可能性[rd.Next( 0 ,可能性.长度)]; 如果(ss!= sp) { 字符 a = sp; } 密码[ 1 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 2 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 3 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 4 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 5 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 6 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 7 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 8 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 9 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 10 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 11 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 12 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 13 ] =可能[rd.Next( 0 ,可能.长度)]; passwords [ 14 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 15 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 16 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 17 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 18 ] =可能[rd.Next( 0 ,可能.长度)]; passwords [ 19 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 20 ] =可能[rd.Next( 0 ,可能.长度)]; passwords [ 21 ] =可能[rd.Next( 0 ,可能.长度)]; passwords [ 22 ] =可能[rd.Next( 0 ,可能.长度)]; 密码[ 23 ] =可能[rd.Next( 0 ,可能.长度)]; } 捕获(IndexOutOfRangeException e) { for ( int j = 0 ; j < pl; j ++) { comb [j] =密码[j]; } } 返回 新 字符串(comb); }
解决方案
hello, Im using this code for password generation, but i want unique character password. character must be unique in a password.plz help me
private static string MakePassword(int pl) { char[] comb = new char[pl]; string possibles = "ZYXWVUTSRQPONMLKJIHGFEDCAB"; char[] passwords = new char[pl]; Random rd = new Random(); try { char ss = possibles[rd.Next(0, possibles.Length)]; char sp= possibles[rd.Next(0, possibles.Length)]; if (ss != sp) { char a = sp; } passwords[1] = possibles[rd.Next(0, possibles.Length)]; passwords[2] = possibles[rd.Next(0, possibles.Length)]; passwords[3] = possibles[rd.Next(0, possibles.Length)]; passwords[4] = possibles[rd.Next(0, possibles.Length)]; passwords[5] = possibles[rd.Next(0, possibles.Length)]; passwords[6] = possibles[rd.Next(0, possibles.Length)]; passwords[7] = possibles[rd.Next(0, possibles.Length)]; passwords[8] = possibles[rd.Next(0, possibles.Length)]; passwords[9] = possibles[rd.Next(0, possibles.Length)]; passwords[10] = possibles[rd.Next(0, possibles.Length)]; passwords[11] = possibles[rd.Next(0, possibles.Length)]; passwords[12] = possibles[rd.Next(0, possibles.Length)]; passwords[13] = possibles[rd.Next(0, possibles.Length)]; passwords[14] = possibles[rd.Next(0, possibles.Length)]; passwords[15] = possibles[rd.Next(0, possibles.Length)]; passwords[16] = possibles[rd.Next(0, possibles.Length)]; passwords[17] = possibles[rd.Next(0, possibles.Length)]; passwords[18] = possibles[rd.Next(0, possibles.Length)]; passwords[19] = possibles[rd.Next(0, possibles.Length)]; passwords[20] = possibles[rd.Next(0, possibles.Length)]; passwords[21] = possibles[rd.Next(0, possibles.Length)]; passwords[22] = possibles[rd.Next(0, possibles.Length)]; passwords[23] = possibles[rd.Next(0, possibles.Length)]; } catch (IndexOutOfRangeException e) { for (int j = 0; j < pl; j++) { comb[j] = passwords[j]; } } return new string(comb); }
解决方案
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