请教PB加密代码问题。 - PowerBuilder - 编程论坛 | 网站首页 | 业界新闻 | 小组 | 技术文章 | 下载频道 | 博客 | 代码贴 | 编程论坛 | 登录 注册 平板 帮助 编程论坛 → 数据库技术 → 『 PowerBuilder 』 → 请教PB加密代码问题。 我的收件箱(0) 装机设首页 每台赚2元 登录论坛，同大家一起切磋吧 ：） 用户名： 注册 忘记密码 密 码： 共有 618 人关注过本帖 标题：请教PB加密代码问题。 [只看楼主] pwin237 等　级：新手上路 帖　子：1 专家分：0 注　册：2010-10-22 新人--> 楼主 问题点数：0 回复次数：0 请教PB加密代码问题。 最近因为工作需要，用C#开发一个功能，普通发票打印后，导出一个DBF文件，里面有一项为JM的项，不知如何加密，后来在网上找到破解PB的软件，破解他原来的程序，看到加密部分如下，由于本人对PB不懂，烦各位帮手看看。 文件一：这个加密方式应该看得懂，但第二个就不清楚了。 $PBExportHeader$f_encrypt.srf $PBExportComments$By PBKILLER v1.3(kivens@21cn.com) global type f_encrypt from function_object end type forward prototypes global function string f_encrypt (string a_ymm) end prototypes global function string f_encrypt (string a_ymm);integer i integer li_kllen string ls_zmkl = "" li_kllen = len(a_ymm) for i = 1 to li_kllen ls_zmkl = ls_zmkl + char(asc(mid(a_ymm,i,1)) + i + li_kllen) next return ls_zmkl end function 文件二：其中这一句不知道是否破解后有问题？好像语句不完整。choose case mod(integer(mid(ls_zmkl,li_i = mod(integer(mid(ls_zmkl,li_i * 2 + 1,2)),26 烦请高人指点。 $PBExportHeader$f_encrypt2.srf $PBExportComments$By PBKILLER v1.3(kivens@21cn.com) global type f_encrypt2 from function_object end type forward prototypes global function string f_encrypt2 (string as_string) end prototypes global function string f_encrypt2 (string as_string);integer li_count integer li_kllen integer li_i string ls_zmkl string ls_zmkl2 string ls_my string ls_zmkl3 string ls_temp li_kllen = len(as_string) ls_zmkl = "" for li_i = 1 to li_kllen choose case mid(as_string,li_i,1) case "0" ls_zmkl = ls_zmkl + "52" continue case "1" ls_zmkl = ls_zmkl + "63" continue case "2" ls_zmkl = ls_zmkl + "45" continue case "3" ls_zmkl = ls_zmkl + "78" continue case "4" ls_zmkl = ls_zmkl + "90" continue case "5" ls_zmkl = ls_zmkl + "27" continue case "6" ls_zmkl = ls_zmkl + "14" continue case "7" ls_zmkl = ls_zmkl + "36" continue case "8" ls_zmkl = ls_zmkl + "81" continue case "9" ls_zmkl = ls_zmkl + "09" end choose next li_kllen = len(ls_zmkl) ls_zmkl2 = "" ls_my = "8754039126" for li_i = 1 to li_kllen ls_zmkl2 = ls_zmkl2 + string(mod(integer(mid(ls_zmkl,li_i,1)) + integer(mid(ls_my,integer(mid(ls_zmkl,li_i,1)),1)),10)) next ls_zmkl3 = "" for li_i = 0 to li_kllen / 2 - 1 choose case mod(integer(mid(ls_zmkl,li_i = mod(integer(mid(ls_zmkl,li_i * 2 + 1,2)),26 case 0 ls_zmkl3 = ls_zmkl3 + "a" continue case 1 ls_zmkl3 = ls_zmkl3 + "b" continue case 2 ls_zmkl3 = ls_zmkl3 + "c" continue case 3 ls_zmkl3 = ls_zmkl3 + "d" continue case 4 ls_zmkl3 = ls_zmkl3 + "e" continue case 5 ls_zmkl3 = ls_zmkl3 + "f" continue case 6 ls_zmkl3 = ls_zmkl3 + "g" continue case 7 ls_zmkl3 = ls_zmkl3 + "h" continue case 8 ls_zmkl3 = ls_zmkl3 + "i" continue case 9 ls_zmkl3 = ls_zmkl3 + "j" continue case 10 ls_zmkl3 = ls_zmkl3 + "k" continue case 11 ls_zmkl3 = ls_zmkl3 + "l" continue case 12 ls_zmkl3 = ls_zmkl3 + "m" continue case 13 ls_zmkl3 = ls_zmkl3 + "n" continue case 14 ls_zmkl3 = ls_zmkl3 + "o" continue case 15 ls_zmkl3 = ls_zmkl3 + "p" continue case 16 ls_zmkl3 = ls_zmkl3 + "q" continue case 17 ls_zmkl3 = ls_zmkl3 + "r" continue case 18 ls_zmkl3 = ls_zmkl3 + "s" continue case 19 ls_zmkl3 = ls_zmkl3 + "t" continue case 20 ls_zmkl3 = ls_zmkl3 + "u" continue case 21 ls_zmkl3 = ls_zmkl3 + "v" continue case 22 ls_zmkl3 = ls_zmkl3 + "w" continue case 23 ls_zmkl3 = ls_zmkl3 + "x" continue case 24 ls_zmkl3 = ls_zmkl3 + "y" continue case 25 ls_zmkl3 = ls_zmkl3 + "z" end choose next ls_temp = mid(ls_zmkl3,3,10) ls_zmkl3 = replace(ls_zmkl3,3,10,mid(ls_zmkl3,29,36)) ls_zmkl3 = ls_zmkl3 + ls_temp ls_temp = mid(ls_zmkl3,14,21) ls_zmkl3 = replace(ls_zmkl3,14,21,mid(ls_zmkl3,5,12)) ls_zmkl3 = ls_zmkl3 + ls_temp ls_temp = mid(ls_zmkl3,1,5) ls_zmkl3 = replace(ls_zmkl3,1,5,mid(ls_zmkl3,31,35)) return ls_zmkl3 end function 搜索更多相关主题的帖子: 代码 2010-10-22 17:16:24 --> 11/1页1 关于我们 | 广告合作 | 编程中国 | 清除Cookies | TOP 编程中国 版权所有，并保留所有权利。 Powered by Discuz, Processed in 0.036613 second(s), 7 queries. Copyright©2004-2014, BCCN.NET, All Rights Reserved