PHP加密算法转换delphi出问题
function encrypt($string,$operation,$key='') { $key=md5($key); $key_length=strlen($key); $string=$operation=='D'?base64_decode($string):substr(md5($string.$key),0,8).$string; $string_length=strlen($string); $rndkey=$box=array(); $result=''; for($i=0;$i<=255;$i++) { $rndkey[$i]=ord($key[$i%$key_length]); $box[$i]=$i; } for($j=$i=0;$i<256;$i++) { $j=($j+$box[$i]+$rndkey[$i])%256; $tmp=$box[$i]; $box[$i]=$box[$j]; $box[$j]=$tmp; } for($a=$j=$i=0;$i<$string_length;$i++) { $a=($a+1)%256; $j=($j+$box[$a])%256; $tmp=$box[$a]; $box[$a]=$box[$j]; $box[$j]=$tmp; $result.=chr(ord($string[$i])^($box[($box[$a]+$box[$j])%256])); } if($operation=='D') { if(substr($result,0,8)==substr(md5(substr($result,8).$key),0,8)) { return substr($result,8); } else { return''; } } else { return str_replace('=','',base64_encode($result)); } }$id = '123sdfsdf';$key = 'aasd1234567985fdgdfdfgdg';$token = encrypt($id, 'E', $key);echo '加密:'.encrypt($id, 'E', $key);echo '<br />';echo '解密:'.encrypt($token, 'D', $key);
function encrypt(str, operation, key: string): string; function StrToMD5(s: string): string; var Md5Encode: TIdHashMessageDigest5; begin Md5Encode := TIdHashMessageDigest5.Create; Result := Md5Encode.AsHex(Md5Encode.HashValue(S)); Md5Encode.Free; end;var key_length: Integer; string_length: Integer; I, j, A: Integer; rndkey: array[0..255] of Byte; Box: array[0..255] of Byte; tmp: Byte;begin key := LowerCase(StrToMD5(key)); key_length := Length(key); if operation = 'D' then str := DecodeString(str) else str := Copy(lowerCase(StrToMD5(str + key)), 1, 8) + str; string_length := Length(str); result := ''; for i := 0 to 255 do begin rndkey[i] := ord(key[(i mod key_length) + 1]); Box[i] := i; end; j := 0; for i := 0 to 255 do begin j := (j + Box[i] + rndkey[i]) mod 256; tmp := Box[i]; Box[i] := Box[j]; Box[j] := tmp; end; a := 0; j := 0; for i := 1 to string_length do begin A := (A + 1) mod 256; j := (j + Box[A]) mod 256; tmp := Box[A]; Box[A] := Box[j]; Box[j] := tmp; result := result + chr(ord(str[i]) xor (Box[(Box[A] + Box[j]) mod 256])); end; if (operation = 'D') then begin if (copy(result, 1, 8) = copy(lowerCase(StrToMd5(copy(result, 9, Length(Result) - 8) + key)), 1, 8)) then begin Result := copy(result, 9, Length(Result) - 8); end else begin Result := ''; end; end else begin Result := StringReplace(EncodeString(result), '=', '', [rfReplaceAll]); end; end;
自己网上找的PHP加密算法,但是转成DELPHI 时候只有加密正确 解密不正确
求前辈帮忙完美转换
回复讨论(解决方案)
这个完全没什么技术难度,你只是不会调试而已。
既然能把代码转过去了,哪行对应哪行就清楚了
取一行php输出一下变量,delphi输出一下变量,如果变量值有不同就是这行错了。
程序就在那里慢慢查肯定能查出是哪行错了,实在不行一行行对比也行啊。
所以找到问题只是时间问题,这个完全没有难度只是费时费力而已。
你这个是什么加密算法?
delphi 的 DecodeString 是对应 php 的 base64_decode 吗?
那么 DecodeString 函数是残废的 base64 串是如何处理的?
建议你补齐尾部的 =
这个完全没什么技术难度,你只是不会调试而已。
既然能把代码转过去了,哪行对应哪行就清楚了
取一行php输出一下变量,delphi输出一下变量,如果变量值有不同就是这行错了。
程序就在那里慢慢查肯定能查出是哪行错了,实在不行一行行对比也行啊。
所以找到问题只是时间问题,这个完全没有难度只是费时费力而已。
调试过了 DecodeString 后解密出来的字符串 delphi 比 php 少了2位
delphi 的 DecodeString 是对应 php 的 base64_decode 吗?
那么 DecodeString 函数是残废的 base64 串是如何处理的?
建议你补齐尾部的 =
我试试吧后尾补=看看,
delphi 的 DecodeString 是对应 php 的 base64_decode 吗?
那么 DecodeString 函数是残废的 base64 串是如何处理的?
建议你补齐尾部的 =
补齐尾部的=也不行,现在发现DecodeString解密出来的字符出错了
按你给出的参数,$token 尾部有一个 = 被删去
这与你 delphi 比 php 少了2位 的说法是吻合的
base64 将 3 个字解扩展成 4 个字节 ,尾部的 2 个字符扩展后需用 = 补齐
解码时如果因残废了而舍弃,就正好少了2个字符
你怎么补的?贴出代码
按你给出的参数,$token 尾部有一个 = 被删去
这与你 delphi 比 php 少了2位 的说法是吻合的
base64 将 3 个字解扩展成 4 个字节 ,尾部的 2 个字符扩展后需用 = 补齐
解码时如果因残废了而舍弃,就正好少了2个字符
按照你的方法我的delphi 版后面加了 =
然后解密出来 字符再转成 Asc码比对了下delphi 和 PHP 分别为
delphi 版
1571920612038416510212118647183621641859132545612373206113391
php 版
157192061203841651021211864718362164185913254561237320611339
后面少了个1
if operation = 'D' then
在这里给 str 加上 (4 - Length(str)%4)%4 个 =
str := DecodeString(str)
else
或者你看看
DecodeString("YWI=")
和
DecodeString("YWI")
的结果分别是什么
if operation = 'D' then
在这里给 str 加上 (4 - Length(str)%4)%4 个 =
str := DecodeString(str)
else
按照你的方法可以了,原来要补齐4的整数=谢谢版主

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