请各位帮忙,怎么用JAVA实现对一个文件进行加密传输!

2004-04-01 10:40:28 399 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

simon11111122222 墨衍会员 · AI 创作全网分发关注 + 加入社区

Java如何实现大文件上传过程中的断点续传？要求：开源，免费，技术支持技术：分片,分块,断点续传,加密传输,加密存储需求：大文件上传，批量上传，断点续传，文件夹上传，大文件下载，批量下载，断点下载，文件夹下载文件夹：上传下载需要支持层级结构，采用非压缩方式文件大小：100G前端：vue2,vue3,vue-cli,jquery,html,webuploader后端：java,JSP,springboot,struts服务器：linux,信创国产化数据库：mysql,oracle,达梦，国产化数据库。阅读详情

用 MD5/SHA1，DSA，DESede/DES，Diffie-Hellman算法怎么样实现文件进行加密传输,请各位帮帮忙!

...全文

Java加密传输速度提升300%？揭秘交通数据隐私保护的5大核心技术！摘要： Java在智能交通数据隐私保护中展现出核心优势，通过HTTPS、AES加密、差分隐私、k-匿名化和联邦学习5大技术，实现高效安全的数据处理。对比Python和Go，Java在加密速度（3s/10MB）、生态系统（Spring Security+H2）及合规性（满足GDPR）方面表现突出。3个实战案例显示，Java方案可将数据泄露风险降至0.001%，联邦学习提升模型共享效率90%。未来将融合量子加密与边缘AI，强化隐私保护。开发者需警惕密钥管理、压力测试等10大常见陷阱。（149字）阅读详情

simon11111122222

Java SE 62,621 成员 307,251 内容

+ 关注 + 加入社区

切换为时间正序

zlhlj2000 2004-06-18

打赏
举报

Md4,Md5 有javascript pls download

guishuanglin 2004-06-18

打赏
举报

我提供给你md5算法源程序：
========================================
public class MD5
{
/*
* A Java implementation of the RSA Data Security, Inc. MD5 Message
* Digest Algorithm, as defined in RFC 1321.
* Based on the JavaScript implementation of Paul Johnston
* Copyright (C) Paul Johnston 1999 - 2000.
* See http://pajhome.org.uk/site/legal.html for details.
* Java Version by Thomas Weber (Orange Interactive GmbH)
*/

/*
* Convert a 32-bit number to a hex string with ls-byte first
*/
String hex_chr = "0123456789abcdef";
private String rhex(int num)
{
String str = "";
for(int j = 0; j <= 3; j++)
str = str + hex_chr.charAt((num >> (j * 8 + 4)) & 0x0F) + hex_chr.charAt((num >> (j * 8)) & 0x0F);
return str;
}

/*
* Convert a string to a sequence of 16-word blocks, stored as an array.
* Append padding bits and the length, as described in the MD5 standard.
*/
private int[] str2blks_MD5(String str)
{
int nblk = ((str.length() + 8) >> 6) + 1;
int[] blks = new int[nblk * 16];
int i = 0;
for(i = 0; i < nblk * 16; i++) {
blks[i] = 0;
}
for(i = 0; i < str.length(); i++) {
blks[i >> 2] |= str.charAt(i) << ((i % 4) * 8);
}
blks[i >> 2] |= 0x80 << ((i % 4) * 8);
blks[nblk * 16 - 2] = str.length()*8;

return blks;
}

/*
* Add integers, wrapping at 2^32
*/
private int add(int x, int y)
{
return ((x&0x7FFFFFFF) + (y&0x7FFFFFFF)) ^ (x&0x80000000) ^ (y&0x80000000);
}

/*
* Bitwise rotate a 32-bit number to the left
*/
private int rol(int num, int cnt)
{
return (num << cnt) | (num >>> (32 - cnt));
}

/*
* These functions implement the basic operation for each round of the
* algorithm.
*/
private int cmn(int q, int a, int b, int x, int s, int t)
{
return add(rol(add(add(a, q), add(x, t)), s), b);
}
private int ff(int a, int b, int c, int d, int x, int s, int t)
{
return cmn((b & c) | ((~b) & d), a, b, x, s, t);
}
private int gg(int a, int b, int c, int d, int x, int s, int t)
{
return cmn((b & d) | (c & (~d)), a, b, x, s, t);
}
private int hh(int a, int b, int c, int d, int x, int s, int t)
{
return cmn(b ^ c ^ d, a, b, x, s, t);
}
private int ii(int a, int b, int c, int d, int x, int s, int t)
{
return cmn(c ^ (b | (~d)), a, b, x, s, t);
}

/*
* Take a string and return the hex representation of its MD5.
*/
public String calcMD5(String str)
{
int[] x = str2blks_MD5(str);
int a = 0x67452301;
int b = 0xEFCDAB89;
int c = 0x98BADCFE;
int d = 0x10325476;

for(int i = 0; i < x.length; i += 16)
{
int olda = a;
int oldb = b;
int oldc = c;
int oldd = d;

a = ff(a, b, c, d, x[i+ 0], 7 , 0xD76AA478);
d = ff(d, a, b, c, x[i+ 1], 12, 0xE8C7B756);
c = ff(c, d, a, b, x[i+ 2], 17, 0x242070DB);
b = ff(b, c, d, a, x[i+ 3], 22, 0xC1BDCEEE);
a = ff(a, b, c, d, x[i+ 4], 7 , 0xF57C0FAF);
d = ff(d, a, b, c, x[i+ 5], 12, 0x4787C62A);
c = ff(c, d, a, b, x[i+ 6], 17, 0xA8304613);
b = ff(b, c, d, a, x[i+ 7], 22, 0xFD469501);
a = ff(a, b, c, d, x[i+ 8], 7 , 0x698098D8);
d = ff(d, a, b, c, x[i+ 9], 12, 0x8B44F7AF);
c = ff(c, d, a, b, x[i+10], 17, 0xFFFF5BB1);
b = ff(b, c, d, a, x[i+11], 22, 0x895CD7BE);
a = ff(a, b, c, d, x[i+12], 7 , 0x6B901122);
d = ff(d, a, b, c, x[i+13], 12, 0xFD987193);
c = ff(c, d, a, b, x[i+14], 17, 0xA679438E);
b = ff(b, c, d, a, x[i+15], 22, 0x49B40821);

a = gg(a, b, c, d, x[i+ 1], 5 , 0xF61E2562);
d = gg(d, a, b, c, x[i+ 6], 9 , 0xC040B340);
c = gg(c, d, a, b, x[i+11], 14, 0x265E5A51);
b = gg(b, c, d, a, x[i+ 0], 20, 0xE9B6C7AA);
a = gg(a, b, c, d, x[i+ 5], 5 , 0xD62F105D);
d = gg(d, a, b, c, x[i+10], 9 , 0x02441453);
c = gg(c, d, a, b, x[i+15], 14, 0xD8A1E681);
b = gg(b, c, d, a, x[i+ 4], 20, 0xE7D3FBC8);
a = gg(a, b, c, d, x[i+ 9], 5 , 0x21E1CDE6);
d = gg(d, a, b, c, x[i+14], 9 , 0xC33707D6);
c = gg(c, d, a, b, x[i+ 3], 14, 0xF4D50D87);
b = gg(b, c, d, a, x[i+ 8], 20, 0x455A14ED);
a = gg(a, b, c, d, x[i+13], 5 , 0xA9E3E905);
d = gg(d, a, b, c, x[i+ 2], 9 , 0xFCEFA3F8);
c = gg(c, d, a, b, x[i+ 7], 14, 0x676F02D9);
b = gg(b, c, d, a, x[i+12], 20, 0x8D2A4C8A);

a = hh(a, b, c, d, x[i+ 5], 4 , 0xFFFA3942);
d = hh(d, a, b, c, x[i+ 8], 11, 0x8771F681);
c = hh(c, d, a, b, x[i+11], 16, 0x6D9D6122);
b = hh(b, c, d, a, x[i+14], 23, 0xFDE5380C);
a = hh(a, b, c, d, x[i+ 1], 4 , 0xA4BEEA44);
d = hh(d, a, b, c, x[i+ 4], 11, 0x4BDECFA9);
c = hh(c, d, a, b, x[i+ 7], 16, 0xF6BB4B60);
b = hh(b, c, d, a, x[i+10], 23, 0xBEBFBC70);
a = hh(a, b, c, d, x[i+13], 4 , 0x289B7EC6);
d = hh(d, a, b, c, x[i+ 0], 11, 0xEAA127FA);
c = hh(c, d, a, b, x[i+ 3], 16, 0xD4EF3085);
b = hh(b, c, d, a, x[i+ 6], 23, 0x04881D05);
a = hh(a, b, c, d, x[i+ 9], 4 , 0xD9D4D039);
d = hh(d, a, b, c, x[i+12], 11, 0xE6DB99E5);
c = hh(c, d, a, b, x[i+15], 16, 0x1FA27CF8);
b = hh(b, c, d, a, x[i+ 2], 23, 0xC4AC5665);

a = ii(a, b, c, d, x[i+ 0], 6 , 0xF4292244);
d = ii(d, a, b, c, x[i+ 7], 10, 0x432AFF97);
c = ii(c, d, a, b, x[i+14], 15, 0xAB9423A7);
b = ii(b, c, d, a, x[i+ 5], 21, 0xFC93A039);
a = ii(a, b, c, d, x[i+12], 6 , 0x655B59C3);
d = ii(d, a, b, c, x[i+ 3], 10, 0x8F0CCC92);
c = ii(c, d, a, b, x[i+10], 15, 0xFFEFF47D);
b = ii(b, c, d, a, x[i+ 1], 21, 0x85845DD1);
a = ii(a, b, c, d, x[i+ 8], 6 , 0x6FA87E4F);
d = ii(d, a, b, c, x[i+15], 10, 0xFE2CE6E0);
c = ii(c, d, a, b, x[i+ 6], 15, 0xA3014314);
b = ii(b, c, d, a, x[i+13], 21, 0x4E0811A1);
a = ii(a, b, c, d, x[i+ 4], 6 , 0xF7537E82);
d = ii(d, a, b, c, x[i+11], 10, 0xBD3AF235);
c = ii(c, d, a, b, x[i+ 2], 15, 0x2AD7D2BB);
b = ii(b, c, d, a, x[i+ 9], 21, 0xEB86D391);

a = add(a, olda);
b = add(b, oldb);
c = add(c, oldc);
d = add(d, oldd);
}
return rhex(a) + rhex(b) + rhex(c) + rhex(d);
}

}

bigcrazy 2004-04-28

打赏
举报

MD5/SHA1是数字签名吧，是单向无法还原的加密。
我给一段例程你，自己看看吧：
片段1：生成密钥
Security.addProvider(new cryptix.provider.Cryptix()); //我用了第三方提供的算法
KeyGenerator kg = KeyGenerator.getInstance( “DES” );
Key key = kg.generateKey();
byte[] desKeyData=key.getEncoded();
writeToFile(desKeyData); //把密钥写到文件中

片段2：加密
Security.addProvider(new cryptix.provider.Cryptix());//我用了个第三方的加密算法提供者

byte[] desKeyData=getDesKey(); //这里我把密钥存在文件里了，读出来字节数组
DESKeySpec dks = new DESKeySpec( desKeyData );
SecretKeyFactory kf = SecretKeyFactory.getInstance( “DES” );
Key key=kf.generateSecret(dks);
System.out.println("Generate DES Key successful...");
mycipher = Cipher.getInstance( “DES” );
mycipher.init( Cipher.ENCRYPT_MODE, key);
System.out.println("Encrypt file: "+file.getName());
byte[] fileData=getFileData(file); //读要加密的文件

byte[] encryptFileData=mycipher.doFinal(fileData);
writeFileData(file,encryptFileData); //写回去

simon11111122222 2004-04-28