com.sun.crypto.provider.SunJCE
Could not instantiate bean class [com.lz.monitor.alert.service.ServiceImp]: Constructor threw exception; nested exception is java.lang.NoClassDefFoundError: com.sun.crypto.provider.SunJCE
Caused by: java.lang.NoClassDefFoundError: com.sun.crypto.provider.SunJCE
3DES加密解密调用示例
文章分类:Java编程 关键字: 3des加密解密调用示例
在java中调用sun公司提供的3DES加密解密算法时,需要使用到$JAVA_HOME/jre/lib/目录下如下的4个jar包:
jce.jar
security/US_export_policy.jar
security/local_policy.jar
ext/sunjce_provider.jar
Java运行时会自动加载这些包,因此对于带main函数的应用程序不需要设置到CLASSPATH环境变量中。对于WEB应用,需要把这些包加到WEB-INF/lib目录下。
以下是java中调用sun公司提供的3DES加密解密算法的样本代码:
/*字符串 DESede(3DES) 加密*/
import java.security.*;
import javax.crypto.*;
import javax.crypto.spec.SecretKeySpec;
public class ThreeDes {
private static final String Algorithm = "DESede"; //定义 加密算法,可用 DES,DESede,Blowfish
//keybyte为加密密钥,长度为24字节
//src为被加密的数据缓冲区(源)
public static byte[] encryptMode(byte[] keybyte, byte[] src) {
try {
//生成密钥
SecretKey deskey = new SecretKeySpec(keybyte, Algorithm);
//加密
Cipher c1 = Cipher.getInstance(Algorithm);
c1.init(Cipher.ENCRYPT_MODE, deskey);
return c1.doFinal(src);
}
catch (java.security.NoSuchAlgorithmException e1) {
e1.printStackTrace();
}
catch (javax.crypto.NoSuchPaddingException e2) {
e2.printStackTrace();
}
catch (java.lang.Exception e3) {
e3.printStackTrace();
}
return null;
}
//keybyte为加密密钥,长度为24字节
//src为加密后的缓冲区
public static byte[] decryptMode(byte[] keybyte, byte[] src) {
try {
//生成密钥
SecretKey deskey = new SecretKeySpec(keybyte, Algorithm);
//解密
Cipher c1 = Cipher.getInstance(Algorithm);
c1.init(Cipher.DECRYPT_MODE, deskey);
return c1.doFinal(src);
}
catch (java.security.NoSuchAlgorithmException e1) {
e1.printStackTrace();
}
catch (javax.crypto.NoSuchPaddingException e2) {
e2.printStackTrace();
}
catch (java.lang.Exception e3) {
e3.printStackTrace();
}
return null;
}
//转换成十六进制字符串
public static String byte2hex(byte[] b) {
String hs="";
String stmp="";
for (int n=0;n<b.length;n++) {
stmp=(java.lang.Integer.toHexString(b[n] & 0XFF));
if (stmp.length()==1) hs=hs+"0"+stmp;
else hs=hs+stmp;
if (n<b.length-1) hs=hs+":";
}
return hs.toUpperCase();
}
public static void main(String[] args){
//添加新安全算法,如果用JCE就要把它添加进去
Security.addProvider(new com.sun.crypto.provider.SunJCE());
final byte[] keyBytes = {0x11, 0x22, 0x4F, 0x58,
(byte)0x88, 0x10, 0x40, 0x38, 0x28, 0x25, 0x79, 0x51,
(byte)0xCB, (byte)0xDD, 0x55, 0x66, 0x77, 0x29, 0x74,
(byte)0x98, 0x30, 0x40, 0x36, (byte)0xE2
}; //24字节的密钥
String szSrc = "This is a 3DES test. 测试";
System.out.println("加密前的字符串:" + szSrc);
byte[] encoded = encryptMode(keyBytes, szSrc.getBytes());
System.out.println("加密后的字符串:" + new String(encoded));
byte[] srcBytes = decryptMode(keyBytes, encoded);
System.out.println("解密后的字符串:" + (new String(srcBytes)));
}
}
com.sun.crypto.provider.SunJCE的更多相关文章
- Maven打包编译找不到com.sun.crypto.provider.SunJCE类
Maven配置 <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>mav ...
- pom里引入lib下的包后编译报 package com.sun.crypto.provider does not exist问题解决
最近正在迭代开发的一个项目编译安装时出现报“package com.sun.crypto.provider does not exist”的错误,由于本人能力水平有限,也是第一次遇到该问题,来来回回折 ...
- 解决 java 使用ssl过程中出现"PKIX path building failed: sun.security.provider.certpath.SunCertPathBuilderException: unable to find valid certification path to requested target"
今天,封装HttpClient使用ssl时报一下错误: javax.net.ssl.SSLHandshakeException: sun.security.validator.ValidatorExc ...
- javax.net.ssl.SSLHandshakeException: sun.security.validator.ValidatorException: PKIX path building failed: sun.security.provider.certpath.SunCertPathBuilderException: unable to find valid certificatio
场景:Java调用PHP接口,代码部署在服务器上后,调用报错,显示PHP服务器那边证书我这边服务器不信任(我猜的). 异常信息: 2019-08-06 14:00:09,102 [http-nio-4 ...
- java程序中访问https时,报 PKIX path building failed: sun.security.provider.certpath.SunCertPathBuilderException: unable to find valid certification path to requested target
在java中使用https访问数据时报异常: Caused by: sun.security.validator.ValidatorException: PKIX path building fail ...
- Maven:sun.security.validator.ValidatorException: PKIX path building failed: sun.security.provider.certpath.SunCertPathBuilderException: unable to find valid certification path to requested target
还是记录使用 maven 时遇到的问题. 一.maven报错 maven package 进行打包时出现了以下报错: Non-resolvable parent POM for com.wpbxin: ...
- mvn 编译报错mavn sun.security.validator.ValidatorException: PKIX path building failed: sun.security.provider.certpath.SunCertPathBuilderException: unable to find valid certification path to requested targ
mavn 编译报错: mavn sun.security.validator.ValidatorException: PKIX path building failed: sun.security.p ...
- Flutter配置环境报错“PKIX path building failed: sun.security.provider.certpath.SunCertPathBuilderException: unable to find valid certification path to requested target”
背景:最近看了很多Flutter漂亮的项目,想要尝试一下.所有环境都搭建好之后,按照文档一步一步配置(抄袭),但始终报如下图错误. PKIX path building failed: sun.sec ...
- PKIX path building failed: sun.security.provider.certpath.SunCertPathBuilderException: unable to find valid certification path to requested target
注:网上搜来的快照,暂未验证 在java代码中请求https链接的时候,可能会报下面这个错误javax.net.ssl.SSLHandshakeException: sun.security.vali ...
随机推荐
- (spring-第8回【IoC基础篇】)BeanDefinition在IoC容器中的注册
在spring中,所有的bean都是由BeanFactory进行管理的.下面是BeanFactory的类体系结构: 我们清楚的看到,DefaultListableBeanFactory继承了BeanF ...
- LeetCode----Word Ladder 2
Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from ...
- cout 计算顺序问题
cout输出流的执行顺序 下面是IBM的一道笔试题 #include <iostream> using namespace std; int fun( ) { cout << ...
- python类的定义和使用
python中类的声明使用关键词class,可以提供一个可选的父类或者说基类,如果没有合适的基类,那就用object作为基类. 定义格式: class 类名(object): "类的说明文档 ...
- 多个storyboard开发应用程序,封装.bundle和.a不用xib使用storyboard!!!
一,封装.bundle和.a使用xib的方式前面已经说过了,具体方式不再赘述,简单介绍于下: 静态库加.h bundle:删plist,改sdk,加xib 简称psx三步 引用库的项目,加.a .b ...
- Android沉浸式(侵入式)标题栏(状态栏)Status(一)
Android沉浸式(侵入式)标题栏(状态栏)Status(一) 现在越来越多的APP设计采用这种称之为沉浸式状态栏(Status)的设计,这种沉浸式状态栏又称之"侵入式"状 ...
- magneto创建运费模板
Magento系统自带了大概7种运费方式:平价.运费表.免运费.ups.usps.fedex.dhl等.不过这些依然无法满足我们的需求,这时候就需要创建一个shipping module 来实现了.创 ...
- 转:Enterprise Library 4.0缓存应用程序块
英文原文:http://msdn.microsoft.com/zh-cn/library/cc511588(en-us).aspx Enterprise Library 缓存应用程序块允许开发人员在应 ...
- JavaScript中递归函数用法需要注意的
<script> function sum(num){ if(num<=1){ return 1; }else{ return num*sum(num-1);//return num ...
- 合并两个有序数组a和b到c
问题:两个有序数组a和b,合并成一个有序数组c. // 合并两个有序数组a和b到c void Merge_Array(int a[], int n, int b[], int m, int c[]) ...