简述一下项目中手写的Token验证服务设计过程java
这里直接展现整个项目中用到的算法库,其中涉及位运算的可无论
直接应用到的方法是hash(str)
大概流程以下
1.构造一个大素数表并随机打乱
2.提供足够快的快速幂
3.哈希规则:\sum 下标对应byte^^randomPrimes[下标 % 素数表长度] % 128
为了更快的hash过程其实能够把下标进一步转为其bitcount,这样算幂会把log的复杂度略降一点web
package com.noresp.oj.utils; /** * 方便OJ搭建的简易算法库 * 目前可提供: * 随机大素数表 * 随机打乱 * 哈希(注意:特定用途) * 整型交换、bitcount、fastPow * 随机数 */ public class AlgsUtils { private static final int[] bitmasks = new int[0x100]; private static final int[] randomPrimes = new int[1<<10]; public static final long magicNumber = 19260817L; public static class SimpleRandom { long seed = 1L; public void setSeed(long seed) { this.seed = seed; } /** * 简易高效的手写随机数 * 大概比Math.random快20倍(2^^26数量级下) * @return 随机数 */ public long next() { seed = seed*1103515245+12345 & 0xffffffffL; // 模拟unsigned int // 切记0xffffffff没有L会翻车。。 return seed >> 16; } public int next(int mod) { return (int)(next()%mod); } } static { initializeBitmasks(); initializePrimeTable(); randomShuffle(randomPrimes,magicNumber); } /** * O(n)打长度为n的二进制表 * 测试经过 */ private static void initializeBitmasks() { for(int i = 0xff; i > 0; --i) { if(bitmasks[i] != 0) continue; for(int j = i; j > 0; j -= j&-j) { bitmasks[i]++; } for(int j = i, k = 0; j > 0; j -= j&-j, k++) { bitmasks[j] = bitmasks[i]-k; } } } /** * 计算二进制1的个数 * 测试经过 * @param value * @return */ public static int bitCount(int value) { int result = 0; for(; value > 0; value >>>= 8) { result += bitmasks[value & 0xff]; } return result; } /** * 经过固定的随机素数进行哈希/加密 * 哈希串 = \sum 下标对应byte^^randomPrimes[下标 % 素数表长度] % 128 * 时间复杂度O(n log2(log2m)),其中n为字符串长度,m为最大的随机素数大小 * @param string * @return */ public static String hash(String string) { if(string == null || string.length() == 0) return null; byte[] b = string.getBytes(); for(int i = 0; i < b.length; i++) { int j = b[i]; int k = randomPrimes[i & randomPrimes.length-1]; // 仿java.util思路 二进制长用&优化取代% k = bitCount(k); b[i] = (byte)fastPow(j,k,0x7f); } return new String(b); //注意不要b.toString() } /** * 快速幂求解a^^n%mod,n不支持负数 * @param a * @param n * @param mod * @return */ public static int fastPow(long a,long n,int mod) { long res = 1; //long防相乘溢出 while(n > 0) { if((n&1) == 1) { res = res*a; if(res >= mod) res %= mod; } a *= a; if(a >= mod) a %= mod; n >>= 1; } return (int)res; } /** * 简单的筛法初始化素数表 */ private static void initializePrimeTable() { int n = randomPrimes.length << 8; // 一个大概的打表估值,预计素数的大小在1e5数量级 boolean[] notPrime = new boolean[n]; for(int i = 2; i*i < n; i++) { if(!notPrime[i]) { for(int j = i; j < n; j += i) { notPrime[j] = true; } } } // 优先选取大素数,所以倒序处理+插入两个极大的素数 randomPrimes[0] = (int)1e9+7; randomPrimes[1] = 998244353; for(int i = n-1, j = 2; true; --i) { if(!notPrime[i]) randomPrimes[j++] = i; if(j == randomPrimes.length) return; } } /** * 随机打乱一个整型数组 * @param toRandom */ public static void randomShuffle(int[] toRandom,long seed) { SimpleRandom roll = new AlgsUtils.SimpleRandom(); roll.setSeed(seed); for(int i = toRandom.length-1; i > 0; --i) { swap(toRandom,i,roll.next(i+1)); } } /** * 交换两个数,注意安全使用 */ public static void swap(int[] arr,int i,int j) { if(SafeUtils.isOutOfBound(arr,i)) return; if(SafeUtils.isOutOfBound(arr,j)) return; int t = arr[i]; arr[i] = arr[j]; arr[j] = t; } public static void swapRange(int[] arr,int lo,int hi) { while(lo < hi) swap(arr,lo++,hi--); } }
由PART A能够看到任意编码的字符串都把char限制在0-127范围内,但可能存在特殊的转义符影响面向文本的协议算法
所以须要把0-127映射到ASCII中a-z A-Z 0-9的范围内数组
为了知足尽量的均匀分布,又乱写了一个算法(其实ch+i是多余的)安全
public static String visualizableHash(String str) { StringBuilder sb = new StringBuilder(""); for(int i = 0; i < str.length(); i++) { char ch = str.charAt(i); if(isVisualChar(ch)) sb.append(ch); else { char curChar = 'a'; long factor = (int)(ch)*17+i*23; int pos = (int)(factor % (26+26+10)); if(pos < 26) curChar = (char)('a'+pos); else if(pos-26 < 26) curChar = (char)('A'+pos-26); else curChar = (char)('0'+pos-26-26); sb.append(curChar); } } return sb.toString(); }
这样调用visualizableHash(hash(str))就能得到一个还能够的文本哈希了cookie
其中payload就是我要负载的内容app
Sign做为签名校验dom
目前是使用简单的String,也提供了简单的Map转换
格式见doc说明测试
package com.noresp.oj.utils; import java.util.*; /** * 使用Token,解放Session * 注:一个Token的格式 * [encode(key1).encode(val1).encode(key2).encode(val2).....mySign] * 目前encode默认是base64 */ public class TokenUtils { private static String encode(String str) { if(str == null || str.length() == 0) return ""; return Base64.getEncoder().encodeToString(str.getBytes()); } private static String decode(String str) { if(str == null || str.length() == 0) return ""; return new String(Base64.getDecoder().decode(str.getBytes())); } public static String getTokenPayload(String key) { return encode(key); } public static String getTokenSign(String... base64Payloads) { StringBuilder sb = new StringBuilder(""); for(String payload : base64Payloads) { sb.append(StringUtils.visualizableHash(AlgsUtils.hash(payload))); } return sb.toString(); } public static String getToken(String... payloads) { StringBuilder sb = new StringBuilder(""); String[] encodedPayloads = new String[payloads.length]; for(int i = 0; i < payloads.length; i++) { encodedPayloads[i] = getTokenPayload(payloads[i]); sb.append(encodedPayloads[i]+"."); } sb.append(getTokenSign(encodedPayloads)); return sb.toString(); } /** * 解密和校验Token * @param token * @return 若是校验失败,会返回null,不然返回Token解密内容 */ public static String[] decodeTokenAndValidate(String token) { if(token == null) return null; List<String> result = new LinkedList<>(); for(int i = 0, len = 1; i < token.length(); i++,len++) { if(token.charAt(i) == '.') { String payload = (token.substring(i-len+1,i)); result.add(payload); len = 0; } if(i == token.length()-1) { String salt = token.substring(i-len+1,i+1); len = 0; String[] encodedPayloads = new String[result.size()]; Iterator<String> itor = result.iterator(); while(itor.hasNext()) { encodedPayloads[len++] = itor.next(); } String comp = getTokenSign(encodedPayloads); if(!salt.equals(comp)) { return null; } String[] decodedPayloads = encodedPayloads; // 引用是同样的 for(len = 0; len < encodedPayloads.length; len++) { decodedPayloads[len] = decode(encodedPayloads[len]); } return decodedPayloads; } } return null; } public static Map<String,String> tokenMap(String[] decodedToken) { Map<String,String> result = new HashMap<>(); if(decodedToken == null) return result; for(int i = 0; i < decodedToken.length; i+=2) { result.put(decodedToken[i],decodedToken[i+1]); } return result; } public static String getTokenAttribute(String token,String key) { Map<String,String> tokenMap = tokenMap(decodeTokenAndValidate(token)); return tokenMap.getOrDefault(key,null); } }
目前用于token的payload有userid和ip,后者是为了进一步提升安全性优化
而且token是直接放在Cookie里头,方便管理生命周期
写得比较杂乱,先贴部分感觉一下吧
@PostMapping("/register") public @ResponseBody String registerPost( HttpServletRequest request, HttpServletResponse response, @RequestParam(value = "email") String email, @RequestParam(value = "username") String username, @RequestParam(value = "password") String password) throws IOException { Boolean isCreated = userService.createUser(username,password,email,userService.getDefaultUserGroup()); Map<String,Boolean> result = new HashMap<>(); result.put("isCreated",isCreated); if(isCreated) { String token = TokenUtils.getToken( "userID", String.valueOf(userService.getUserByUsername(username).getUserID()), "ip",controllerUtils.getRemoteAddr(request) ); Cookie cookie = new Cookie("token",token); cookie.setMaxAge(60*60*24*7); cookie.setHttpOnly(true); response.addCookie(cookie); } return JSONUtils.toJSON(result); }
其中getRemoteAddr的实现为
public String getRemoteAddr(HttpServletRequest request) { if ( request.getHeader("X-Real-IP") != null ) { return request.getHeader("X-Real-IP"); } return request.getRemoteAddr(); }
校验过程太繁琐了,固然要用到AOP,这里采用注解的方式来实现
1.先给一个注解标记
@Retention(RetentionPolicy.RUNTIME) @Target(ElementType.METHOD) public @interface NeedLogin { }
2.接着就是AOP
(请无视直接println
@Component @Aspect public class LoginAspect { @Autowired ControllerUtils controllerUtils; /** * 【约定大于配置】 * 当须要使用@NeedLogin时,token需做为入参的第一个保证AOP成功拦截 * Token校验包括了加盐的检验和IP的对比,以及开启HttpOnly安全设置 * 若是有错会及时把劫持的Cookie删除 // PS.有点小瑕疵 * @param proceedingJoinPoint * @param token * @return * @throws Throwable */ @Around(value = "@annotation(com.noresp.oj.annotations.NeedLogin) && args(token,request,response,..)") public ModelAndView loginCheck( ProceedingJoinPoint proceedingJoinPoint, String token, HttpServletRequest request, HttpServletResponse response) throws Throwable { if(token == null) { System.out.println("没有token"); return ViewUtils.redirect( "/",new ErrorInfo("login required")); } Map<String,String> tokenMap = TokenUtils.tokenMap( TokenUtils.decodeTokenAndValidate(token)); Integer userID = StringUtils.safeStringToInteger(tokenMap.get("userID")); String recordedIP = tokenMap.get("ip"); System.out.println(userID+" "+recordedIP); boolean tokenIllegal = userID == null || !controllerUtils.getRemoteAddr(request).equals(recordedIP); if(tokenIllegal) { System.out.println("token错误"); Cookie fakeToken = controllerUtils.getCookie(request,"token"); if(fakeToken != null) { fakeToken.setMaxAge(0); } return ViewUtils.redirect( "/",new ErrorInfo("login required")); } return (ModelAndView)proceedingJoinPoint.proceed(); } }
3.使用样例
须要注意AOP没有很好的arg通配方法,这里使用的规约见上面定义
@NeedLogin @GetMapping("/{problemID}/submit") public ModelAndView submitView( @CookieValue(value = "token",required = false) String token, HttpServletRequest request, HttpServletResponse response, @PathVariable("problemID") int problemID) { ModelAndView view = new ModelAndView("/problems/submit"); Problem problem = problemService.getProblem(problemID); if(problem == null) { return ViewUtils.redirect("/",new ErrorInfo("No Such Problem.")); } view.addObject("problem",problem); return view; }
目前的不足 1.token长度受限于Cookie 2.校验的复杂度仍是大了点