发 微信公众号作为强大的自媒体工具，对接一下是很正常的了。不过这不是本文的方向，本文的方向公众号接入的排序问题。
最近接了一个重构的小项目，需要将原有的php的公众号后台系统，转换为java系统。当然，也很简单的了。
不过，在接入的时候，遇到有一个有趣的问题，可以分享下。
大家知道，要将微信在接到用户的请求之后，可以将消息转发给咱们在公众号后台指定的 server 地址，而在指定这个地址的时候，又需要先校验下这个地址是否连通的，是不是开发者自己的用来处理微信转发消息的地址。因此有一个服务器 token 的校验过程。
校验token的过程，算法很简单，引用微信文档原文如下:
开发者通过检验signature对请求进行校验（下面有校验方式）。若确认此次GET请求来自微信服务器，请原样返回echostr参数内容，则接入生效，成为开发者成功，否则接入失败。加密/校验流程如下：
1）将token、timestamp、nonce三个参数进行字典序排序
2）将三个参数字符串拼接成一个字符串进行sha1加密
3）开发者获得加密后的字符串可与signature对比，标识该请求来源于微信
php 示例如下：
private function checkSignature() { _GET["signature"]; _GET["timestamp"]; _GET["nonce"]; tmpArr = array(timestamp, $nonce); sort($tmpArr); // 官方最新版demo已经修复该排序问题了 sort($tmpArr, SORT_STRING); $tmpStr = implode( $tmpArr ); $tmpStr = sha1( $tmpStr ); if( signature ){ return true; }else{ return false; } }
而且，下方demo也给的妥妥的。好吧，对接是不会有问题了！
我也按照java版的demo，给整了个接入进来！java 的验证样例如下：
public String validate(@ModelAttribute WxValidateBean validateBean) { String myValidToken = signatureWxReq(validateBean.getToken(), validateBean.getTimestamp(), validateBean.getNonce()); if(myValidToken.equals(validateBean.getSignature())) { return validateBean.getEchostr(); } return ""; } public String signatureWxReq(String token, String timestamp, String nonce) { try { String[] array = new String[] { token, timestamp, nonce }; StringBuffer sb = new StringBuffer(); // 字符串排序 Arrays.sort(array); for (int i = 0; i < 4; i++) { sb.append(array[i]); } String str = sb.toString(); // SHA1签名生成 MessageDigest md = MessageDigest.getInstance("SHA-1"); md.update(str.getBytes()); byte[] digest = md.digest(); StringBuffer hexstr = new StringBuffer(); String shaHex = ""; for (int i = 0; i < digest.length; i++) { shaHex = Integer.toHexString(digest[i] & 0xFF); if (shaHex.length() < 2) { hexstr.append(0); } hexstr.append(shaHex); } return hexstr.toString(); } catch (Exception e) { e.printStackTrace(); throw new RuntimeException("加密sign异常!"); } }
按理肯定也不会有问题了。不过为了保险起见，我还是写了个测试用例！自测一下！
private final String oldSysWxAddress = "http://a.com/wx"; private final String newSysWxAddress = "http://localhost:8080/wx"; @Test public void testWxValidateToken() throws IOException { String token = "abc123"; String timestamp = new Date().getTime() / 1000 + ""; String nonce = "207665"; // 这个就随便一写的数字 String echoStr = "1kjdslfj"; String signature = signatureWxReq(token, timestamp, nonce); Map<String, String> params = new HashMap<>(); params.put("token", token); params.put("timestamp", timestamp); params.put("nonce", nonce); params.put("signature", signature); params.put("echostr", echoStr); String oldSysResponse = HttpClientOp.doGet(oldSysWxAddress, params); Assert.assertEquals("老系统验证不通过，请检查加密算法！", oldSysResponse, echoStr); String newSysResponse = HttpClientOp.doGet(newSysWxAddress, params); Assert.assertEquals("新系统验证不通过，出bug了！", newSysResponse, echoStr); Assert.assertEquals("新老返回不一致，测试不通过！", newSysResponse, oldSysResponse); System.out.println("OK"); }
想着吧，也就走个过场得了。结果，还真不是这样！出问题了，"老系统验证不通过，请检查加密算法！" 。
按理不应该啊！但是代码是理智的，咱们得找到bug不是。
最后，通过一步步排查，终于发现了，原来是 php 的排序结果，与 java 的排序结果不一致，因此得到的加密串就不对了。
为啥呢？sort($tmpArr); php是弱类型语言，我们请求的虽然看起来是字符串，但是解析后，因为得到的是一整串数字，因此就认为是可以用整型或这种比较数字大小方式了。
所以，一比较时间和 nonce 随机数，因为随机数位数小，因此自然就应该排在时间戳的前面了。
而对于 java 的排序呢？ Arrays.sort(Object[] a), 我们来看一下源码！
public static void sort(Object[] a) { if (LegacyMergeSort.userRequested) legacyMergeSort(a); else // 默认使用 ComparableTimSort 排序 ComparableTimSort.sort(a, 0, a.length, null, 0, 0); } // ComparableTimSort.sort() /** * Sorts the given range, using the given workspace array slice * for temp storage when possible. This method is designed to be * invoked from public methods (in class Arrays) after performing * any necessary array bounds checks and expanding parameters into * the required forms. * * @param a the array to be sorted * @param lo the index of the first element, inclusive, to be sorted * @param hi the index of the last element, exclusive, to be sorted * @param work a workspace array (slice) * @param workBase origin of usable space in work array * @param workLen usable size of work array * @since 1.8 */ static void sort(Object[] a, int lo, int hi, Object[] work, int workBase, int workLen) { assert a != null && lo >= 0 && lo <= hi && hi <= a.length; int nRemaining = hi - lo; if (nRemaining < 2) return; // Arrays of size 0 and 1 are always sorted // If array is small, do a "mini-TimSort" with no merges if (nRemaining < MIN_MERGE) { // 二分插入排序 int initRunLen = countRunAndMakeAscending(a, lo, hi); binarySort(a, lo, hi, lo + initRunLen); return; } /** * March over the array once, left to right, finding natural runs, * extending short natural runs to minRun elements, and merging runs * to maintain stack invariant. */ ComparableTimSort ts = new ComparableTimSort(a, work, workBase, workLen); int minRun = minRunLength(nRemaining); do { // Identify next run int runLen = countRunAndMakeAscending(a, lo, hi); // If run is short, extend to min(minRun, nRemaining) if (runLen < minRun) { int force = nRemaining <= minRun ? nRemaining : minRun; binarySort(a, lo, lo + force, lo + runLen); runLen = force; } // Push run onto pending-run stack, and maybe merge ts.pushRun(lo, runLen); ts.mergeCollapse(); // Advance to find next run lo += runLen; nRemaining -= runLen; } while (nRemaining != 0); // Merge all remaining runs to complete sort assert lo == hi; ts.mergeForceCollapse(); assert ts.stackSize == 1; } /** * Returns the length of the run beginning at the specified position in * the specified array and reverses the run if it is descending (ensuring * that the run will always be ascending when the method returns). * * A run is the longest ascending sequence with: * * a[lo] <= a[lo + 1] <= a[lo + 2] <= ... * * or the longest descending sequence with: * * a[lo] > a[lo + 1] > a[lo + 2] > ... * * For its intended use in a stable mergesort, the strictness of the * definition of "descending" is needed so that the call can safely * reverse a descending sequence without violating stability. * * @param a the array in which a run is to be counted and possibly reversed * @param lo index of the first element in the run * @param hi index after the last element that may be contained in the run. It is required that {@code lo < hi}. * @return the length of the run beginning at the specified position in * the specified array */ @SuppressWarnings({"unchecked", "rawtypes"}) private static int countRunAndMakeAscending(Object[] a, int lo, int hi) { assert lo < hi; int runHi = lo + 1; if (runHi == hi) return 1; // Find end of run, and reverse range if descending // 调用的是 XXObject.compareTo() 方法，找出第一个比后续值小的index, 作为快排的基点 if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { // Descending while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) < 0) runHi++; reverseRange(a, lo, runHi); } else { // Ascending while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) >= 0) runHi++; } return runHi - lo; } /** * 反转元素 * Reverse the specified range of the specified array. * * @param a the array in which a range is to be reversed * @param lo the index of the first element in the range to be reversed * @param hi the index after the last element in the range to be reversed */ private static void reverseRange(Object[] a, int lo, int hi) { hi--; while (lo < hi) { Object t = a[lo]; a[lo++] = a[hi]; a[hi--] = t; } } /** * 二分插入排序 * Sorts the specified portion of the specified array using a binary * insertion sort. This is the best method for sorting small numbers * of elements. It requires O(n log n) compares, but O(n^2) data * movement (worst case). * * If the initial part of the specified range is already sorted, * this method can take advantage of it: the method assumes that the * elements from index {@code lo}, inclusive, to {@code start}, * exclusive are already sorted. * * @param a the array in which a range is to be sorted * @param lo the index of the first element in the range to be sorted * @param hi the index after the last element in the range to be sorted * @param start the index of the first element in the range that is * not already known to be sorted ({@code lo <= start <= hi}) */ @SuppressWarnings({"fallthrough", "rawtypes", "unchecked"}) private static void binarySort(Object[] a, int lo, int hi, int start) { assert lo <= start && start <= hi; if (start == lo) start++; for ( ; start < hi; start++) { Comparable pivot = (Comparable) a[start]; // Set left (and right) to the index where a[start] (pivot) belongs int left = lo; int right = start; assert left <= right; /* * Invariants: * pivot >= all in [lo, left). * pivot < all in [right, start). */ while (left < right) { int mid = (left + right) >>> 1; if (pivot.compareTo(a[mid]) < 0) right = mid; else left = mid + 1; } assert left == right; /* * The invariants still hold: pivot >= all in [lo, left) and * pivot < all in [left, start), so pivot belongs at left. Note * that if there are elements equal to pivot, left points to the * first slot after them -- that‘s why this sort is stable. * Slide elements over to make room for pivot. */ int n = start - left; // The number of elements to move // Switch is just an optimization for arraycopy in default case switch (n) { case 2: a[left + 2] = a[left + 1]; case 1: a[left + 1] = a[left]; break; default: System.arraycopy(a, left, a, left + 1, n); } a[left] = pivot; } } // String.compareTo() 方法，比较 char大小，即字典顺序 public int compareTo(String anotherString) { int len1 = value.length; int len2 = anotherString.value.length; int lim = Math.min(len1, len2); char v1[] = value; char v2[] = anotherString.value; int k = 0; while (k < lim) { char c1 = v1[k]; char c2 = v2[k]; if (c1 != c2) { return c1 - c2; } k++; } return len1 - len2; }
可以看到，Arrays.sort(), 对于小数的排序，是使用二分插入排序来做的，而具体排序先后，则是调用具体类的 compareTo() 方法，也就是说要进行比较的类，须实现 Comparator 接口。另外，从String的比较方法中，我们也可以看出 char 其实能保存很多东西而不只是 1-128。比如 char s = (char)"中";
而对于排序算法，则针对不同情况，选择不同的合适算法，从而提高运行效率！
而对于String.compareTo() 则是比较字符的ascii顺序！