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。而修复方式自然就是，指明是使用字符串方式来排序就行了，如上官方修正。
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Comparable 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！
算法说明：
1. 对于小数的排序，使用二分插入排序，原理：
1. 先从最开始出发，找出最大的递增或者递减的个数，如果递减，则倒序处理排列，从而使前 n 个元素呈排列状态；
2. 从之前排好序之后的元素开始，使用二分查找法，得到需要移动的个数，进行相应元素的转换。因为 lo 一直是初始值，所以，在 start 循环到 hi 操作完成交换之后，就完成了所有的排序了。
2. 而对大些的数组排序，则使用分而治之的方法，拆分处理，原理：
1. 先算出一个合适的大小，在将输入按其升序和降序特点进行了分区，拆分的原则是大小需要小 MIN_MERGE,也就是32。
2. 针对这些第个分区序列，先按照小数的排序方式排好，然后将结果按规则进行合并。
3. 每次合并会将两个run合并成一个 run。合并的结果保存到栈中。合并直到消耗掉所有的run，这时将栈上剩余的 run合并到只剩一个 run 为止。这时这个仅剩的 run 便是排好序的结果。