微信公众号接入之排序问题小记 Arrays.sort()

　　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。

微信公众号接入之排序问题小记 Arrays.sort()的相关教程结束。