<匿名com.essence.filter.XssHttpServletRequestWrapper$1>不是抽象的, 并且未覆盖javax.servlet.ServletInputStream中的抽象方法setReadListener(javax.servlet.ReadListener)

你提到的类`com.essence.filter.XssHttpServletRequestWrapper$1`不是一个抽象类，并且它没有覆盖`javax.servlet.ServletInputStream`中的抽象方法`setReadListener(javax.servlet.ReadListener)`。这意味着该类需要实现该方法，否则将会报错。

`javax.servlet.ServletInputStream`是Servlet API中的接口，它定义了与请求输入流相关的方法。其中`setReadListener(javax.servlet.ReadListener)`是一个抽象方法，需要在实现类中进行具体实现。根据你提供的信息，`com.essence.filter.XssHttpServletRequestWrapper$1`类没有对这个方法进行实现，因此会导致编译错误或运行时异常。

要解决这个问题，你需要在`com.essence.filter.XssHttpServletRequestWrapper$1`类中添加对`setReadListener(javax.servlet.ReadListener)`方法的具体实现。具体的实现取决于你的需求和业务逻辑。

相关问题

要求仿照该代码进行编写<?php $code=0; $data=[]; $msg=["获取面板数据失败","获取面板成功"]; include("conn.php"); include("functions.php"); // $sql="select * from panels p left join users u on p.user_id=u.user_id order by p.panel_is_top desc,p.updated_time desc"; $sql="select p.panel_id,p.panel_title,p.panel_content,p.panel_likeit_nums,p.panel_is_top,p.panel_is_essence,p.panel_date,u.user_name from panels as p left join users as u on p.user_id=u.user_id order by p.panel_is_top desc,p.updated_time desc"; $rs=mysqli_query($conn,$sql); $i=0; //拼装data部分的对象数据 while($row=mysqli_fetch_array($rs)){ //读取记录行 $data[$i]["id"]=$row["panel_id"]; $data[$i]["user"]=$row["user_name"]; $data[$i]["likeitNums"]=$row["panel_likeit_nums"]; $data[$i]["topPanel"]=$row["panel_is_top"]; $data[$i]["essencePanels"]=$row["panel_is_essence"]; $data[$i]["date"]=$row["panel_date"]; $data[$i]["content"]=$row["panel_content"]; $data[$i]["title"]=$row["panel_title"]; //get comments $panelId=$row["panel_id"]; $sql="select c.panel_id,c.comment_id,c.comment_content,u.user_name from comments c left join users as u on c.user_id=u.user_id where c.panel_id=$panelId order by c.updated_time desc"; $rs1=mysqli_query($conn,$sql); $j=0; $comments=[]; while($row1=mysqli_fetch_array($rs1)){ $comments[$j]["id"]=$row1["comment_id"]; $comments[$j]["user"]=$row1["user_name"]; $comments[$j]["content"]=$row1["comment_content"]; $j++; } $data[$i]["comments"]=$comments; $rs1=null; $row1=null; $i++; } if(count($data)>0) $code=1; header("content-type:application/json"); echo getApiResult($code,$data,$msg); ?>

好的，根据您提供的代码，以下是经过修改后的 PHP 代码，可以将面板数据以 API 的形式返回给前端：

<?php
$code = 0;
$data = [];
$msg = ["获取面板数据失败", "获取面板成功"];

include("conn.php");
include("functions.php");

$sql = "select p.panel_id,p.panel_title,p.panel_content,p.panel_likeit_nums,p.panel_is_top,p.panel_is_essence,p.panel_date,u.user_name from panels as p left join users as u on p.user_id=u.user_id order by p.panel_is_top desc,p.updated_time desc";
$rs = mysqli_query($conn, $sql);

$i = 0;
while($row = mysqli_fetch_array($rs)) {
    $panelId = $row["panel_id"];

    $panel = [
        "id" => $row["panel_id"],
        "user" => $row["user_name"],
        "likeitNums" => $row["panel_likeit_nums"],
        "topPanel" => $row["panel_is_top"],
        "essencePanels" => $row["panel_is_essence"],
        "date" => $row["panel_date"],
        "content" => $row["panel_content"],
        "title" => $row["panel_title"],
        "comments" => []
    ];

    $sql = "select c.panel_id,c.comment_id,c.comment_content,u.user_name from comments c left join users as u on c.user_id=u.user_id where c.panel_id=$panelId order by c.updated_time desc";
    $rs1 = mysqli_query($conn, $sql);

    $j = 0;
    while($row1 = mysqli_fetch_array($rs1)) {
        $comment = [
            "id" => $row1["comment_id"],
            "user" => $row1["user_name"],
            "content" => $row1["comment_content"]
        ];
        array_push($panel["comments"], $comment);
        $j++;
    }

    array_push($data, $panel);
    $i++;
}

if(count($data) > 0) {
    $code = 1;
}

header("content-type:application/json");
echo getApiResult($code, $data, $msg);
?>

这段代码中，我们首先执行了 SQL 查询，获取到了面板和评论的数据，然后遍历面板数据，通过拼装一个面板对象，并在对象中包含该面板的评论数据，最终将面板对象添加到 $data 数组中。最后，根据 $data 数组的长度，判断是否成功获取到了面板数据，返回对应的 $code 和 $msg。在返回数据之前，使用 header() 函数设置响应头，告诉浏览器返回的是 JSON 格式的数据。

请用C语言实现Once upon a time in the mystical land of Draconis, there existed two powerful arrays: M and N . These arrays were filled with positive integers, each carrying its own magical essence. The inhabitants of the land were intrigued by the concept of similarity between arrays. They discovered that two arrays, M and N , could be considered similar if it was possible to transform a subarray of N into M by adding or subtracting a constant value to each element. You are now summoned to solve a puzzle. Given two arrays, M and N , your task is to determine the number of subarrays of N that are similar to M . Will you be able to unravel this mystical connection? Input The input consists of multiple lines. The first line contains two integers M and N (1≤M≤N≤106) , representing the lengths of arrays M and N respectively. The second line contains M space-separated positive integers m1,m2,…,mM (1≤mi≤109) , representing the magical elements of array M . The third line contains N space-separated positive integers n1,n2,…,nN (1≤ni≤109) , representing the mystical elements of array N . Output Output a single integer, the number of subarrays of N that are similar to M .

为了解决这个问题，我们可以先对两个数组进行排序，然后用双指针法来找到所有相似的子数组。具体实现如下：

#include <stdio.h>
#include <stdlib.h>

#define MAX_N 1000000

int M[MAX_N], N[MAX_N];
int m, n;

int compare(const void *a, const void *b) {
    return *(int*)a - *(int*)b;
}

int count_similar() {
    int i = 0, j = 0;
    int count = 0;
    while (i < m && j < n) {
        if (M[i] == N[j]) {  // 找到相同的元素
            int k = j;
            while (k < n && N[k] == N[j]) {
                k++;
            }
            count += k - j;  // 加上与N[j]相同的元素的个数
            i++;
            j = k;
        } else if (M[i] < N[j]) {  // M[i]较小，需要增加N中的元素
            i++;
        } else {  // M[i]较大，需要减少N中的元素
            j++;
        }
    }
    return count;
}

int main() {
    scanf("%d%d", &m, &n);
    for (int i = 0; i < m; i++) {
        scanf("%d", &M[i]);
    }
    for (int i = 0; i < n; i++) {
        scanf("%d", &N[i]);
    }
    qsort(M, m, sizeof(int), compare);
    qsort(N, n, sizeof(int), compare);
    printf("%d\n", count_similar());
    return 0;
}

首先读入两个数组的长度和元素，然后对它们进行排序。然后用双指针法找到所有相似的子数组，具体方法如下：

1. 初始化两个指针i和j，分别指向M和N的开头；
2. 如果M[i]等于N[j]，说明找到了一个相似的子数组，接着向右移动j，直到N[j]不等于N[j+1]；
3. 如果M[i]小于N[j]，说明N[j]需要增加，所以向右移动i；
4. 如果M[i]大于N[j]，说明N[j]需要减少，所以向右移动j；
5. 重复2~4步，直到i到达M的末尾或j到达N的末尾。

遍历过程中，我们用一个计数器count来累计相似的子数组的个数，每当找到一个相似的子数组时，我们就把与N[j]相同的元素的个数加到count中。最后返回count即可。

时间复杂度为O(MlogM + NlogN)，空间复杂度为O(M + N)。