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How do I solve SQL Injection Lab #11?

Zephyr | Last updated: Aug 20, 2021 02:29AM UTC

Hi, I am working through lab #11: https://portswigger.net/web-security/sql-injection/blind/lab-conditional-responses After finding out how many characters there are in the admin's password, the final stage is to crack the password one character at a time. I am using Burp Suite, but for some reason none of my SQL injections are working when I put them through the intruder. I know this because the webpage is not returning any "Welcome back!" messages. Below is my SQL statement, but I cannot see anything wrong with it. 'x' is a placeholder. I use the intruder to test characters from 0-9 and a-z and get no results. ' AND (SELECT SUBSTRING(password,1,1) FROM users WHERE username='administrator')='x'--

Michelle, PortSwigger Agent | Last updated: Aug 23, 2021 08:49AM UTC

Thanks for your message. You're on the right track with your SQL statement. Have you had a look through the community solution video solution to check for any differences in the setup of the Intruder attack?

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