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gunitinug

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Everything posted by gunitinug

  1. sorry APP.accounts table with user column
  2. correction. it should be APP.contacts table with user column.
  3. Correction. This logs in admin' and (select count(user) from app.accounts where user not like '%')=0-- But not admin' and (select count(user) from app.accounts where user not like '%')>0--
  4. Hi. I have to pen test altoro mutual site(https://demo.testfire.net) for a project. The site uses DERBY DB. I have discovered that its login page is vulnerable to blind boolean sqli. I have discovered that there is a table called user under schemaname of APP (ie. APP.user). I typed in Username: admin' and (select count(user) from app.accounts where user like '%a%')>0-- Password: anything This tests whether there is a user that contains a letter 'a'. If the test succeeds altoro mutual site logs in. Otherwise it says "Login Failed: We're sorry, but this username or password was not found in our system. Please try again." I've tried the same test but this time iterated from a-zA-Z. But it never succeeds in logging in which tells me that maybe Username is not English alphabet. But this is unlikely. So my problem is I don't know why LIKE operator doesn't return a result that is expected. I also tried Username: admin' and (select count(user) from app.accounts where user not like '%a%')>0-- Password: anything And this time every iteration of a-zA-Z logs in. So this result also tells me Username does not contain a letter. Lastly this one works (it logs in) Username: admin' and (select count(user) from app.accounts where user not like '%')>0-- Password: anything Can you help me why LIKE operator fails when user LIKE '%a%' and so on? THX
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