SQL Blind Injection Wirkungsweise: Unterschied zwischen den Versionen
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| (4 dazwischenliegende Versionen desselben Benutzers werden nicht angezeigt) | |||
| Zeile 10: | Zeile 10: | ||
1 row in set (0.00 sec) | 1 row in set (0.00 sec) | ||
*select username from users where username='hans.will' and length(password) > 10; | *select username from users where username='hans.will' and length(password) > 10; | ||
| − | + | Empty set (0.00 sec) | |
| − | + | ||
| − | + | *select username from users where username='hans.will' and length(password) = 6; | |
| − | + | +-----------+ | |
| − | + | | username | | |
| − | + | +-----------+ | |
| − | *select | + | | hans.will | |
| − | +-------+ | + | +-----------+ |
| − | | | ||
| − | +-------+ | ||
| − | | | ||
| − | +-------+ | ||
1 row in set (0.00 sec) | 1 row in set (0.00 sec) | ||
;Erkenntnis das Passwort ist genau 6 Zeichen lang. | ;Erkenntnis das Passwort ist genau 6 Zeichen lang. | ||
| − | *select | + | *select username from users where username='hans.will' and substring(password,1,1) = 'g'; |
Empty set (0.00 sec) | Empty set (0.00 sec) | ||
*select user from my_auth where user='erwin' and substring(password,1,1) > 3 ; | *select user from my_auth where user='erwin' and substring(password,1,1) > 3 ; | ||
Aktuelle Version vom 30. August 2023, 09:55 Uhr
Wir wollen das Passwort herausfinen ohne es zu sehen
- In diesem Beispiel benutzen wir zur Vereinfachung nur Zahlen
Länge der Passworts ermitteln
- select username from users where username='hans.will' and length(password) > 5;
+-----------+ | username | +-----------+ | hans.will | +-----------+ 1 row in set (0.00 sec)
- select username from users where username='hans.will' and length(password) > 10;
Empty set (0.00 sec)
- select username from users where username='hans.will' and length(password) = 6;
+-----------+ | username | +-----------+ | hans.will | +-----------+ 1 row in set (0.00 sec)
- Erkenntnis das Passwort ist genau 6 Zeichen lang.
- select username from users where username='hans.will' and substring(password,1,1) = 'g';
Empty set (0.00 sec)
- select user from my_auth where user='erwin' and substring(password,1,1) > 3 ;
Empty set (0.00 sec)
- select user from my_auth where user='erwin' and substring(password,1,1) > 2 ;
+-------+ | user | +-------+ | erwin | +-------+ 1 row in set (0.00 sec)
- select user from my_auth where user='erwin' and substring(password,1,1) = 3 ;
+-------+ | user | +-------+ | erwin | +-------+ 1 row in set (0.00 sec)
- Erkenntnis Erstes Zeichen gleich 3