SQL Blind Injection Wirkungsweise: Unterschied zwischen den Versionen
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| Zeile 25: | Zeile 25: | ||
;Erkenntnis das Passwort ist genau 6 Zeichen lang. | ;Erkenntnis das Passwort ist genau 6 Zeichen lang. | ||
| − | + | *select user from my_auth where user='erwin' and substring(password,1,1) > 5 ; | |
Empty set (0.00 sec) | Empty set (0.00 sec) | ||
| − | + | *select user from my_auth where user='erwin' and substring(password,1,1) > 3 ; | |
| − | |||
Empty set (0.00 sec) | 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 | ||
Version vom 12. März 2023, 14:02 Uhr
Wir wollen das Passwort herausfinen ohne es zu sehen
- In diesem Beispiel benutzen wir zur Vereinfachung nur Zahlen
Länge der Passworts ermitteln
- select user from my_auth where user='erwin' and length(password) > 5;
+-------+ | user | +-------+ | erwin | +-------+ 1 row in set (0.00 sec)
- select user from my_auth where user='erwin' and length(password) < 7;
+-------+ | user | +-------+ | erwin | +-------+ 1 row in set (0.00 sec)
- select user from my_auth where user='erwin' and length(password) = 6;
+-------+ | user | +-------+ | erwin | +-------+ 1 row in set (0.00 sec)
- Erkenntnis das Passwort ist genau 6 Zeichen lang.
- select user from my_auth where user='erwin' and substring(password,1,1) > 5 ;
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