[NISACTF 2022]babyupload 详解

 

 试图传文件，但是失败了，有过滤于是查看了下源代码，发现了一个/source

 

 然后输入对应地址，下载下来了源代码

from flask import Flask, request, redirect, g, send_from_directory
import sqlite3
import os
import uuid

app = Flask(__name__)

SCHEMA = """CREATE TABLE files (
id text primary key,
path text
);
"""


def db():
    g_db = getattr(g, '_database', None)
    if g_db is None:
        g_db = g._database = sqlite3.connect("database.db")
    return g_db


@app.before_first_request
def setup():
    os.remove("database.db")
    cur = db().cursor()
    cur.executescript(SCHEMA)


@app.route('/')
def hello_world():
    return """<!DOCTYPE html>
<html>
<body>
<form action="/upload" method="post" enctype="multipart/form-data">
    Select image to upload:
    <input type="file" name="file">
    <input type="submit" value="Upload File" name="submit">
</form>
<!-- /source -->
</body>
</html>"""


@app.route('/source')
def source():
    return send_from_directory(directory="/var/www/html/", path="www.zip", as_attachment=True)


@app.route('/upload', methods=['POST'])
def upload():
    if 'file' not in request.files:
        return redirect('/')
    file = request.files['file']
    if "." in file.filename:
        return "Bad filename!", 403
    conn = db()
    cur = conn.cursor()
    uid = uuid.uuid4().hex
    try:
        cur.execute("insert into files (id, path) values (?, ?)", (uid, file.filename,))
    except sqlite3.IntegrityError:
        return "Duplicate file"
    conn.commit()

    file.save('uploads/' + file.filename)
    return redirect('/file/' + uid)


@app.route('/file/<id>')
def file(id):
    conn = db()
    cur = conn.cursor()
    cur.execute("select path from files where id=?", (id,))
    res = cur.fetchone()
    if res is None:
        return "File not found", 404

    # print(res[0])

    with open(os.path.join("uploads/", res[0]), "r") as f:
        return f.read()


if __name__ == '__main__':
    app.run(host='0.0.0.0', port=80)

进行代码审计

@app.route('/upload', methods=['POST'])
def upload():
    if 'file' not in request.files:
        return redirect('/')
    file = request.files['file']
    if "." in file.filename:
        return "Bad filename!", 403
    conn = db()
    cur = conn.cursor()
    uid = uuid.uuid4().hex
    try:
        cur.execute("insert into files (id, path) values (?, ?)", (uid, file.filename,))
    except sqlite3.IntegrityError:
        return "Duplicate file"
    conn.commit()

    file.save('uploads/' + file.filename)
    return redirect('/file/' + uid)

 

 上述代码首先一个个分析，

首先是POST方式传递文件，然后是过滤的地方

if "." in file.filename:
        return "Bad filename!", 403

 

这段代码过滤了带文件名中带  .   的文件。

然后接下来是后面

try:
        cur.execute("insert into files (id, path) values (?, ?)", (uid, file.filename,))
    except sqlite3.IntegrityError:
        return "Duplicate file"

 

上面这个代码是进行一个sql语句，表示增加一个数据，数据为uid和文件名。

因此我们只需要传输一个文件名为/flag的文件就可以得到flag的uid。然后再访问对应路径就可以得到flag

 

 

 

posted @ 2022-07-21 14:16  FPointmaple  阅读(810)  评论(6)  编辑  收藏  举报