我正在编写一个脚本，使用python中的RSA算法加密和解密视频。现在，我已经从视频中提取帧并单独加密每个图像，然后组合图像以创建视频。然后我再次读取加密视频的帧，当我应用解密密钥时，我没有取回原始图像。但是，当我在制作视频的任何图像上应用相同的键时，我会取回原始图像。假设我们有 image1，它是加密的，当我在此图像上应用密钥时，我将用于制作加密视频，我正在取回原始图像。现在我有从加密视频中读取的 image2，如果应用了密钥，那么它会提供更加密的图像。这是代码：
import cv2
import numpy
import os
import imageio
import time
from tkinter.filedialog import askopenfilename
from tkinter.ttk import *
from tkinter import *
from tkinter import filedialog
from tqdm import tqdm
from tkinter import messagebox
import subprocess
def load_image_decrypt(folder):
videofile = 'envid.avi'
try:
if not os.path.exists('Dedata'):
os.makedirs('Dedata')
except OSError:
messagebox.showinfo('Error Occured', 'Error: Creating directory of decrypted data')
vid_to_image(videofile)
for filename1 in tqdm(os.listdir(folder)):
imgV = imageio.imread(os.path.join(folder, filename1), format='PNG-FI')
if imgV is not None:
RGBdecryption(imgV, filename1)
else:
break
vidname = 'devid.avi'
image_to_vid(dedata2, vidname)
messagebox.showinfo('Finish!', 'Decryption Done succesfully!')
def RGBdecryption(img, filename):
img1 = img
img = img.astype(numpy.uint16)
img1= img1.tolist()
for i1 in tqdm(range(len(img1))):
for j1 in (range(len(img1[i1]))):
for k1 in (range(len(img1[i1][j1]))):
x1 = img1[i1][j1][k1]
x1 = pow(x1,16971,25777)
img1[i1][j1][k1] = x1
img1 = numpy.array(img1).astype(numpy.uint16)
name = './Dedata/'+str(filename)
imageio.imwrite(name, img1, format='PNG-FI')
def vid_to_image(filename):
# Playing video from file:
cap = cv2.VideoCapture(filename)
try:
if not os.path.exists('data'):
os.makedirs('data')
messagebox.showinfo('Info!', 'Data directory is created where the frames are stored')
except OSError:
print ('Error: Creating directory of data')
currentFrame = 0
while(True):
# Capture frame-by-frame
ret, frame = cap.read()
if not ret:
break
# Saves image of the current frame in jpg file
name = './data/frame' + str(currentFrame) + '.png'
print ('Creating...' + name)
imageio.imwrite(name, frame,format='PNG-FI')
# To stop duplicate images
currentFrame += 1
# When everything done, release the capture
cap.release()
cv2.destroyAllWindows()
def image_to_vid(folder, vidname): #the code which is creating a video out of images stored in the folder
image_folder = folder
video_name = vidname
sort_image = []
images = [img for img in os.listdir(image_folder) if img.endswith(".png")]
print(images)
print('nn')
for i in range(0,1000):
for j in range(len(images)):
name = 'frame' + str(i) + '.png'
if ((str(images[j])) == str(name)):
sort_image.append(images[j])
print(sort_image)
frame = cv2.imread(os.path.join(image_folder, sort_image[0]))
height, width, layers = frame.shape
video = cv2.VideoWriter(video_name, 0, 29, (width,height)) #29 is the fs of the original video and I don't know what the 0 is for
for image in sort_image:
video.write(cv2.imread(os.path.join(image_folder, image)))
cv2.destroyAllWindows()
video.release()
data = './data'
load_image_decrypt(data)
我不知道我哪里弄错了。我是opencv和视频处理的新手。 任何帮助将不胜感激。谢谢。
视频帧会受到有损压缩的影响。因此，您不能以图像为幌子向编解码器提供一些二进制数据，对其进行编码，并期望在播放生成的视频时获得完全相同的二进制数据。
您最好的选择是按照视频文件加密将视频文件作为一个整体进行加密？或如何实时加密视频？。它需要解密才能播放;这显然是OSX的"内容保护"所做的，透明地加密和解密数据。
IEEE的一篇(付费)文章《基于OpenCV的视频加密 - IEEE会议出版物》称，他们将阿诺德变换应用于图像数据。它是一种换位密码，因此可以被破解。它的主要优势似乎是它使内容在常规播放中难以理解，并且它保留了对视频编解码器至关重要的图像特征(照明、帧差异)，并且不需要确切的密文进行解密，因此它不会因有损压缩而损坏到无法修复。