85 lines
2.7 KiB
Python
85 lines
2.7 KiB
Python
# -*- coding: utf-8 -*-
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"""
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Created on Sun May 19 11:20:05 2019
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@author: Administrator
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"""
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from PIL import Image
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def plus(string):
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#Python zfill() 方法返回指定长度的字符串，原字符串右对齐，前面填充0。
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return string.zfill(8)
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def get_key(strr):
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#获取要隐藏的文件内容
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with open(strr,"rb") as f:
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s = f.read()
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string=""
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for i in range(len(s)):
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#逐个字节将要隐藏的文件内容转换为二进制，并拼接起来
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#1.先用ord()函数将s的内容逐个转换为ascii码
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#2.使用bin()函数将十进制的ascii码转换为二进制
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#3.由于bin()函数转换二进制后，二进制字符串的前面会有"0b"来表示这个字符串是二进制形式，所以用replace()替换为空
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#4.又由于ascii码转换二进制后是七位，而正常情况下每个字符由8位二进制组成，所以使用自定义函数plus将其填充为8位
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string=string+""+plus(bin(s[i]).replace('0b',''))
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#print(string)
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return string
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def mod(x,y):
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return x%y
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#str1为载体图片路径，str2为隐写文件，str3为加密图片保存的路径
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def func(str1,str2,str3):
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im = Image.open(str1)
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#获取图片的宽和高
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width,height= im.size[0],im.size[1]
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print("width:"+str(width))
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print("height:"+str(height))
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count = 0
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#获取需要隐藏的信息
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key = get_key(str2)
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keylen = len(key)
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for h in range(height):
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for w in range(width):
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pixel = im.getpixel((w,h))
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a=pixel[0]
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b=pixel[1]
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c=pixel[2]
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if count == keylen:
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break
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#下面的操作是将信息隐藏进去
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#分别将每个像素点的RGB值余2，这样可以去掉最低位的值
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#再从需要隐藏的信息中取出一位，转换为整型
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#两值相加，就把信息隐藏起来了
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a= a-mod(a,2)+int(key[count])
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count+=1
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if count == keylen:
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im.putpixel((w,h),(a,b,c))
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break
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b =b-mod(b,2)+int(key[count])
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count+=1
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if count == keylen:
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im.putpixel((w,h),(a,b,c))
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break
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c= c-mod(c,2)+int(key[count])
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count+=1
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if count == keylen:
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im.putpixel((w,h),(a,b,c))
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break
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if count % 3 == 0:
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im.putpixel((w,h),(a,b,c))
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im.save(str3)
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def main():
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#原图
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old = "old.png"
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#处理后输出的图片路径
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new = "new.png"
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#需要隐藏的信息
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enc = "flag.txt"
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func(old,enc,new)
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if __name__=='__main__':
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main() |