python - 如何解决UnicodeDecodeError : 'utf-8' codec can't decode byte 0xa1 in position 3: invalid start byte?

Question

print(cipher.decrypt(pad(base64.b64decode(encrypted_string),16)).decode('utf-8'))UnicodeDecodeError:“utf-8”编解码器无法解码位置 3 中的字节 0xa1:起始字节无效

数据流:JavaScript:

1. 输入 = {"名称":"abcd","密码":"abcd"}
2. JSON.stringify(输入)
3. CryptoJS 使用 CBC 模式加密 AES
4. Base64 编码
5. AJAX 到 Django

Python3/Django/PyCryptodome

1. base64 解码
2. Crypto.pad 16B
3. Crypto.cipher.decrypt [成功]

收到的输出是

b'\xf5R\x0ck_\x90\xca\xcc\n\xe6S:\xed\xf1\x0f\x14\x1f\x8c_\x00\xd8m\x82\x96\xa3\x19K\xca\xc2\xfe\x14\x80\x8bD8\xcc\xdb\x91~\xa2~\xbf\xe5U\xe4\x12E`\xb5\r\xd8t\xbd=\xc0\x862\x837W6|\x07i\x8b\xb1"\xe9\xac}\xef\xf0\xf5\xa8\x0e\xf5z\xdc\xb0\xb5'

如何解码该字符串以获得输出。

<小时/>

代码:

function toWordArray(str){
    return CryptoJS.enc.Utf8.parse(str);
}

function toString(words){
    return CryptoJS.enc.Utf8.stringify(words);
}

function toBase64String(words){
    return CryptoJS.enc.Base64.stringify(words);
}

function encrypt(input, key){
    console.log("Input: " + input);  
    var PROTOCOL_AES256 = 2;
    var secret_key = CryptoJS.SHA256(key);
    var header = toWordArray("AMAZON" + String.fromCharCode(PROTOCOL_AES256));
    var iv = CryptoJS.lib.WordArray.random(16);
    var body = CryptoJS.AES.encrypt(input, secret_key, {mode:CryptoJS.mode.CBC,iv: iv,padding:CryptoJS.pad.Pkcs7});

    // construct the packet
    // HEADER + IV + BODY
    header.concat(iv);
    header.concat(body.ciphertext);

    console.log("Bytes before Base64 encoding: " + header);
    // encode in base64
    return toBase64String(header);
}



$("#submitid").click(function(event) {
            event.preventDefault()
            // var encrypted = CryptoJS.AES(...);
            // var encrypted = CryptoJS.SHA256(...);
            var data = {};
            data["name"] = $("#nameid").val();
            data["password"] = $("#passwordid").val();
            var json_payload = JSON.stringify(data);

            var payload = encrypt(json_payload, key);
            console.log("Payload: " + payload);

            $.ajax({
                type: 'GET',
                url: "{% url 'Shenzen:actsignin' %}",
                data: {encrypted_string: payload},
                success: function(data,status,xhr){
                    console.log("AES sucessful withs status: "+status);
                },
                error: function(xhr,status,e) {
                    console.log("AES failed.");
                },
                async: true,
                datatype: 'json'
            })
        })

    },
    error: function(xhr,status,e) {
        console.log("error");
    },
    async: true,
    datatype: 'json'
});

Python3

def actsignin(request):
    global key
    global prime
    global base_num

    encrypted_string = request.GET['encrypted_string']

    print("Encrypted string decoded: ",base64.b64decode(encrypted_string).hex())
    print("----")
    protocol_aes = str(2)
    header_skip = 'AMAZON'+protocol_aes

    sha256_key = SHA256.new(data=bytes(key))
    cipher = AES.new(sha256_key.digest(),AES.MODE_CBC)
    print(cipher.decrypt(pad(base64.b64decode(encrypted_string),16)))
    print('in signin')
    # student = +Student.objects.get(name = name)
    # salt = student.salt
    # key = hashlib.pbkdf2_hmac('sha256',password.encode('utf-8'),salt,100000)
    # # Hash the password here to check
    # password_to_check_hashed = key
    # print(key)
    # print(student.password)
    # if student.password == key:
    return render(request,"Shenzen/display.html",{'student':'student'})
    # else:
    #     return render(request,"Shenzen/display.html",{'status':'failed'})

Answer 1

base64 decode
Crypto.pad 16B
Crypto.cipher.decrypt [successful]

实现者

cipher = AES.new(sha256_key.digest(),AES.MODE_CBC)
print(cipher.decrypt(pad(base64.b64decode(encrypted_string),16)))

这没有道理。顺序是先填充然后加密，然后解密然后取消填充。取消填充通常由 decrypt 方法本身执行。

解密可能在取消填充期间失败，但如果您跳过取消填充，那么您只会收到垃圾。如果你填充密文，那么最后一部分肯定是垃圾。解码字符只是您注意到结果是垃圾的第一个地方。

还要注意，Python 代码默认使用全零 IV，而 CryptoJS 显然使用全随机 IV，并以密文为前缀。

如果没有身份验证标签(由 MAC 或经过身份验证的密码生成)，则始终可以解密密文。尽管人们普遍认为，分组密码解密永远不会失败。