使用这个公钥是否能解密此加密信息？

使用私钥 rsa 加密内容：

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

对应的公钥：

如果解密出的结果像是下面这样，应该是对的：

{"2658844581501957149":{"file_name":"wwdc2023-10117_hd.mp4","file_size":"100597894","pick_code":"b12x4iiejvkkq83a","url":{"url":"http:\/\/xxx.xxx.com\/xxx\/wwdc2023-10117_hd.mp4?t=1690607362&u=101248331&s=1048576&d=658578066-b12x4iiejvkkq83a-0&c=2&f=1&k=ddb3d920081771b2f40ee75164c5bb1b&us=52428800&uc=10&v=1","client":2,"desc":null,"isp":null,"oss_id":"fhnfile\/xxx","ooid":""}}}

注意由于加密内容数据比较大，需要分段解密。

不管用什么方法，python/go/java 还是网上的在线解密工具，只要能解密出来即可。

我用 swift 写了分段解密，结果解密失败，用网上在线的解密工具也是失败，不知道网上的工具是不是由于不支持分段解密才失败的。

下面是我用 Swift 写的方法：

import Security
func convertPublicKey(pemString: String) -> SecKey? {
    // 1. Remove headers, newlines and spaces
    let keyString = pemString
        .replacingOccurrences(of: "-----BEGIN PUBLIC KEY-----", with: "")
        .replacingOccurrences(of: "-----END PUBLIC KEY-----", with: "")
        .replacingOccurrences(of: "\n", with: "")
        .replacingOccurrences(of: " ", with: "")

    // 2. Convert base64 string to Data
    guard let keyData = Data(base64Encoded: keyString) else {
        return nil
    }

    // 3. Create key dictionary
    let keyDict: [CFString: Any] = [
        kSecAttrKeyType: kSecAttrKeyTypeRSA,
        kSecAttrKeyClass: kSecAttrKeyClassPublic,
        kSecAttrKeySizeInBits: NSNumber(value: 2048),
        kSecReturnPersistentRef: true
    ]

    // 4. Create SecKey from Data
    var error: Unmanaged<CFError>?
    guard let key = SecKeyCreateWithData(keyData as CFData, keyDict as CFDictionary, &error) else {
        // Handle the error here
        return nil
    }

    return key
}
let rsaKeyInSwift = convertPublicKey(pemString: rsaPublicKey)
func rsaDecrypt(data: Data) -> Data? {
    let blockSize = SecKeyGetBlockSize(rsaKeyInSwift!)
    var decryptedData = Data()
    for i in stride(from: 0, to: data.count, by: blockSize) {
        let endIndex = min(i + blockSize, data.count)
        let chunk = data[i..<endIndex]
        print(chunk)
        var error: Unmanaged<CFError>?
        guard let decryptedChunk = SecKeyCreateDecryptedData(
            rsaKeyInSwift!,
            .rsaEncryptionPKCS1,
            chunk as CFData,
            &error
        ) else {
            print("Decryption error:", error.debugDescription)
            return nil
        }
        decryptedData.append(decryptedChunk as Data)
    }

    return decryptedData
}

报错：

Decryption error: Optional(Swift.Unmanaged<__C.CFErrorRef>(_value: Error Domain=NSOSStatusErrorDomain Code=-50 "RSAdecrypt wrong input (err -27)" UserInfo={numberOfErrorsDeep=0, NSDescription=RSAdecrypt wrong input (err -27)}))

在网上找到了一个开源的项目，它用 go 语言自己实现了 rsa 解密算法，不清楚为什么它不用标准库来解密。我用标准库实现的（ Swift 版本）无法解密。所以不清楚是不是我的代码问题还是其他问题。

下面是 go 语言开源版本：

func rsaDecrypt(input []byte) []byte {
	output := make([]byte, 0)
	cipherSize, blockSize := len(input), rsaServerKey.Size()
	for offset := 0; offset < cipherSize; offset += blockSize {
		sliceSize := blockSize
		if offset+sliceSize > cipherSize {
			sliceSize = cipherSize - offset
		}

		n := big.NewInt(0).SetBytes(input[offset : offset+sliceSize])
		m := big.NewInt(0).Exp(n, big.NewInt(int64(rsaServerKey.E)), rsaServerKey.N)
		b := m.Bytes()
		index := bytes.IndexByte(b, '\x00')
		if index < 0 {
			return nil
		}
		output = append(output, b[index+1:]...)
	}
	return output
}

由于不熟悉 go 语言，大概看了下算法，也是计算 (n^e) mod N。

所以请大家用这个公钥解密上面的内容，能否解密出来？

sankooc

307 天前

const crypto = require("crypto")
const fs = require('fs')
const path = require('path')

const data = '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'
const publicKey = fs.readFileSync(path.resolve("./k.pub"), "utf8");
const raw = Buffer.from(data, "base64")
const len = raw.length
const mod = 128
const buffer = [];
for(let i = 0 ; i <= ~~(len/mod);i += 1){
const start = i * mod;
const end = Math.min(len, start + mod)
if(end == start) break;
const r1 = raw.subarray(start, end)
const rt = crypto.publicDecrypt({
key: publicKey,
}, r1)
buffer.push(rt)
}
const buf = Buffer.concat(buffer)
console.log(buf);
好像就是你要的结果你可以试试

tuwulin365

307 天前

N 是
8686980c0f5a24c4b9d43020cd2c22703ff3f450756529058b1cf88f09b8602136477198a6e2683149659bd122c33592fdb5ad47944ad1ea4d36c6b172aad6338c3bb6ac6227502d010993ac967d1aef00f0c8e038de2e4d3bc2ec368af2e9f10a6f1eda4f7262f136420c07c331b871bf139f74f3010e3c4fe57df3afb71683

密文是
825F44F727B969C55D9F932EB098F4A5ACBACBE8BAA839DAF034D23479243E306C104F43277E1C2A1AC288DB91588C58E3E0D09ABAD03CCCE26E00AF3FEDEE9D67A5C709B3311CE64904034E6F59FA80621C609CA965EDCF8EC7FDBD6EA02E13E995E44EBB6E6003959510E7FB6C4A3DC3BDAA9AB9812F9C53A48F4917113FF7
00B61AA72D0D052498D8DAF3AAB2829D6B9212B712E345224F5ECE739891551FF8A2EC9FEED6FA7C47AB1450D2E12666A116538380EA043C34D6E7493A3B92D2F0C6278992EF23F0B38FD6EDD73F39D2B5C6977C07BDFC26848587F2E991E4DA9A3C9F3F9032571A4E5E8C7A258D1376189E222F8F21BC96B0B2808D1E992DAE
5A91B44EA079017BE2007984BCD2BD8D9951C468D5FC7B0F1A4F9D1A9FC7B72D7FA9BB9C2D2FA216B1E0DBDD609ED35A72BEA0318224514944D83E2FF0C995C7248D698BDE71A815880749F5706577D32CE70A3ADC7C3EF77651D8F1FBDB76FD3E03A89366855CC1BD7FED12A065BFE75E01B93C0DA1B1E057015A6D5BE63DAA
314A80DF73952182CECE0145CDD305F0EA84BC03F0AB2835A478EDF623F35975958F3140D99D72568BB64F067A67D42AAE3686C8AE8A7E94FE876E35160095AD981B83C83F25653BC051D08E212325349366EE857B0790A5152E532FC0F0B4B459A6E1D1E650D2170F7110707F68063A599B7C2A23ADF64D1DB8D1921D11CFA6
00AB74F9AB955D7EC2970543097138A57D4191CC88AF5401663392F1E56CE057B2EC3D222E12AE6C345132A64E6BACAA654E2AEADEA13A99A237F8A29B51C700B366A4EB318BB5F16725AB178AFC183EB8AF5274C3CFD715D5D54560A7BBDBF6869CD6F2D93C56BCDA5B0EEEB44FC0655F5085C058FE250C83A85BBE12F8014D