[英]RSA/ECB/PKCS1Padding iOS encryption
我目前陷入一个涉及iOS加密的问题。
我的客户给了我公钥,
"-----BEGIN PUBLIC KEY-----
xxxx
-----END PUBLIC KEY-----"
需要使用的填充策略是RSA / ECB / PKCS1Padding。 使用android似乎很简单
cipher = Cipher.getInstance("RSA/ECB/PKCS1Padding");
cipher.init(Cipher.ENCRYPT_MODE, publicKey);
encryptedBytes = cipher.doFinal(plain.getBytes());
return encryptedBytes;
我在iOS中看不到任何直接方法可以做到这一点。 像Commoncrypto一样使用的任何常见吊舱都不允许我强制执行PKCS1填充方案。 作为一个没有RSA和加密功能的经验不足的人,如果您能帮助我了解如何实现这一目标并指导我完成这一过程,将不胜感激。
使用标准的安全框架- SecKeyEncrypt与kSecPaddingPKCS1
参数
我的问题使用非填充解决:
kSecPaddingNone
-(SecKeyRef)getPublicKeyForEncryption
{
NSString *thePath = [MAuthBundle pathForResource:@"certificate" ofType:@"der"];
//2. Get the contents of the certificate and load to NSData
NSData *certData = [[NSData alloc]
initWithContentsOfFile:thePath];
//3. Get CFDataRef of the certificate data
CFDataRef myCertData = (__bridge CFDataRef)certData;
SecCertificateRef myCert;
SecKeyRef aPublicKeyRef = NULL;
SecTrustRef aTrustRef = NULL;
//4. Create certificate with the data
myCert = SecCertificateCreateWithData(NULL, myCertData);
//5. Returns a policy object for the default X.509 policy
SecPolicyRef aPolicyRef = SecPolicyCreateBasicX509();
if (aPolicyRef) {
if (SecTrustCreateWithCertificates((CFTypeRef)myCert, aPolicyRef, &aTrustRef) == noErr) {
SecTrustResultType result;
if (SecTrustEvaluate(aTrustRef, &result) == noErr) {
//6. Returns the public key for a leaf certificate after it has been evaluated.
aPublicKeyRef = SecTrustCopyPublicKey(aTrustRef);
}
}
}
return aPublicKeyRef;
}
-(NSString*) rsaEncryptString:(NSString*) string
{
SecKeyRef publicKey = [self getPublicKeyForEncryption];
NSData* strData = [string dataUsingEncoding:NSUTF8StringEncoding];
CFErrorRef err ;
NSData * data = CFBridgingRelease(SecKeyCreateEncryptedData(publicKey, kSecKeyAlgorithmRSAEncryptionPKCS1, ( __bridge CFDataRef)strData, &err));
NSString *base64EncodedString = [data base64EncodedStringWithOptions:0];
return base64EncodedString;
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.