找到两者并更新

时间:2019-04-27 08:41:42

标签: c# oop

我正在尝试比较两组并进行更新

1 个答案:

答案 0 :(得分:2)

当您要更新extension String { func aesEncrypt(key:String, iv:String, options:Int = kCCOptionPKCS7Padding) -> String? { if let keyData = key.data(using: String.Encoding.utf8), let data = self.data(using: String.Encoding.utf8), let cryptData = NSMutableData(length: Int((data.count)) + kCCBlockSizeAES128) { let keyLength = size_t(kCCKeySizeAES128) let operation: CCOperation = UInt32(kCCEncrypt) let algoritm: CCAlgorithm = UInt32(kCCAlgorithmAES128) let options: CCOptions = UInt32(options) var numBytesEncrypted :size_t = 0 let cryptStatus = CCCrypt(operation, algoritm, options, (keyData as NSData).bytes, keyLength, iv, (data as NSData).bytes, data.count, cryptData.mutableBytes, cryptData.length, &numBytesEncrypted) if UInt32(cryptStatus) == UInt32(kCCSuccess) { cryptData.length = Int(numBytesEncrypted) let base64cryptString = cryptData.base64EncodedString(options: .lineLength64Characters) return base64cryptString } else { return nil } } return nil } func aesDecrypt(key:String, iv:String, options:Int = kCCOptionPKCS7Padding) -> String? { if let keyData = key.data(using: String.Encoding.utf8), let data = NSData(base64Encoded: self, options: .ignoreUnknownCharacters), let cryptData = NSMutableData(length: Int((data.length)) + kCCBlockSizeAES128) { let keyLength = size_t(kCCKeySizeAES128) let operation: CCOperation = UInt32(kCCDecrypt) let algoritm: CCAlgorithm = UInt32(kCCAlgorithmAES128) let options: CCOptions = UInt32(options) var numBytesEncrypted :size_t = 0 let cryptStatus = CCCrypt(operation, algoritm, options, (keyData as NSData).bytes, keyLength, iv, data.bytes, data.length, cryptData.mutableBytes, cryptData.length, &numBytesEncrypted) if UInt32(cryptStatus) == UInt32(kCCSuccess) { cryptData.length = Int(numBytesEncrypted) let unencryptedMessage = String(data: cryptData as Data, encoding:String.Encoding.utf8) return unencryptedMessage } else { return nil } } return nil } } 时,最好的做法是普通的listB循环。 Linq不是用于更新,而是用于查询。另外,考虑将foreach作为字典存储,以便在listA中进行访问:

O(1)