来自NSNetService的Swift IP地址

我能够在我的Swift项目中使用Objective-C代码从NSNetService获取IP地址。有没有可以做同样的Swift代码（以避免有一个桥头）？否则，我会保持现在的样子 - 但希望它可以在Swift中完成。来自NSNetService的Swift IP地址

-(NSString*)IPAddressesFromData:(NSNetService *)service { 
    for (NSData *address in [service addresses]) { 
     struct sockaddr_in *socketAddress = (struct sockaddr_in *) [address bytes]; 
     //NSLog(@"Service name: %@ , ip: %s , port %i", [service name], inet_ntoa(socketAddress->sin_addr), [service port]); 
     NSString *retString = [NSString stringWithFormat:@"%s", inet_ntoa(socketAddress->sin_addr)]; 
     return retString; 
    } 
    return @"Unknown"; 
}

更新代码的工作原理：

func netServiceDidResolveAddress(sender: NSNetService) { 
    let theAddress = sender.addresses!.first! as NSData 
    var hostname = [CChar](count: Int(NI_MAXHOST), repeatedValue: 0) 
    if getnameinfo(UnsafePointer(theAddress.bytes), socklen_t(theAddress.length), 
        &hostname, socklen_t(hostname.count), nil, 0, NI_NUMERICHOST) == 0 { 
     if let numAddress = String.fromCString(hostname) { 
      print("Resolved IP address: \(numAddress)") 
     } 
    } 
}

来源

2016-06-16 Radagast the Brown

看一看http://stackoverflow.com/questions/25890533/how-can-i -get-a-real-ip-address-from-dns-query-in-swift查看如何从sockaddr结构获取IP字符串。 –

SHazaam！非常感谢 - 我将用我更新的代码编辑我的原始问题。 –

您可以使用：

func netServiceDidResolveAddress(_ sender: NetService) { 

    var hostname = [CChar](repeating: 0, count: Int(NI_MAXHOST)) 
    guard let data = sender.addresses?.first else { return } 
    do { 
     try data.withUnsafeBytes { (pointer:UnsafePointer<sockaddr>) -> Void in 
      guard getnameinfo(pointer, socklen_t(data.count), &hostname, socklen_t(hostname.count), nil, 0, NI_NUMERICHOST) == 0 else { 
       throw NSError(domain: "domain", code: 0, userInfo: ["error":"unable to get ip address"]) 
      } 
     } 
    } catch { 
     print(error) 
     return 
    } 
    let address = String(cString:hostname) 
    print("\(sender.name) IP:\(address)") 

}

来源

2017-08-04 05:29:31 LaiBit

来自NSNetService的Swift IP地址

回答

相关问题