And*_*dan 1 ip external objective-c ip-address
在我的应用程序中,我需要获取一个NSString对象值等于用户公共/互联网IP地址的值.我试图解决这个问题,但两者都返回本地IP地址而不是公开.以下是我的两种方法.一个更精确,并始终返回数组中的正确项.另一个没有.(因为只选一个随机索引)...
- (NSString *)getPublicIP {
NSHost *publicIP = [[[NSHost currentHost] addresses] objectAtIndex:0];
return publicIP;
}
Run Code Online (Sandbox Code Playgroud)
其他更精确:(但不获得公共IP)
//start get ip
- (NSString *)getIPWithNSHost {
NSArray *addresses = [[NSHost currentHost] addresses];
NSString *stringAddress;
for (NSString *anAddress in addresses) {
if (![anAddress hasPrefix:@"127"] && [[anAddress componentsSeparatedByString:@"."] count] == 4) {
stringAddress = anAddress;
break;
}
else {
stringAddress = @"IPv4 address not available" ;
}
//NSLog(stringAddress);
}
NSLog (@"getIPWithNSHost: stringAddress = %@ ",stringAddress);
stringAddress = (@"getIPWithNSHost: stringAddress = %@ ",stringAddress);
return stringAddress;
}
Run Code Online (Sandbox Code Playgroud)
无论哪种方式,我只需要一种方法来获取外部/公共/互联网IP地址.(只是为了澄清外部/公共/互联网IP是可以从whatsmyip.org检索的)
这是iOS友好版的joshbillions答案:
+(void)getPublicIP:(void(^)(NSString *))block {
NSURL *url = [NSURL URLWithString:@"http://checkip.dyndns.org"];
NSURLRequest *request = [NSURLRequest requestWithURL:url];
[[[NSURLSession sharedSession]dataTaskWithRequest:request completionHandler:^(NSData *data, NSURLResponse *response, NSError *error) {
if (error) {
// consider handling error
} else {
NSString *html = [[NSString alloc]initWithData:data encoding:NSUTF8StringEncoding];
NSCharacterSet *numbers = [NSCharacterSet characterSetWithCharactersInString:@"0123456789."];
NSString *ipAddr = [[html componentsSeparatedByCharactersInSet:numbers.invertedSet]componentsJoinedByString:@""];
if (block) {
block(ipAddr);
}
}
}]resume];
}
Run Code Online (Sandbox Code Playgroud)