通过 URL 获取 IP 地址 [英] Get IP-address by URL

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本文介绍了通过 URL 获取 IP 地址的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

这行得通

target.sin_addr.s_addr = inet_addr("127.0.0.1");

但我想从网站 URL 中输入 IP

but I want to put the IP from a website URL

我试过了

const char host[] = "http://www.google.com/";
struct hostent *host_ip;
host_ip = gethostbyaddr(host, strlen(host), 0);

我在使用 gethostbyaddr(); 之前做了 WSAStartup;

I of corse did WSAStartup before I used gethostbyaddr();

我试过了

target.sin_addr.s_addr = inet_addr(host_ip);

我也尝试了一些类似的方法,但它不起作用.有人可以告诉我如何正确地做到这一点.

I've tried a few simmilar ones too but it isn't working. Could someone show me how to do this correctly.

谢谢!

当我这样做时

host_ip = gethostbyaddr((char *)&host, strlen(host), 0);
std::cout << host_ip->h_addr;

它给了我

httpa104-116-116-112.deploy.static.akamaitechnologies.com

推荐答案

inet_addr() 接受 IPv4 地址字符串作为输入并返回该地址的二进制表示.在这种情况下,这不是您想要的,因为您没有 IP 地址,而是有一个主机名.

inet_addr() accepts an IPv4 address string as input and returns the binary representation of that address. That is not what you want in this situation, since you don't have a IP address, you have a hostname instead.

您使用 gethostby...() 走在正确的轨道上,但您需要使用 gethostbyname()(按主机名查找)而不是 gethostbyaddr()(按 IP 地址查找)1.而且您不能将完整的 URL 传递给其中任何一个.gethostbyname() 只接受一个主机名作为输入,所以你需要解析 URL 并提取它的主机名,然后你可以这样做:

You are on the right track by using gethostby...(), but you need to use gethostbyname() (lookup by hostname) instead of gethostbyaddr() (lookup by IP address) 1. And you cannot pass a full URL to either of them. gethostbyname() takes only a hostname as input, so you need to parse the URL and extract its hostname, and then you can do something like this:

const char host[] = ...; // an IP address or a hostname, like "www.google.com" by itself
target.sin_addr.s_addr = inet_addr(host);
if (target.sin_addr.s_addr == INADDR_NONE)
{
    struct hostent *phost = gethostbyname(host);
    if ((phost) && (phost->h_addrtype == AF_INET))
        target.sin_addr = *(in_addr*)(phost->h_addr);
    ...
}
else
    ...

1 顺便说一句,gethostby...() 函数已弃用,请使用 getaddrinfo()getnameinfo() 代替.

1 BTW, the gethostby...() functions are deprecated, use getaddrinfo() and getnameinfo() instead.

const char host[] = ...; // an IP address or a hostname, like "www.google.com" by itself

addrinfo hints = {0};
hints.ai_flags = AI_NUMERICHOST;
hints.ai_family = AF_INET;
hints.ai_socktype = SOCK_STREAM;
hints.ai_protocol = IPPROTO_TCP;

addrinfo *addr = NULL;

int ret = getaddrinfo(host, NULL, &hints, &addr);
if (ret == EAI_NONAME) // not an IP, retry as a hostname
{
    hints.ai_flags = 0;
    ret = getaddrinfo(host, NULL, &hints, &addr);
}
if (ret == 0)
{
    target = *(sockaddr_in*)(addr->ai_addr);
    freeaddrinfo(addr);
    ...
}
else
    ...

这篇关于通过 URL 获取 IP 地址的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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