Linux 获取所有网络接口名称
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Linux getting all network interface names
提问by tez
I need to collect all the interface names, even the ones that aren't up at the moment. Like ifconfig -a.
我需要收集所有的接口名称，即使是那些目前没有启动的名称。喜欢ifconfig -a。
getifaddrs()is iterating through same interface name multiple times. How can I collect all the interface names just once using getifaddrs()?
getifaddrs()多次迭代相同的接口名称。如何仅使用一次就收集所有接口名称getifaddrs()？
采纳答案by brm
You could check which entries from getifaddrs belong to the AF_PACKET family. On my system that seems to list all interfaces:
您可以检查 getifaddrs 中的哪些条目属于 AF_PACKET 系列。在我的系统上，似乎列出了所有接口：
struct ifaddrs *addrs,*tmp;
getifaddrs(&addrs);
tmp = addrs;
while (tmp)
{
if (tmp->ifa_addr && tmp->ifa_addr->sa_family == AF_PACKET)
printf("%s\n", tmp->ifa_name);
tmp = tmp->ifa_next;
}
freeifaddrs(addrs);
回答by aet
You are on the right track (it is getifaddrs). It returns each interface once per family, so you get eth0 for ipv4 and one for ipv6. If you just want each interface once you will have to uniq the output yourself.
你在正确的轨道上（它是 getifaddrs）。它为每个系列返回每个接口一次，因此您将获得用于 ipv4 的 eth0 和用于 ipv6 的 eth0。如果您只想要每个接口一次，您将不得不自己 uniq 输出。
回答by ldav1s
It seems that ifconfig -aonly lists active interfaces (at least on Fedora 19). I know I have at least one more network card that I'm not seeing. Anyway, I get the same list as:
似乎ifconfig -a只列出活动接口（至少在 Fedora 19 上）。我知道我至少还有一张我没有看到的网卡。无论如何，我得到与以下相同的列表：
ls -1 /sys/class/net
Which could easily be done programatically.
这可以很容易地以编程方式完成。
回答by Makaro
I thing this show you all interface, at least for me
我认为这会向您展示所有界面，至少对我而言
ip link show
ip链接显示
ls -1 /sys/class/net
only show interface name
只显示接口名称
lo
p4p1
回答by Mike Q
I need the main device that is being used by the system (assuming there is only one) such as eth0 wlan0 or whatever RHEL7 is trying to do...
我需要系统正在使用的主要设备（假设只有一个），例如 eth0 wlan0 或 RHEL7 试图做的任何事情......
The best I hacked together was this:
我一起破解的最好的是：
#!/bin/bash
# -- Get me the interface for the main ip on system
for each in $(ls -1 /sys/class/net) ;do
result=$(ip addr show $each | awk ' == "inet" {gsub(/\/.*$/, "", ); print }' | grep "$(hostname -I | cut -d' ' -f1)")
if [ ! -z "${result// }" ] && [ -d /sys/class/net/${each// } ] ;then
echo "Device: $each IP: $result"
break;
fi
done
Sample output:
示例输出：
./maineth.sh
Device: enp0s25 IP: 192.168.1.6
回答by Hugues
getifaddrs() will only return your interfaces addresses, not the interfaces themselves.
getifaddrs() 只会返回您的接口地址，而不是接口本身。
What if any of your interface has no address, or no address of the requested family, as suggested with the 'AF_PACKET' one ?
如果您的任何接口没有地址，或者没有所请求系列的地址，如 'AF_PACKET' 所建议的那样怎么办？
Here, an example where I've got a tunnel interface (with an OpenVPN connexion), and where I'm listing all entries from getifaddrs() for each of my network interfaces:
在这里，我有一个隧道接口（带有 OpenVPN 连接）的示例，我列出了每个网络接口的 getifaddrs() 中的所有条目：
[0] 1: lo address family: 17 (AF_PACKET) b4:11:00:00:00:01
address family: 2 (AF_INET) address: <127.0.0.1>
address family: 10 (AF_INET6) address: <::1>
[...]
[5] 10: tun0 address family: 2 (AF_INET) address: <172.16.0.14>
[EOF]
Bam. No AF_PACKET on the "tun0" interface, but it DOES exist on the system.
砰。“tun0”接口上没有 AF_PACKET，但它确实存在于系统中。
You should, instead, use if_nameindex() syscall, which does exactly what you want. In other words, with no arguments, it returns a list of all interfaces on your system:
相反，您应该使用 if_nameindex() 系统调用，它完全符合您的要求。换句话说，没有参数，它返回系统上所有接口的列表：
#include <net/if.h>
#include <stdio.h>
int main (void)
{
struct if_nameindex *if_nidxs, *intf;
if_nidxs = if_nameindex();
if ( if_nidxs != NULL )
{
for (intf = if_nidxs; intf->if_index != 0 || intf->if_name != NULL; intf++)
{
printf("%s\n", intf->if_name);
}
if_freenameindex(if_nidxs);
}
return 0;
}
And voilà.
瞧。
