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Better IPv4 detection (#278)

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On some systems like Hetzner VM cloud i have a Point-to-Point interface so i have a peer address on the same line as my public IPv4 (look at peer here : https://linux.die.net/man/8/ip )

An example of `ip a` with peer is : 
```
2: eth0: <BROADCAST,MULTICAST,UP,LOWER_UP> mtu 1500 qdisc pfifo_fast state UP group default qlen 1000
    link/ether 96:00:00:a2:88:c2 brd ff:ff:ff:ff:ff:ff
    altname enp0s3
    inet XX.XX.XX.XX peer XX.XX.XX.XX/32 brd XX.XX.XX.XX scope global eth0
       valid_lft forever preferred_lft forever
    inet6 fe80::9400:ff:fea2:88c2/64 scope link 
       valid_lft forever preferred_lft forever
```

With a peer, the output of the command line 74 is : `XX.XX.XX.XX peer XX.XX.XX.XX`

I just modify this line with awk to print only the first field which is always the IPv4.
I think it's correct and it's work like a charm when there is a peer or not now. But tell me if it's not good for you :)

Thanks for your work !
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Ravinou 2021-10-15 18:13:06 +02:00 committed by GitHub
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commit e05e633014
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wireguard-install.sh
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@ -71,7 +71,7 @@ function installQuestions() {
echo ""
# Detect public IPv4 or IPv6 address and pre-fill for the user
SERVER_PUB_IP=$(ip -4 addr | sed -ne 's|^.* inet \([^/]*\)/.* scope global.*$|\1|p' | head -1)
SERVER_PUB_IP=$(ip -4 addr | sed -ne 's|^.* inet \([^/]*\)/.* scope global.*$|\1|p' | awk '{print $1}' | head -1)
if [[ -z ${SERVER_PUB_IP} ]]; then
# Detect public IPv6 address
SERVER_PUB_IP=$(ip -6 addr | sed -ne 's|^.* inet6 \([^/]*\)/.* scope global.*$|\1|p' | head -1)