Get IPv4 out of a Ruby string that contains both IPv4 and IPv6?

I have a string from the X-Forwarded-For header that contains both IPv4 and IPv6 addresses.

X-Forwarded-For

I need to pull just the IPv4 address from the string.

It's comma-separated, but the order of them changes so I can't just split and pull the second item.

Example: header = 2600:1740:8540:cff9:1c50:617:c9c5:63f7, 165.154.107.112

header = 2600:1740:8540:cff9:1c50:617:c9c5:63f7, 165.154.107.112

I ultimately just want 165.154.107.112.

165.154.107.112

I'm using Ruby 2.5.1 (and this happens to be inside a Rails 5.2.0 app, for what it's worth).

2 Answers
2

Assuming your header is always as you've posted:

require 'ipaddr'

header = "165.154.107.112, 2600:1740:8540:cff9:1c50:617:c9c5:63f7"

ip = header.split(', ').select ip if IPAddr.new(ip).ipv4? .pop

# => "165.154.107.112"


header = "2600:1740:8540:cff9:1c50:617:c9c5:63f7, 165.154.107.112, 166.155.108.113"

header.split(/s?,s?/).find IPAddr.new(s).ipv4?
#=> "165.154.107.112"


or

header.split(/,s+/).select IPAddr.new(s).ipv4?
#=> ["165.154.107.112", "166.155.108.113"]


See IPAddr::new and IPAddr#ipv4?.

If "header = " is part of the string str, replace header.split with str[/d.+/].split.

"header = "

str

header.split

str[/d.+/].split

If the string may contain text that is not a valid IP address, you could write the following.

header.split(/s?,s?/).find (IPAddr.new(s) rescue nil)&.ipv4?


IPAddr.new('cat'), for example, raises the exception IPAddr::InvalidAddressError (invalid address). & is Ruby's safe navigation operator, which made it's debut in v2.3.

IPAddr.new('cat')

IPAddr::InvalidAddressError (invalid address)

&

require 'ipaddr'

ipaddr

require 'ipaddr'

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