Need to grep /etc/hosts with a known hostname, and then capture the ip address for the hostname from /etc/hosts

Question

Need to grep /etc/hosts with a known hostname, and then capture the ip address for the hostname from /etc/hosts.

I am not a programmer, and don't know how to go about this. I have very limited experience with regex, but think that this might work somehow. I am not using DNS, just managing with /etc/hosts file.

I need to grep the /etc/hosts file with the known hostname, and then capture the IP address for the hosts entry. The host file is standard format:

Please help!

UPDATE:

# Maintenance Network

192.168.80.192  testsrv01-maint
192.168.80.193  testsrv02-maint
192.168.80.194  testsrv03-maint

# Lights Out Network

192.168.120.192  testsrv01-ilo
192.168.120.193  testsrv02-ilo
192.168.120.194  testsrv03-ilo

# Primary Data Network

192.168.150.192  testsrv01-pri
192.168.150.193  testsrv02-pri
192.168.150.194  testsrv03-pri

# Secondary Data Network

192.168.200.192  testsrv01-sec
192.168.200.193  testsrv02-sec
192.168.200.194  testsrv03-sec

I need to be able to capture the ip address and full host name entry for every machine into a variable that I can use. For instance run through the file looking to match " testsrv01* ", and capture all of the ip addresses and name for that search. Then same for " testsrv02* " , and so on.

Solution 1

Simple answer

ip=$(grep 'www.example.com' /etc/hosts | awk '{print $1}')

---

Better answer The simple answer returns all matching IP, even those on comment lines. You probably only want the first non-comment match, in which case just use awk outright:

ip=$(awk '/^[[:space:]]*($|#)/{next} /www.example.com/{print $1; exit}' /etc/hosts)

---

One other thing If you, at some point, care to resolve www.example.com whether your system is configured to use hosts, dns, etc, then consider the lesser known getent command:

ip=$(getent hosts 'www.example.com' | awk '{print $1}')

---

Edit in response to update

$ cat script.sh
#!/bin/bash

host_to_find=${1:?"Please tell me what host you want to find"}

while read ip host; do
    echo "IP=[$ip] and host=[$host]"
done < <(awk "/^[[:space:]]*($|#)/{next} /$host_to_find/{print \$1 \" \" \$2}" /etc/hosts)

$ ./script.sh testsrv01
IP=[192.168.80.192] and host=[testsrv01-maint]
IP=[192.168.120.192] and host=[testsrv01-ilo]
IP=[192.168.150.192] and host=[testsrv01-pri]
IP=[192.168.200.192] and host=[testsrv01-sec]

Solution 2

You could use grep with a Perl regex to output the IP of your target hostname.

grep -oP '^\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}(?=.*hostname)' /etc/hosts

Explanation:

• ^ finds the start of a line
• \d{1,3} finds one through three digits
• \. finds a dot
• (?=something) finds something but doesn't include it in the match ("zero-width positive look-ahead assertion")
• . without a preceding backslash finds any character
• * repeats the preceding expression (in this case "any character") zero or more times

In other words, this will find a series of four one through three-digit numbers separated by dots, and print them (grep -o) if they are followed by any string and then hostname), all on this on one line.

Comments

[tripleee]

grep|awk is almost always useless. awk '/www.example.com/{print $1}' /etc/hosts

[tripleee]

Might want to add $1 ~ /^#/ { next } as the first line of the Awk script to skip any comments.

[bishop]

@tripleee: Oh, I know, but OP so deliberately asked to grep. :) Good call on the comments, too.

[bishop]

+1 for using -o and positive look-ahead. I'd recommend -m 1 to limit to the first match. Also, note that this won't work for IPv6 addresses which might be in /etc/hosts (though the OP didn't state he needed v6).