Bash replace substring after first colon

Question

I am trying to build a connection string that requires pulling 3 IP addresses from another config file. When I get those values, I need to replace the port on each. I plan to replace each port using simple Bash find and replace ${string/pattern/replacement} but my problem is I'm stuck on the best way to parse the pattern out of the IP.

Here is what I have so far:

myFile.config:

ip.1=ip-ip-1-address:1234:5678
ip.2=ip-ip-2-address:1234:5678
ip.3=ip-ip-3-address:1234:5678

Copying some other simple process, I found I can pull the value of each IP like this:

IP1=`grep "ip.1=" /path/to/conf/myFile.config  | awk -F "=" '{print $2}'`

which gives me ip.1=ip-ip-1-address:1234:5678. However, I need to replace 1234:5678 with 6543 for example. I've been looking around and I found this awesome answer that detailed using Bash prefix substitution but that relies on knowing the parameter. for example, I would have to do it this way:

test=${ip1##ip-ip-1-address:}

which results in $test being 1234:5678. That's fine but maybe I don't know the IP address as the parameter, so I'm back to considering regex unless there's a way for me to use * as the parameter or something, but I have been unsuccessful so far. For regex, I have tried a bunch such as test=${ip1/(?<=:).*/}.

Maybe sed -i 's/:.*/:6543/' myFile.config? It will replace all text starting with the first : on each line of the file with :6543. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jul 29, 2020 at 19:30
Can you elaborate @WiktorStribiżew. Is this instead of my grep? — johnny_mac
– johnny_mac, Commented Jul 29, 2020 at 19:32
Yes, instead of grep, since you say you need to replace. grep is used to extract. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jul 29, 2020 at 19:33
while IFS="=:" read -r var ip n1 n2; do echo "$var $ip $n1 $n2"; done < file? — Cyrus
– Cyrus, Commented Jul 29, 2020 at 19:54

Wiktor Stribiżew · Accepted Answer · 2020-07-29 19:53:05Z

1

Note that ${ip1/(?<=:).*/} you tried is an example of string manipulation syntax that does not support regex, only specific patterns.

You seem to want

x='ip.1=ip-ip-1-address:1234:5678'
echo "${x%%:*}:6543" # => ip.1=ip-ip-1-address:6543

The ${x%%:*} takes the value of x and removes all chars from the end till the first : including it. :6543 is added to the result of this manipulation using "${x%%:*}:6543".

To extract that value, you may also use

awk '/^ip\.1=/{sub("^[^:]+:", "");print}' myFile.config

The awk command finds lines starting with ip.1= and then removes all text from the start till the first colon including the colon and only prints these values.

edited Jul 29, 2020 at 19:53

answered Jul 29, 2020 at 19:36

Wiktor Stribiżew

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2 Comments

johnny_mac Over a year ago

Thanks for the detailed answer but I am not looking to modify the input file. I need to take values from it and then replace the port for another dependency, hence my attempt at grep and replace in my new variable.

Wiktor Stribiżew Over a year ago

@johnny_mac You may use sed to modify strings, too. ip="$(sed 's/:.*/:6543' file)". I re-wrote the answer with a string manipulation example.

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