Grab a decimal between 2 strings regex

Question

I'm still pretty new to regex and I have this string:
Client Version: openshift-clients-4.3.0-201910250623-88-g6a937dfe Server Version: 4.3.0 Kubernetes Version: v1.16.2:q
And I wanted to grab 4.3.0, which is between Server Version:  and Kubernetes
I thought I could do something like: (d*.?d+.d), which grabs every decimal number that has a length of 3, but I just want it to return a single number.
So I tried (Server Version: )+(d*.?d+.d) but this gives me 2 capture groups and I want to store the number 4.3.0 in a variable
I'm playing around here
Wanted to use this regex with grep or sed
Any help is appreciated!

The fourth bird · Answer

Use an outer capturing group, and repeat the dot and the digits inside the group using another group if you want to use sed.
Server Version: (d+(.d+)+) Kubernetes

Regex demo | Sed demo
If you can use grep -P you could only get the match.
Server Version: Kd+(?:.d+)+(?= Kubernetes)

Regex demo | Bash demo
For example
echo "Client Version: openshift-clients-4.3.0-201910250623-88-g6a937dfe Server Version: 4.3.0 Kubernetes Version: v1.16.2:q" | grep -oP 'Server Version: Kd+(?:.d+)+(?= Kubernetes)'

Output
4.3.0

Grab a decimal between 2 strings regex

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