如何在 bash 中的特定单词之前打印字符串？

Question

I want to learn how to extract only the Kernel Version from this specific output:

3.10.0-1127.18.2.el7.x86_64Repository rhel-7-server-optional-rpms is listed more than once in the configuration

This is the output that I am aiming for: 3.10.0-1127.18.2.el7.x86_64

Answer 1

bash:

var="3.10.0-1127.18.2.el7.x86_64Repository rhel-7-server-optional-rpms"
echo "${var%%Repository*}"

See 3.5.3 Shell Parameter Expansion in the manual.

Answer 2

There are several ways.
A "simple" is using sed with a regex to replace the part that you want to strip.
Ex:

echo "3.10.0-1127.18.2.el7.x86_64Repository rhel-7-server-optional-rpms" | sed -E "s/Repository.*//"

3.10.0-1127.18.2.el7.x86_64

To explain the sed command used: sed -E "s/Repository.*//" :

E` is for extended regular expression.
And sed syntax substitution is:

s/regexp/replacement/

Attempt to match regexp against the pattern space. If successful, replace that portion matched with replacement

Here we replace the string found by nothing.