Extract pattern from a string

Question

In bash script, what is the easy way to extract a text pattern from a string?

For example, I want to extract X followed by 2 digits in the end of the st

Answer 1

There's a nifty =~ regex operator when you use double square brackets. Captured groups are made available in the $BASH_REMATCH array.

if [[ $STRING =~ (X[0-9]{2})$ ]]; then
    echo "matched part is ${BASH_REMATCH[1]}"
fi

Answer 2

You can also use parameter expansion:

V="abcX23"
PREFIX=${V%%X[0-9][0-9]} # abc
SUFFIX=${V:${#PREFIX}}   # X23

Answer 3

Lets take your input as

Input.txt

ASD123
GHG11D3456
FFSD11dfGH
FF87SD54HJ

And the pattern I want to find is "SD[digit][digit]"

Code

grep -o 'SD[0-9][0-9]' Input.txt

Output

SD12
SD11
SD54

And if you want to use this in script...then you can assign the above code in a variable/array... that's according to your need.

Answer 4

$ foo="abcX23"
$ echo "$(echo "$foo" | sed 's/.*\(X[0-9][0-9]\)$/\1/')"
X23

or

if [[ "$foo" =~ X[0-9][0-9]$ ]]; then
  echo "${foo:$((${#foo}-3))}"
fi

Answer 5

I need to extract the host port from this string: NIC 1 Rule(0): name = guestssh, protocol = tcp, host ip = , host port = 2222, guest ip = , guest port = 22

That string is obtained by using: vboxmanage showvminfo Mojave | grep 'host port', I mean is filtered and I need to extract whatever number be in the host port; in this case is 2222 but it can be different.