How to extract last part of string in bash?

Question

I have this variable:

A="Some variable has value abc.123" 

I need to extract this value i.e abc.123. Is this possible in bash?

Answer 1

Simplest is

echo $A | awk '{print $NF}' 

Edit: explanation of how this works...

awk breaks the input into different fields, using whitespace as the separator by default. Hardcoding 5 in place of NF prints out the 5th field in the input:

echo $A | awk '{print $5}' 

NF is a built-in awk variable that gives the total number of fields in the current record. The following returns the number 5 because there are 5 fields in the string "Some variable has value abc.123":

echo $A | awk '{print NF}' 

Combining $ with NF outputs the last field in the string, no matter how many fields your string contains.

Answer 2

Yes; this:

A="Some variable has value abc.123" echo "${A##* }" 

will print this:

abc.123

(The ${parameter##word} notation is explained in §3.5.3 "Shell Parameter Expansion" of the Bash Reference Manual.)