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 i

Answer 1

How do you know where the value begins? If it's always the 5th and 6th words, you could use e.g.:

B=$(echo $A | cut -d ' ' -f 5-)

This uses the cut command to slice out part of the line, using a simple space as the word delimiter.

Answer 2

Some examples using parameter expansion

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

abc.123

Longest match on " " space

echo "${A% *}"

Some variable has value

Longest match on . dot

echo "${A%.*}"

Some variable has value abc

Shortest match on " " space

echo "${A%% *}"

some

Read more Shell-Parameter-Expansion

Answer 3

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.)

Answer 4

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 5

As pointed out by Zedfoxus here. A very clean method that works on all Unix-based systems. Besides, you don't need to know the exact position of the substring.

A="Some variable has value abc.123"

echo $A | rev | cut -d ' ' -f 1 | rev                                                                                            

# abc.123

Answer 6

The documentation is a bit painful to read, so I've summarised it in a simpler way.

Note that the '*' needs to swap places with the '' depending on whether you use # or %. (The * is just a wildcard, so you may need to take off your "regex hat" while reading.)

• ${A%% *} - remove longest trailing * (keep only the first word)
• ${A% *} - remove shortest trailing * (keep all but the last word)
• ${A##* } - remove longest leading * (keep only the last word)
• ${A#* } - remove shortest leading * (keep all but the first word)

Of course a "word" here may contain any character that isn't a literal space.