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

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

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 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

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 5

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 shortest trailing * (strip the last word)
• ${A%% *} - remove longest trailing * (strip the last words)
• ${A#* } - remove shortest leading * (strip the first word)
• ${A##* } - remove longest leading * (strip the first words)

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

You might commonly use this syntax to trim filenames:

• ${A##*/} removes all containing folders, if any, from the start of the path, e.g.
  /usr/bin/git -> git
/usr/bin/ -> (empty string)

• ${A%/*} removes the last file/folder/trailing slash, if any, from the end:
  /usr/bin/git -> /usr/bin
/usr/bin/ -> /usr/bin

• ${A%.*} removes the last extension, if any (just be wary of things like my.path/noext):
  archive.tar.gz -> archive.tar

Answer 6

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 7

More ways to do this:

(Run each of these commands in your terminal to test this live.)

For all answers below, start by typing this in your terminal:

A="Some variable has value abc.123"

The array example (#3 below) is a really useful pattern, and depending on what you are trying to do, sometimes the best.

1. with awk, as the main answer shows

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

2. with grep:

echo "$A" | grep -o '[^ ]*$'

1. the -o says to only retain the matching portion of the string
2. the [^ ] part says "don't match spaces"; ie: "not the space char"
3. the * means: "match 0 or more instances of the preceding match pattern (which is [^ ]), and the $ means "match the end of the line." So, this matches the last word after the last space through to the end of the line; ie: abc.123 in this case.

3. via regular bash "indexed" arrays and array indexing

Convert A to an array, with elements being separated by the default IFS (Internal Field Separator) char, which is space:

1. Option 1 (will "break in mysterious ways", as @tripleee put it in a comment here, if the string stored in the A variable contains certain special shell characters, so Option 2 below is recommended instead!):

   # Capture space-separated words as separate elements in array A_array
   A_array=($A)
   

2. Option 2 [RECOMMENDED!]. Use the read command, as I explain in my answer here, and as is recommended by the bash shellcheck static code analyzer tool for shell scripts, in ShellCheck rule SC2206, here.

   # Capture space-separated words as separate elements in array A_array, using
   # a "herestring". 
   # See my answer here: https://stackoverflow.com/a/71575442/4561887
   IFS=" " read -r -d '' -a A_array <<< "$A"
   

Then, print only the last elment in the array:

# Print only the last element via bash array right-hand-side indexing syntax
echo "${A_array[-1]}"  # last element only

Output:

abc.123

Going further:

What makes this pattern so useful too is that it allows you to easily do the opposite too!: obtain all words except the last one, like this:

array_len="${#A_array[@]}"
array_len_minus_one=$((array_len - 1))
echo "${A_array[@]:0:$array_len_minus_one}"

Output:

Some variable has value

For more on the ${array[@]:start:length} array slicing syntax above, see my answer here: Unix & Linux: Bash: slice of positional parameters, and for more info. on the bash "Arithmetic Expansion" syntax, see here:

1. https://www.gnu.org/savannah-checkouts/gnu/bash/manual/bash.html#Arithmetic-Expansion
2. https://www.gnu.org/savannah-checkouts/gnu/bash/manual/bash.html#Shell-Arithmetic

Answer 8

You can use a Bash regex:

A="Some variable has value abc.123"
[[ $A =~ [[:blank:]]([^[:blank:]]+)$ ]] && echo "${BASH_REMATCH[1]}" || echo "no match"

Prints:

abc.123

That works with any [:blank:] delimiter in the current local (Usually [ \t]). If you want to be more specific:

A="Some variable has value abc.123"
pat='[ ]([^ ]+)$'
[[ $A =~ $pat ]] && echo "${BASH_REMATCH[1]}" || echo "no match"

Answer 9

echo "Some variable has value abc.123"| perl -nE'say $1 if /(\S+)$/'