How to find the last field using 'cut'

Without awk ?... But it's so simple with awk:

echo 'maps.google.com' | awk -F. '{print $NF}'

AWK is a way more powerful tool to have in your pocket. -F if for field separator NF is the number of fields (also stands for the index of the last)

You could try something like this:

echo 'maps.google.com' | rev | cut -d'.' -f 1 | rev

Explanation

rev reverses "maps.google.com" to be moc.elgoog.spam
cut uses dot (ie '.') as the delimiter, and chooses the first field, which is moc
lastly, we reverse it again to get com

It is not possible using just cut. Here is a way using grep:

grep -o '[^,]*$'

Replace the comma for other delimiters.

Explanation:

-o (--only-matching) only outputs the part of the input that matches the pattern (the default is to print the entire line if it contains a match).
[^,] is a character class that matches any character other than a comma.
* matches the preceding pattern zero or more time, so [^,]* matches zero or more non‑comma characters.
$ matches the end of the string.
Putting this together, the pattern matches zero or more non-comma characters at the end of the string.
When there are multiple possible matches, grep prefers the one that starts earliest. So the entire last field will be matched.

Full example:

If we have a file called data.csv containing

one,two,three
foo,bar

then grep -o '[^,]*$' < data.csv will output

three
bar

Use a parameter expansion. This is much more efficient than any kind of external command, cut (or grep) included.

data=foo,bar,baz,qux
last=${data##*,}

See BashFAQ #100 for an introduction to native string manipulation in bash.

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