How to cut a string after a specific character in unix

Question

So I have this string:

$var=server@10.200.200.20:/home/some/directory/file

I just want to extract the directory address meaning I only wan

Answer 1

This should do the trick:

$ echo "$var" | awk -F':' '{print $NF}'
/home/some/directory/file

Answer 2

This will also do.

echo $var | cut -f2 -d":"

Answer 3

awk -F: '{print $2}' <<< $var

Answer 4

This might work for you:

echo ${var#*:}

See Example 10-10. Pattern matching in parameter substitution

Answer 5

For completeness, using cut

cut -d : -f 2 <<< $var

And using only bash:

IFS=: read a b <<< $var ; echo $b

Answer 6

You don't say which shell you're using. If it's a POSIX-compatible one such as Bash, then parameter expansion can do what you want:

Parameter Expansion

...

${parameter#word}

Remove Smallest Prefix Pattern.
The word is expanded to produce a pattern. The parameter expansion then results in parameter, with the smallest portion of the prefix matched by the pattern deleted.

In other words, you can write

$var="${var#*:}"

which will remove anything matching *: from $var (i.e. everything up to and including the first :). If you want to match up to the last :, then you could use ## in place of #.

This is all assuming that the part to remove does not contain : (true for IPv4 addresses, but not for IPv6 addresses)