Excluding the first 3 characters of a string using regex - regex

Question

Given any string in bash, e.g flaccid, I want to match all characters in the string but the first 3 (in this case I want to exclude "fla" and match only "ccid"). The regex also needs to work in sed.
I have tried positive look behind and the following regex expressions (as well as various other unsuccessful ones):
^.{3}+([a-z,A-Z]+)
sed -r 's/(?<=^....)(.[A-Z]*)/,/g'
Google hasn't been very helpful as it only produce results like "get first 3 characters .."
Thanks in advance!

Answer 1

If you want to get all characters but the first 3 from a string, you can use cut:
str="flaccid"
cut -c 4- <<< "$str"
or bash variable subsitution:
str="flaccid"
echo "${str:3}"
That will strip the first 3 characters out of your string.

Answer 2

You may just use a capturing group within an expression like ^.{3}(.*) / ^.{3}([a-zA-Z]+) and grab the ${BASH_REMATCH[1]} contents:
#!/bin/bash
text="flaccid"
rx="^.{3}(.*)"
if [[ $text =~ $rx ]]; then
echo ${BASH_REMATCH[1]};
fi
See online Bash demo
In sed, you should also be using capturing groups / backreferences to get what you need. To just keep the first 3 chars, you may use a simple:
echo "flaccid" | sed 's/.\{3\}//'
See this regex demo. The .\{3\} matches exactly any 3 chars and will remove them from the beginning only, since g modifier is not used.
Now, both the solutions above will output ccid, returning the first 3 chars only.

Answer 3

Using sed, just remove them
echo string | sed 's/^...//g'

Answer 4

How is it that no-one has named the most simple and portable solution:
shell "Parameter expansions":
str="flacid"
echo "${str#???}
For a regex (bash):
$ str="flaccid"
$ regex='^.{3}(.*)$'
$ [[ $str =~ $regex ]] && echo "${BASH_REMATCH[1]}"
ccid
Same regex in sed:
$ echo "flaccid" | sed -E "s/$regex/\1/"
ccid
Or sed (Basic Regex):
$ echo "flaccid" | sed 's/^.\{3\}\(.*\)$/\1/'
ccid