Bash Regex - string should not start and end with a dot

Question

I have a script that takes a string input from users. I am looking to check that the string input should have exactly 2 dots. The relevance is only to the dots. The string should not start and end with a dot.There should be no consecutive dots.

This is the pattern I am using:

^[^\.]*\.[^\.]*\.[^\.]*$

This is the string I am looking for:

abc.def.xyz

But in the pattern above, if dots are in front or at the end, then that string gets selected - which I don't want. There should be only two dots in the string.

Not wanted:

.abc.xyz # no dot at the start   
abc.xyz. # no dot at the end   
abc.def.ced.xyz # only two dots not more than that

I have tried using (?!\.) for the dot at the start, but it didn't work.

Answer 1

You're not saying how the string is input from the user, but note that if it may contain newline characters, you can't use grep to filter them (unless you use the --null extension) as grep works on one line at a time. Also note that the [^\.] regex matches on characters other than backslash and . and the . regex operator (or [...]) in many regex implementations will not match on bytes that don't form valid characters in the locale.

Here, to check that $string contains 2 and only 2 dots, but not at the start nor end and not next to each other, you can use the standard sh:

case $string in
  (*.*.*.* | .* | *. | *..* ) echo not OK;;
  (*.*.*) echo OK;;
  (*) echo not OK;;
esac

Or with ksh globs, a subset of which can be made available in the bash shell by doing shopt -s extglob:

case $string in
  ( +([!.]).+([!.]).+([!.]) ) echo OK;;
  (*) echo not OK;;
esac

bash can also do extended regex matching with the =~ operator inside its [[...]] ksh-style construct, but again, you'll want to fix the locale to C:

regex_match_in_C_locale() {
  local LC_ALL=C
  [[ $1 =~ $2 ]]
}

if regex_match_in_C_locale "$string" '^[^.]+\.[^.]+\.[^.]+$'; then
  echo OK
else
  echo not OK
fi

POSIXly, you can do basic regex matching with the expr utility:

if
  LC_ALL=C expr "x$string" : 'x[^.]\{1,\}\.[^.]\{1,\}\.[^.]\{1,\}$' > /dev/null
then
  echo OK
else
  echo not OK
fi

Or extended regex matching with the awk utility:

regex_match_in_C_locale() {
  LC_ALL=C awk -- 'BEGIN {exit(ARGV[1] !~ ARGV[2])}' "$@"
}
if regex_match_in_C_locale "$string" '^[^.]+\.[^.]+\.[^.]+$'; then
  echo OK
else
  echo not OK
fi

Answer 2

I think you are looking for this regex ^[^.]\+\.[^.]\+\.[^.]\+$, in this example we will use grep:

The characters inside brackets are treated literally (except -), so you don't need to escape the dot.

$ echo ".a.b.c." | grep  "^[^.]\+\.[^.]\+\.[^.]\+$"
$ echo ".a.b.c"  | grep  "^[^.]\+\.[^.]\+\.[^.]\+$"
$ echo "a.b.c."  | grep  "^[^.]\+\.[^.]\+\.[^.]\+$"
$ echo "a..c"    | grep  "^[^.]\+\.[^.]\+\.[^.]\+$"
$ echo "a.b.c"   | grep  "^[^.]\+\.[^.]\+\.[^.]\+$"
a.b.c

The regex says

• The string must start with one or more characters that are not dots, followed by a dot ^[^.]\+\., followed by one or more characters that are not dots and a dot [^.]\+\., followed by one or more characters that are not dots [^.]\+$ until the end of the line.

Answer 3

In awk, we can do this:

$ awk '  $0"."  ~   /^([^.]+\.){3}$/  ' file

a.b.c
abc.def.xyz

Adding a dot at the end, makes the pattern repetitive, just three times of not-dot followed by a dot. Like a. -- b. -- c. or abc. -- def. -- xyz.

Or, in regex parlance: ([^.]\.){3}

Only accept if the regex could match the whole line.

Answer 4

If you want to do this in bash, this is one approach:

IFS="." read -ra words <<<"$input"
if ((${#words[@]} == 3)) && [[ $input != .* && $input != *. ]]; then
    echo "valid input"
fi

This one actually use the values in the words array:

IFS="." read -ra words <<<"$input"
# 3 dot-separated fields, and the first and last cannot be empty
if ((${#words[@]} == 3)) && [[ -n ${words[0]} && -n ${words[2]} ]]; then
    echo "valid input"
fi