Concatenating digits from a string in sh - shell

Question

Assuming that I have a string like this one:
string="1 0 . # 1 1 ? 2 2 4"
Is it possible to concatenate digits that are next to each other?
So that string be like: 10 . # 11 ? 224 ?
I found only basic things how to distinguish integers from other characters and how to "connect" them. But I have no idea how to iterate properly.
num=""
for char in $string; do
if [ $char -eq $char 2>/dev/null ] ; then
num=$num$char

Answer 1

Here's an almost pure-shell implementation -- transforming the string into a character per line and using a BashFAQ #1 while read loop.
string="1 0 . # 1 1 ? 2 2 4"
output=''
# replace spaces with newlines for easier handling
string=$(printf '%s\n' "$string" | tr ' ' '\n')
last_was_number=0
printf '%s\n' "$string" | {
while read -r char; do
if [ "$char" -eq "$char" ] 2>/dev/null; then # it's a number
if [ "$last_was_number" -eq "1" ]; then
output="$output$char"
last_was_number=1
continue
fi
last_was_number=1
else
last_was_number=0
fi
output="$output $char"
done
printf '%s\n' "$output"
}

Answer 2

To complement Charles Duffy's helpful, POSIX-compliant sh solution with a more concise perl alternative:
Note: perl is not part of POSIX, but it is preinstalled on most modern Unix-like platforms.
$ printf '%s\n' "1 0 . # 1 1 ? 2 2 4" | perl -pe 's/\d( \d)+/$& =~ s| ||gr/eg'
10 . # 11 ? 224
The outer substitution, s/\d( \d)+/.../eg, globally (g) finds runs of at least 2 adjacent digits (\d( \d)+), and replaces each run with the result of the expression (e) specified as the replacement string (represented as ... here).
The expression in the inner substitution, $& =~ s| ||gr, whose result is used as the replacement string, removes all spaces from each run of adjacent digits:
$& represents what the outer regex matched - the run of adjacent digits.
=~ applies the s call on the RHS to the LHS, i.e., $& (without this, the s call would implicitly apply to the entire input string, $_).
s| ||gr replaces all (g) instances of <space> from the value of the value of $& and returns (r) the result, effectively removing all spaces.
Note that | is used arbitrarily as the delimiter character for the s call, so as to avoid a clash with the customary / delimiter used by the outer s call.

Answer 3

POSIX compliant one-liner with sed:
string="1 0 . # 1 1 ? 2 2 4"
printf '%s\n' "$string" | sed -e ':b' -e ' s/\([0-9]\) \([0-9]\)/\1\2/g; tb'
It just iteratively removes the any space between two digits until there aren't any more, resulting in:
10 . # 11 ? 224

Answer 4

Here is my solution:
string="1 0 . # 1 1 ? 2 2 4"
array=(${string/// })
arraylength=${#array[#]}
pattern="[0-9]"
i=0
while true; do
str=""
start=$i
if [ $i -eq $arraylength ]; then
break;
fi
for (( j=$start; j<${arraylength}; j++ )) do
curr=${array[$j]}
i=$((i + 1))
if [[ $curr =~ $pattern ]]; then
str="$str$curr"
else
break
fi
done
echo $str
done