how to capture only digits in array using if statement in bash

Having:

value=(cat dog 1234 pl345 567ab 12234)

You could print "only numbers":

$ printf "%s\n" "${value[@]}" | grep -o "[0-9]*"
1234
345
567
12234

or maybe you want "only lines that contain only numbers":

$ printf "%s\n" "${value[@]}" | grep -x "[0-9]*"
1234
12234

or maybe you want "only lines with a number":

$ printf "%s\n" "${value[@]}" | grep "[0-9]"
1234
pl345
567ab
12234

Your if/else is backwards.

value=(cat dog 1234 pl345 567ab 12234)
len=${#value[*]}
    
for ((i=0;i<${len};i++)); do
    if [[ "${value[$i]}" =~ ^[[:digit:]]+$ ]]; then 
         echo "${value[$i]}"   
    fi
done

This outputs:

1234
12234

There is a non-regex way of doing it in bash using shopt:

# enable extended glob
shopt -s extglob

value=(cat dog 1234 pl345 567ab 12234)

# lop through array and print only it contains 1+ digits
for v in "${value[@]}"; do [[ $v == +([0-9]) ]] && echo "$v"; done

1234
12234

You probably should use two nested loops:

1. iterate over all elements of an array
2. iterate over all chars in the element, check if the char is a digit

(2) would probably be simpler with grep, but since we're speaking of bash implementaton I've came up with this:

#!/bin/bash

a=(cat dog 1234 pl345 567ab 12234) # input array

for s in "${a[@]}"; do # iterate over all elements of our array
        res=''
        for ((i=0;i<${#s};i++)); do # iterate over all chars of element $s
                c="${s:$i:1}" # $c is now char at $i position inside $s
                case "$c" in
                        [0-9]) res="${res}${c}" # append digit to res
                esac
        done
        [ -n "$res" ] && echo "$res" # if $s contains at least 1 digit, print it
                                     # you may add $res to another array here if needed

done

Simply

tr -dc '0-9\n' <getdata.txt

You can use:

$ cat getdata.txt |grep -P '^[0-9]+$'