Create string t from string s, keeping only digit characters 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
t = filter (`elem` ['0'..'9']) s
(require '[clojure.string :as str])
(let [s "1a22b3c4de5f6" t (str/replace s #"[^\d]" "")] (println t))
using System.Linq;
var t = string.Concat(s.Where(c => char.IsDigit(c)));
var t = String.fromCharCodes( s.codeUnits.where((x) => (x ^ 0x30) <= 9) );
import "regexp"
re := regexp.MustCompile("[^\\d]") t := re.ReplaceAllLiteralString(s, "")
t = s.replace(/[^\d]/gm,"");
String t = s.replaceAll("\\D+", "");
import static java.lang.String.valueOf; import static java.util.stream.Collectors.joining;
String t = s.chars() .filter(Character::isDigit) .mapToObj(x -> valueOf((char) x)) .collect(joining());
import static java.lang.Character.isDigit;
String t = ""; char c; int i, n = s.length(); for (i = 0; i < n; ++i) if (isDigit(c = s.charAt(i))) t = t + c;
String t = ""; for (char c : s.toCharArray()) if (isDigit(c)) t = t + c;
StringBuilder tb = new StringBuilder(); for (int i=0; i<s.length(); i++){ if(Character.isDigit(s.charAt(i))) tb.append(s.charAt(i)); } String t = tb.toString();
(let ((*t* (remove-if-not #'digit-char-p *s*))) (format t "~A~%" *t*))
for i := 1 to length(s) do if s[i] in ['0'..'9'] then t := t + s[i];
uses RegExpr;
t := ReplaceRegExpr('[^\d]', s, '');
($t = $s) =~ tr/0-9//cd;
my $t = $s; $t =~ s/\D+//g;
import re
t = re.sub(r"\D", "", s)
from string import digits
f = lambda x: x in digits t = ''.join(filter(f, s))
t = s.delete("^0-9")
let t: String = s.chars().filter(|c| c.is_digit(10)).collect();
No security, no password. Other people might choose the same nickname.