Idiom #168 Trim suffix

Create string t consisting of string s with its suffix w removed (if s ends with w).

String t = s;
int index = s.lastIndexOf(w);
if (index + w.length() == s.length()) {
    t = s.substring(0, index);
}

int i = s.lastIndexOf(w),
    m = s.length(), n = w.length();
t = i == m - n ? s.substring(0, i) : s;

import static java.util.regex.Pattern.quote;

String t = s.replaceAll(quote(w) + '$', "");

using System;

string t = s.TrimEnd(w);

import std.string;

string t = s.chomp(w);

  i = index(s,w,back=.true.) - 1
  if (i == len(s) - len(w) ) then
     allocate (t, source=s(:i))
  else
     allocate (t, source=s)
  end if

import "strings"

t := strings.TrimSuffix(s, w)

import Data.List (isSuffixOf)

t :: String
t = if w `isSuffixOf` s then take (length s - length w) else s

const t = s.endsWith(w) ? s.slice(0, -w.length) : s

local t = s:gsub(w.."$", "")

preg_replace("/{$w}$/u", '', $s);

uses StrUtils;

if AnsiEndsStr(w, s) then
  t := copy(s, 1, length(s) - length(w))
else
  t :=s;

if (length $s == rindex($s, $w) + length $w) {
    my $t = substr $s, 0, rindex $s, $w;
}

t = s.removesuffix(w)

t = s.delete_suffix(w)

t = s.sub(/#{w}\z/, "")

let t = s.trim_end_matches(w);

let t = s.strip_suffix(w).unwrap_or(s);

Do you know the best way to do this in your language ? New implementation...

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Idiom created by programming-idioms.org

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