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Programming-Idioms

  • C#
  • Rust
  • Perl
  • Ruby

Idiom #168 Trim suffix

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

t = s.delete_suffix(w)

Introduced in Ruby 2.5.0
t = s.sub(/#{w}\z/, "")

Like Strings, Regular expressions can be interpolated. \z means "end of string".
using System;
string t = s.TrimEnd(w);
let t = s.strip_suffix(w).unwrap_or(s);

Removes at most 1 occurrence of w
let t = s.trim_end_matches(w);

This may remove several occurrences of w at the end of s.
if (length $s == rindex($s, $w) + length $w) {
    my $t = substr $s, 0, rindex $s, $w;
}
import std.string;
string t = s.chomp(w);
New implementation...
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programming-idioms.org 
t = s.delete_suffix(w)
t = s.sub(/#{w}\z/, "")
string t = s.TrimEnd(w);
let t = s.strip_suffix(w).unwrap_or(s);
let t = s.trim_end_matches(w);
if (length $s == rindex($s, $w) + length $w) {
    my $t = substr $s, 0, rindex $s, $w;
}
string t = s.chomp(w);
  i = index(s,w,back=.true.) - 1
  if (i == len(s) - len(w) ) then
     allocate (t, source=s(:i))
  else
     allocate (t, source=s)
  end if
t := strings.TrimSuffix(s, w)
t :: String
t = if w `isSuffixOf` s then take (length s - length w) else s
const t = s.endsWith(w) ? s.slice(0, -w.length) : s
String t = s;
int index = s.lastIndexOf(w);
if (index + w.length() == s.length()) {
    t = s.substring(0, index);
}
int i = s.lastIndexOf(w),
    m = s.length(), n = w.length();
t = i == m - n ? s.substring(0, i) : s;
String t = s.replaceAll(quote(w) + '$', "");
local t = s:gsub(w.."$", "")
preg_replace("/{$w}$/u", '', $s);
if AnsiEndsStr(w, s) then
  t := copy(s, 1, length(s) - length(w))
else
  t :=s;
t = s.removesuffix(w)