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Programming-Idioms

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  • Perl

Idiom #296 Replace last occurrence of substring

Assign to x2 the value of string x with the last occurrence of y replaced by z.
If y is not contained in x, then x2 has the same value as x.

$x = 'A BB CCC this DDD EEEE this FFF';
$y = 'this';
$z = 'that';

$x2 = $x;
$pos = rindex $x, $y;
substr($x2, $pos, length($y)) = $z
    unless $pos == -1;

print $x2;

perl substr can be used as an lvalue, which is ideal here. We use rindex to find the last occurence of substring $y in $x and use that position in a substr of a copy of $x called $x2. Assigning $z to the substring replaces it -- but we skip that if rindex didn't find the substring.
var position = x.lastIndexOf(y).clamp(0, x.length);
var x2 = x.replaceFirst(y, z, position);
New implementation...
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programming-idioms.org 
$x = 'A BB CCC this DDD EEEE this FFF';
$y = 'this';
$z = 'that';

$x2 = $x;
$pos = rindex $x, $y;
substr($x2, $pos, length($y)) = $z
    unless $pos == -1;

print $x2;

var position = x.lastIndexOf(y).clamp(0, x.length);
var x2 = x.replaceFirst(y, z, position);
  character(len=:), allocatable :: x, x2, y, z

  k = index(x,y,back=.true.)
  if (k > 0) then
     x2 = x(1:k-1) // z // x(k+len(y):)
  else
     x2 = x
  end if

func replaceLast(x, y, z string) (x2 string) {
	i := strings.LastIndex(x, y)
	if i == -1 {
		return x
	}
	return x[:i] + z + x[i+len(y):]
}
String x2 = x;
int i = x.lastIndexOf(y);
if (i != -1) {
    String t = x2.substring(0, i);
    x2 = t + z + x2.substring(i + y.length());
}
StringBuilder x2 = new StringBuilder(x);
int i = x2.lastIndexOf(y);
x2.replace(i, i + y.length(), z);
String p = quote(y), x2;
x2 = x.replaceAll(p + "(?!.*" + p + ')', z);
  x2 := x;
  p := RPos(y, x);
  if (p > 0) then
  begin
    Delete(x2, p, Length(y));
    Insert(z, x2, p);
  end;

  x2 := ReverseString(StringReplace(ReverseString(x), ReverseString(y), ReverseString(z), []));

x2 = z.join(x.rsplit(y, 1))

x2 = x.gsub(/#{y}(?!.*#{y})/, z )