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Str.subst: Add examples for nth(*), etc. #3460

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Jun 6, 2020
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Str.subst: Add examples for nth(*), etc. #3460

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23 changes: 16 additions & 7 deletions doc/Type/Str.pod6
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Expand Up @@ -823,14 +823,23 @@ regex, not the C<subst> method call.
Here are other examples of usage:

my $str = "Hey foo foo foo";
$str.subst(/foo/, "bar", :g); # global substitution - returns Hey bar bar bar

$str.subst(/\s+/, :g); # global substitution - returns Heyfoofoofoo

$str.subst(/foo/, "no subst", :x(0)); # targeted substitution. Number of times to substitute. Returns back unmodified.
$str.subst(/foo/, "bar", :x(1)); #replace just the first occurrence.

$str.subst(/foo/, "bar", :nth(3)); # replace nth match alone. Replaces the third foo. Returns Hey foo foo bar
say $str.subst(/foo/, "bar", :g); # OUTPUT: «Hey bar bar bar␤»
say $str.subst(/\s+/, :g); # OUTPUT: «Heyfoofoofoo␤»

say $str.subst(/foo/, "bar", :x(0)); # OUTPUT: «Hey foo foo foo␤»
say $str.subst(/foo/, "bar", :x(1)); # OUTPUT: «Hey bar foo foo␤»
# Can not match 4 times, so no substitutions made
say $str.subst(/foo/, "bar", :x(4)); # OUTPUT: «Hey foo foo foo␤»
say $str.subst(/foo/, "bar", :x(2..4)); # OUTPUT: «Hey bar bar bar␤»
# Replace all of them, identical to :g
say $str.subst(/foo/, "bar", :x(*)); # OUTPUT: «Hey bar bar bar␤»

say $str.subst(/foo/, "bar", :nth(3)); # OUTPUT: «Hey foo foo bar␤»
# Replace last match
say $str.subst(/foo/, "bar", :nth(*)); # OUTPUT: «Hey foo foo bar␤»
# Replace next-to-last last match
say $str.subst(/foo/, "bar", :nth(*-1)); # OUTPUT: «Hey foo bar foo␤»

The C<:nth> adverb has readable English-looking variants:

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