Result from a regex substitute

[Julia deSilva]

Hi there all,

I sure this is really obvious but what's the syntax for this please:

my $string1 = "mark had a little dog";
$string1 =~ s/dog/lamb/g; # $string1 is "mary had a little lamb"


But ................
my $string1 = "mark had a little dog";
my $string2 = $string1 =~ s/dog/lamb/g;
This returns
$string2 = 1 (presumably a boolean for substitute done)
whereas I want it to return $string2 = "mary had a little lamb";

Thanks in advance

 

[Tassilo v. Parseval]

Also sprach Julia deSilva:

I sure this is really obvious but what's the syntax for this please:

my $string1 = "mark had a little dog";
$string1 =~ s/dog/lamb/g; # $string1 is "mary had a little lamb"


But ................
my $string1 = "mark had a little dog";
my $string2 = $string1 =~ s/dog/lamb/g;
This returns
$string2 = 1 (presumably a boolean for substitute done)

Yes, it returns the number of replacements done. You need to add
parens:

(my $string2 = $string1) =~ s/dog/lamb/g;

Tassilo

 

[Chris]

Hi there all,

I sure this is really obvious but what's the syntax for this please:

my $string1 = "mark had a little dog";
$string1 =~ s/dog/lamb/g; # $string1 is "mary had a little lamb"


But ................
my $string1 = "mark had a little dog";
my $string2 = $string1 =~ s/dog/lamb/g;
This returns
$string2 = 1 (presumably a boolean for substitute done)
whereas I want it to return $string2 = "mary had a little lamb";

If you want to keep the $string1 to remain as "mark had a little dog"
and have $string2 be the changed string do this:

my string1 = my string2 = "mark had a little dog";
$string2 =~ s/dog/lamb/g;

BTW This substitution won't ever give "mary had a little lamb" it will
give you "mark had a little lamb".

 

[Tad McClellan]

my string1 = my string2 = "mark had a little dog";

^^ ^^


Perl variables require a funny character (sigil) before their names.

 

[John Strauss]

Hi there all,

I sure this is really obvious but what's the syntax for this please:

my $string1 = "mark had a little dog";
$string1 =~ s/dog/lamb/g; # $string1 is "mary had a little lamb"


But ................
my $string1 = "mark had a little dog";
my $string2 = $string1 =~ s/dog/lamb/g;
This returns
$string2 = 1 (presumably a boolean for substitute done)
whereas I want it to return $string2 = "mary had a little lamb";

Thanks in advance

(my $string2 = $string1) =~ s/dog/lamb/g;

unless you really want to change "mark" to "mary":
my %lut=qw/mark mary dog lamb/;
(my $string2 = $string1) =~ s/(mark|dog)/$lut{$1}/g;



~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
drop the .thetenant to get me via mail

 

[Julia deSilva]

Thanks everyone,

my $string1 = "mark had a little dog";

Sorry, I did get my nursery rhyme wrong, and for YOUR information, I live 2
miles from Kilmersdon, the village where Jack and Jill fell down the hill,
but that's another story.....