Substitution with parameters and variable don't works/$search/$replace/i

[Joan Interactive Bussiness]

Very easy sample:

# line to translate
$line = 'var varchar2(10)';

# I hope to answer: var number(10)

# With replace Variable, not OK
$find = 'varchar2\((.+)\)';
$replace = 'number($1)';

$line =~ s/$find/$replace/;
print $line."\n";
# result >>>>>> var number($1)


# Without replace Variable, it's OK
$line2 = 'var varchar2(10)';
$line2 =~ s/varchar2\((.+)\)/number($1)/;
print $line2."\n";
# result >>>>>> var number(10)

# ----------------------------------------
P:\DB\WORK\perl>perl test.pl
var number($1)
var number(10)

 

[sln]

# line to translate
$line = 'var varchar2(10)';

# I hope to answer: var number(10)

# With replace Variable, not OK
$find = 'varchar2\((.+)\)';

$replace = '"number($1)"';

$line =~ s/$find/$replace/ee;


Here we go again..
-sln

 

[Kyle T. Jones]

Very easy sample:

# line to translate
$line = 'var varchar2(10)';

# I hope to answer: var number(10)

# With replace Variable, not OK
$find = 'varchar2\((.+)\)';
$replace = 'number($1)';

You're using single quotes '' instead of double quotes "" so the
variable ($1) isn't being interpolated. If you're using variables or
special characters (like \n) you need to use double-quotes.

my $name="Joe";
print "$name is a genius.\n";
print "He loves perl.\n";

Output:

Joe is a genuis.
He loves perl.

my $name='Joe';
print '$name is a genius.\n';
print 'He loves perl.\n';

Output:

$name is a genius.\nHe loves perl.\n

Cheers.