Shell Programming and Scripting

Originally Posted by KevinADC

In the regexp you see ,?. In this context the ? symbol is a quantifier that means zero or one of whatever is to the left of it. So it will match zero or one comma.



In your regexp you have used .* which means to match zero or more of anything as much as possible. This is called greedy matching. So it matches whatever is between the first single quote and the last single quote in the string. What you should have done was used .*? which means to match zero or more of anything but as little as possible. This is called stingy matching. But using the negated character class [^'] is actually more efficient then using .*? to achieve the same result.

Originally Posted by unknown123

try this:

my $str = " c1 ='mmdnn, ,sdm = sm,m,jkjk',c2 = 'chj=kjk, khj ', c3='hhshhj, hsjh'";

print "Before substitution-> $str\n";

$str =~ s/',/':/g;

print "After substitution-> $str\n"; # Getting Wrong o/p

Originally Posted by unknown123

then little change, try this:

try this:

my $str = " c1 ='mmdnn, , sdm = 'sm' ,m ,jkjk' , c2 = 'chj= kjk'' , khj ', c3='hhshhj, hsjh' ";

print "Before substitution-> $str\n";

$str =~ s/('[ ]*),([ ]*c)/$1:$2/g;

print "After substitution-> $str\n"; # Getting Wrong o/p

Originally Posted by unknown123

try this:
$col_strings="string_nm =' 12th dec, 1984, 'op=1 ' , s: 'g' ', name= ' Brian Hogg' , class = '1st, A.C: Field Pt. 0.1 ' ";
print "Before substitution-> $col_strings \n";

$col_strings =~ s/('[ ]*),([ ]*[a-zA-Z0-9\-_]+[ ]*=)/$1:$2/g;
print "After substitution-> $col_strings \n";