The question simply wants me to demonstrate changing different date formats into one one format. To change 3/1/2004 or 3.1.2004 into 3-1-2004.
I believe the date 3-1-2003 matches the following code -
$date =~ /^(0[1-9]¦1[012])[- /.](0[1-9]¦[12][0-9]¦3[01])[- /.](19¦20)\d\d
But i need to use the substitution operator to replace the following dates from 3/1/2004 to 3-1-2004.
Anyone with some suggestions please?
regards
Dave
if there is more text in the string:
$date =~ s/(\d+)[/.](\d+)[/.](\d+)/$1-$2-$3/g;
only the date:
$date =~ tr#/.#-#;
$date =~ s/(\d+)[\/.](\d+)[\/.](\d+)/$1-$2-$3/g;
A dissection of perl diver's code, I've added an x modifier to allow the comments and white space.
#!/usr/bin/perl
$date = '08/05/2008';
print "before $date\n";
#$date =~ s/(\d+)[\/.](\d+)[\/.](\d+)/$1-$2-$3/g;
$date =~ s/ # substitute this match . . .
(\d+) # one or more (+) of any digit. Store this value in $1
[\/.] # followed by ONE slash or dot
(\d+) # followed by one or more digits, store in $2
[\/.] # followed by another slash or dot
(\d+) # followed by one or more digits, store in $3
/$1-$2-$3/gx; # replace with this. Note you can also do $3-$1-$2
# Apply globally (g), allow white space and comments(x)
print "after $date\n";
Except you need to escape the slashes or it gives you an unmatched [ error
oops.... thanks.
