i have this code:
if ( $Mail =~ /$Tld{$_}/ ) {
$TldFound = 1;
}
The variable $Mail has for example the info "mail#mail.com". The variable $Tld has the info ".com". How can i cut the variable $Mail that only the tld .com will remain?
You should use Email::Address to parse email addresses.
To be able to extract a TLD with certainty requires a list of what you consider to be TLDs. For example, do .co.uk, or .com.tr count? Or, do you just want the last string of non-dot characters?
If you restrict your attention to 2 - 3 character TLDs such as .co, .com, .io, .net, .org, .us etc, you can do my ($tld) = ($email =~ /[.] ([a-z]{2,3}) \z/x); and then check with if ($tld and ($tld eq 'com')) { ... } etc, but you really want a good list of acceptable strings that can be TLDs: Net::Domain::TLD, Mozilla::PublicSuffix.
Naive Regex Solutions
The following solutions will solve your problem as posted, but are not intended to address every possible edge case. Parsing email addresses in a comprehensive way is non-trivial, and requires a parser such as Email::Address if you want to handle the full complexity of the RFCs.
Printing Your TLD from a String
Since you already know the string you want to print on success (e.g. ".com"), you don't actually need the result of your regular expression match; you can print the string stored in $Tld when the match is true using a post-statement condition. For example:
$Mail = 'mail#mail.com';
$Tld = '.com';
print "$Tld\n" if $Mail =~ /${Tld}$/;
This will correctly print:
.com
Printing the Match
If you really want the full match, there are a number of ways to do it. One way would be to use the special $& variable:
$Mail = 'mail#mail.com';
$Tld = '.com';
if ($Mail =~ /${Tld}$/) {
print "$&\n";
}
This will also correctly print:
.com
Partitioning the String
All of the previous examples will solve your problem as posted, but the best generic solution short of a parser is really to partition the TLD, and treat the last segment of the domain as an unvalidated TLD. Ruby has the super-handy String#rpartition method, but I'm unaware of a similar function in Perl. However, you can use an anchored match to accomplish much the same thing. For example:
$Mail = 'mail#mail.com';
$Mail =~ /(\.[[:alpha:]]+)$/;
print "$1\n";
If you need to validate the TLD against an expected value such as .com, you can compare it to a string or variable. For example:
$Mail = 'mail#mail.com';
$Tld = '.com';
$Mail =~ /(\.[[:alpha:]]+)$/;
print "$1\n" if $1 eq $Tld
Related
I am new to Perl, so this is basic question. I have a string as shown below. I am interested in taking date out of it, so thinking of splitting it using slash
my $path = "/bla/bla/bla/20160306";
my $date = (split(/\//,$path))[3];#ideally 3 is date position in array after split
print $date;
However, I don't see the expected output, but instead I see 5 getting printed.
Since the path starts with the pattern / itself, split returns a list with an empty string first (to the left of the first /); one element more. Thus the posted code miscounts by one and returns the one before last element (subdirectory) in the path, not the date.
If date is always the last thing in the string you can pick the last element
my $date = (split '/', $path)[-1];
where i've used '' for delimiters so to not have to escape /. (This, however, may confuse since the separator pattern is a regex and // convey that, while '' may appear to merely quote a string.)
This can also be done with regex
my #parts = $path =~ m{([^/]+)}g;
With this there can be no inital empty string. Or, the last part can be picked out of the full list as above, with ($path =~ m{...}g)[-1], but if you indeed only need the last bit then extract it directly
my ($last_part) = $path =~ m{.*/(.*)};
Here the "greedy" .* matches everything in the string up to the last instance of the next subpattern (/ here), thus getting us to the last part of the path, which is then captured. The regex match operator returns its matches only when it is in the list context so parens on the left are needed.
What brings us to the fact that you are parsing a path, and there are libraries dedicated to that.
For splitting a path into its components one tool is splitdir from File::Spec
use File::Spec;
my #parts = File::Spec->splitdir($path);
If the path starts with a / we'll again get an empty string for the first element (by design, see docs). That can then be removed, if there
shift #parts if $parts[0] eq '';
Again, the last element alone can be had like in the other examples.
Simply bind it to the end:
(\d+)$
# look for digits at the end of the string
See a demo on regex101.com. The capturing group is only for clarification though not really needed in this case.
In Perl this would be (I am a PHP/Python guy, so bear with me when it is ugly)
my $path = "/bla/bla/bla/20160306";
$path =~ /(\d+)$/;
print $1;
See a demo on ideone.com.
Try this
Use look ahead for to do it. It capture the / by splitting. Then substitute the data using / for remove the slash.
my $path = "/a/b/c/20160306";
my $date = (split(/(?=\/)/,$path))[3];
$date=~s/^\///;
print $date;
Or else use pattern matching with grouping for to do it.
my $path = "/a/b/c/20160306";
my #data = $path =~m/\/(\w+)/g;
print $data[3];
in my Perl script I want to do conditional search & replace using regular expression: Find a certain pattern, and if the pattern exists in a hash, then replace it with something else.
For example, I want to search for a combination of "pattern1" and "pattern2", and if the latter exists in a hash, then replace the combination with "pattern1" and "replacement". I tried the following, but it just doesn't do anything at all.
$_ =~ s/(pattern1)(pattern2)/$1replacement/gs if exists $my_hash{$2};
I also tried stuff like:
$_ =~ s/(pattern1)(pattern2) && exists $my_hash{$2}/$1replacement/gs;
Also does nothing at all, as if no match is found.
Can anyone help me with this regex problem? Thx~
I would do it a different way. It looks like you have a 'search this, replace that' hash.
So:
#!/usr/bin/env perl
use strict;
use warnings;
#our 'mappings'.
#note - there can be gotchas here with substrings
#so make sure you anchor patterns or sort, so
#you get the right 'substring' match occuring.
my %replace = (
"this phrase" => "that thing",
"cabbage" => "carrot"
);
#stick the keys together into an alternation regex.
#quotemeta means regex special characters will be escaped.
#you can remove that, if you want to use regex in your replace keys.
my $search = join( "|", map {quotemeta} keys %replace );
#compile it - note \b is a zero width 'word break'
#so it will only match whole words, not substrings.
$search = qr/\b($search)\b/;
#iterate the special DATA filehandle - for illustration and a runnable example.
#you probably want <> instead for 'real world' use.
while (<DATA>) {
#apply regex match and replace
s/(XX) ($search)/$1 $replace{$2}/g;
#print current line.
print;
}
##inlined data filehandle for testing.
__DATA__
XX this phrase cabbage
XX cabbage carrot cabbage this phrase XX this phrase
XX no words here
and this shouldn't cabbage match this phrase at all
By doing this, we turn your hash keys into a regex (you can print it - it looks like: (?^:\b(cabbage|this\ phrase)\b)
Which is inserted into the substitution pattern. This will only match if the key is present, so you can safely do the substitution operation.
Note - I've added quotemeta because then it escapes any special characters in the keys. And the \b is a "word boundary" match so it doesn't do substrings within words. (Obviously, if you do want that, then get rid of them)
The above gives output of:
XX that thing cabbage
XX carrot carrot cabbage this phrase XX that thing
XX no words here
and this shouldn't cabbage match this phrase at all
If you wanted to omit lines that didn't pattern match, you can stick && print; after the regex.
What is wrong (as in not working) with
if (exists($h{$patt1)) { $text =~ s/$patt1$patt2/$patt1$1replacement/g; }
If $patt1 exists as a key in a hash then you go ahead and replace $patt1$patt2 with $patt1$replacement. Of course, if $patt1$patt2 is found in $text, otherwise nothing happens. Your first code snippet is circular, while the second one can't work like that at all.
If you want $patt1$patt2 first, and hash key as well then it seems that you'd have to go slow
if ($str =~ /$patt11$patt2/ && exists $h{$patt2}) {
$str =~ s/$patt1$patt2/$patt1$replacement/gs;
}
If this is what you want then it is really simple: you need two unrelated conditions, whichever way you turn it around. Can't combine them since it would be circular.
From the point of view of the outcome these are the same. If either condition fails nothing happens, regardless of the order in which you check them.
NOTE Or maybe you don't have to go slow, see Sobrique's post.
So I'm trying to put all numbered domains into on element of a hash doing this:
### Domanis ###
my $dom = $name;
$dom =~ /(\w+\.\w+)$/; #this regex get the domain names only
my $temp = $1;
if ($temp =~ /(^d+\.\d+)/) { # this regex will take out the domains with number
my $foo = $1;
$foo = "OTHER";
$domain{$foo}++;
}
else {
$domain{$temp}++;
}
where $name will be something like:
something.something.72.154
something.something.72.155
something.something.72.173
something.something.72.175
something.something.73.194
something.something.73.205
something.something.73.214
something.something.abbnebraska.com
something.something.cableone.net
something.something.com.br
something.something.cox.net
something.something.googlebot.com
My code currently print this:
72.175
73.194
73.205
73.214
abbnebraska.com
cableone.net
com.br
cox.net
googlebot.com
lstn.net
but I want it to print like this:
abbnebraska.com
cableone.net
com.br
cox.net
googlebot.com
OTHER
lstn.net
where OTHER is all the numbered domains, so any ideas how?
You really shouldn't need to split the variable into two, e.g. this regex will match the case you want to trap:
/\d{1,3}\.\d{1,3}$/ -- returns true if the string ends with two 1-3 long digits separated by a dot
but I mean if you only need to separate those domains that are not numbered you could just check the last character in the domain whether it is a letter, because TLDs cannot contain numbers, so you would do something like
/\w$/ -- if returns true, it is not a numbered domain (providing you've stripped spaces and new lines)
But I suppose it is better to be more specific in the regex, which also better illustrates the logic you are looking for in your script, so I'd use the former regex.
And actually you could do something like this:
if (my ($domain) = $name =~ /\.(\w+.\w+)$/)
{
#the domain is assigned to the variable $domain
} else {
#it is a number domain
}
Take what it currently puts, and use the regex:
/\d+\.\d+/
if it matches this, then its a pair of numbers, so remove it.
This way you'll be able to keep any words with numbers in them.
Please, please indent your code correctly, and use whitespace to separate out various bits and pieces. It'll make your code so much easier to read.
Interestingly, you mentioned that you're getting the wrong output, but the section of the code you post has no print, printf, or say statement. It looks like you're attempting to count up the various domain names.
If these are the value of $name, there are several issues here:
if ($temp =~ /(^d+\.\d+)/) {
Matches nothing. This is saying that your string starts with one or more letter d followed by a period followed by one or more digits. The ^ anchors your regular expression to the beginning of the string.
I think, but not 100% sure, you want this:
if ( $temp =~ /\d\.\d/ ) {
This will find all cases where there are two digits with a period in between them. This is the sub-pattern to /\d+\.\d+/, so both regular expressions will match the same thing.
The
$dom =~ /(\w+\.\w+)$/;
Is matching anywhere in the entire string $dom where there are two letters, digits. or underscores with a decimal between them. Is that what you want?
I also believe this may indicate an error of some sort:
my $foo = $1;
$foo = "OTHER";
$domain{$foo} ++;
This is setting $foo to whatever $dom is matching, but then immediately resets $foo to OTHER, and increments $domain{OTHER}.
We need a sample of your initial data, and maybe the actual routine that prints your output.
I need to be able to take /calendar/MyCalendar.ics where MyCalendar.ics coudl be about anything with an ICS extention and rewrite it to /feeds/ics/ics_classic.asp?MyCalendar.ics
Thanks
Regular expressions are meant for searching/matching text. Usually you will use regex to define what you search for some text manipulation tool, and then use a tool specific way to tell the tool with what to replace the text.
Regex syntax use round brackets to define capture groups inside the whole search pattern. Many search and replace tools use capture groups to define which part of the match to replace.
We can take the Java Pattern and Matcher classes as example. To complete your task with the Java Matcher you can use the following code:
Pattern p = Pattern.compile("/calendar/(.*\.(?i)ics)");
Matcher m = p.matcher(url);
String rewritenUrl = "";
if(m.matches()){
rewritenUrl = "/feeds/ics/ics_classic.asp?" + url.substring( m.start(1), m.end(1));
}
This will find the requested pattern but will only take the first regex group for creating the new string.
Here is a link to regex replacement information in (imho) a very good regex information site: http://www.regular-expressions.info/refreplace.html
C:\x>perl foo.pl
Before: a=/calendar/MyCalendar.ics
After: a=/feeds/ics/ics_classic.asp?MyCalendar.ics
...or how about this way?
(regex kind of seems like overkill for this problem)
b=/calendar/MyCalendar.ics
index=9
c=MyCalendar.ics (might want to add check for ending with '.ics')
d=/feeds/ics/ics_classic.asp?MyCalendar.ics
Here's the code:
C:\x>type foo.pl
my $a = "/calendar/MyCalendar.ics";
print "Before: a=$a\n";
my $match = (
$a =~ s|^.*/([^/]+)\.ics$|/feeds/ics/ics_classic.asp?$1.ics|i
);
if( ! $match ) {
die "Expected path/filename.ics instead of \"$a\"";
}
print "After: a=$a\n";
print "\n";
print "...or how about this way?\n";
print "(regex kind of seems like overkill for this problem)\n";
my $b = "/calendar/MyCalendar.ics";
my $index = rindex( $b, "/" ); #find last path delim.
my $c = substr( $b, $index+1 );
print "b=$b\n";
print "index=$index\n";
print "c=$c (might want to add check for ending with '.ics')\n";
my $d = "/feeds/ics/ics_classic.asp?" . $c;
print "d=$d\n";
C:\x>
General thoughts:
If you do solve this with a regex, a semi-tricky bit is making sure your capture group (the parens) exclude the path separator.
Some things to consider:
Are your paths separators always forward-slashes?
Regex seems like overkill for this; the simplest thing I can think of of to get the index of your last path separator and do simple string manipulation (2nd part of sample program).
Libraries often have routines for parsing paths. In Java I'd look at the java.io.File object, for example, specifically
getName()
Returns the name of the file or directory denoted by
this abstract pathname. This is just the last name in
the pathname's name sequence
I'm trying to parse my maillog, which contains a number of lines which look similar to the following line:
Jun 6 17:52:06 host sendmail[30794]: p569q3sX030792: to=<person#recipient.com>, ctladdr=<apache#host.com> (48/48), delay=00:00:03, xdelay=00:00:03, mailer=esmtp, pri=121354, relay=gmail-smtp-in.l.google.com. [1.2.3.4], dsn=2.0.0, stat=Sent (OK 1307354043 x8si28599066ict.63)
The rules I'm trying to apply are:
The date is always the first 2 words
The email address always occurs between " to=person#recipient.com, " however the email address might be surrounded by <>
There are some lines in the log which do not relate to a recipient, so I'd like to ignore those lines entirely.
The following code works for either rule individually, however I'm having trouble combining them:
if($_ =~ m/\ to=([<>a-zA-Z0-9\.\#]*),\ /g) {
print "$1\n";
}
if($_ =~ /^+(\S+\s+\S+\s)/g) {
print "$1\n";
}
As always, I'm not sure whether the regex I'm using above is "best practice" so feel free to point out anything I'm doing badly there too :)
Thanks!
print substr($_, 0, 7), "$1\n" if / to=(.+?), /;
Your date is in a fixed-length format, you don't need a regular expression to match it.
For the address, what you need is the part between to= and the next ,, so a non-greedy match is just what you need.
To match either with one regex, or them using syntax (regex1|regex2) together:
((?<\ to=)[<>a-zA-Z0-9\.\#]*(?=,\ )|^\S+\s+\S+\s)
The outer brackets preserve $1 being assigned the match.
The look behind (?<\ to=) and look ahead (?=,\ ) do not capture anything, so these regexes only capture your target string.