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I want to be able to match all of the following strings to my regex below. It doesnt seem to be working. Any suggestions?
Strings to compare :
5878ce43aa3f1e1d713427d118115310 -1 Script Kiddie <perm>
f939f88b50fa5f0099b6751e7be27761 -1 Hacking <perm>
468f6634c5a9b00b5b3872dd6437143f 1356474103 Being Annoying <7day>
This is my perl code. It isnt working at the moment. Any suggestions?
my $bn_re = q{(.+?) (\d+) (.+?)};
If the first two fields are always without whitespace in them, you can use split to great effect, using the LIMIT option to only get three fields:
my ($str, $num, $other) = split ' ', $_, 3;
That is, assuming you read the file something like this:
while (<>) {
... # your code here
}
Also, this:
my $bn_re = q{(.+?) (\d+) (.+?)};
is not a regex. You may be confusing q() with qr(). You may also be confusing the functionality of
$str =~ $bn_re;
Which will automagically include the regex in a match operator m//. But you should use qr(). The q() operator does what the single quote does.
Also, you should be aware that .+? will match a single char if you allow it. As it does at the end of your "regex". At the end of your string, either do
... (.+)/ # matching greedily
... (.+?)$/ # using anchor to end of string
$bn_re =~ /[0-9a-z]+?\s[-0-9]+\s[\w\s]+?[<>a-z0-9]+?/i
Related
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Writing a script to convert a Windows host file into a CSV. Used RegEx to get to this stage:
1.1.1.1,,server1,,,
2.2.2.2,,server2
3.3.3.3,,server3
4.5.6.7,,server4,,server5,,server6,,server7,,
8.8.8.8,,server8
9.9.9.9,server9
I need some RegEx that can remove the duplicate commas (in sequence) so it would look like this:
1.1.1.1,server1,
2.2.2.2,server2
3.3.3.3,server3
4.5.6.7,server4,server5,server6,server7,
8.8.8.8,server8
9.9.9.9,server9
Will also need to remove the comma at the end of each line (if there is one) but think this will be simpler to do.
The regex for your first task of removing duplicate commas was already provided in the comments above, but if you also want to remove trailing commas at the end of the line, you can use this to solve both problems at once:
(?m),(?=,|$)
Explanation:
(?m) # turn on multiline mode ($ matches end-of-line, not just end-of-string)
, # Match a comma
(?= # only if followed by
, # another comma
| # or
$ # the end of the string.
) # End of lookahead assertion
Test it live on regex101.com.
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If i had a string:
my $string = "a/hello/bye/d";
I would like to add in a "\" symbol infront of every "/" symbol found inside the string. Are there any possible ways to do this?
Example:
$string = "a\/hello\/bye\/d";
Change the regular expression delimiter to a | and then substitute all forward slashes / with back-slash forward-slash \/. The back-slash must be escaped, since it is itself the 'escape' character. So \\/; The trailing g means perform the replacement everywhere, the leading s means substitute: s|\|\\/|g.
Have a read of perldoc perlretut for a friendly introduction to regular expressions.
my $string = "a/hello/bye/d";
$string =~ s|/|\\/|g;
print $string . "\n";
output
a\/hello\/bye\/d
You could use quotemeta() to achieve that.
: perl -e 'my $string = "a/hello/bye/d"; print quotemeta($string); print "\n"'
a\/hello\/bye\/d
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I have a groovy value like the following:
"hostname blah blah blah"
When that occurs, I want to keep the hostname, but delete everything after it including the first space.
I'm having difficulty with the regex. Any ideas?
Thanks!
To extract the hostname and/or a part of a string by a regular expression can be done in a straightforward way.
If the values are separated by whitespace then can just split the string by a space character and use the first part. Split() method takes in a regular expression as its first argument.
String s = "hostname blah blah blah"
String[] parts = s.split(' ', 2)
if (parts) {
println parts[0]
s = parts[0] // if want to replace the variale s with just the hostname
}
Here is an alternate example using a regular expression to pull out the hostname part of the string value. The first part of the regular expression (\S+) is looking for a sequence of non-whitespace characters. For a set of possible hostnames, a stricter expression could be something like (\w+(\.\w+)*).
import java.util.regex.Matcher
if (s =~ /^(\S+)\s/) {
println Matcher.lastMatcher.group(1)
}
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I am trying to replace the string which matches the following pattern:
my $string ="+++details of candidate
++name of candidate
+age of candidate
+++idiot
+idi";
That string should be replaced with:
my $string ="hii
how
fine
hii
fine";
How can I accomplish this?
maybe:
$string =~ s/(\++)([^\+\n]*)/(qw(fine how hii))[length($1)-1]/ge;
Use
$string =~ s/([^+]*)[+]+((\w+).*)([^ \r\n]*|$)/$1<div class="$3">$2<\/div>/g;
$string =~ s/ class="details"//g;
Explanation
searches for a sequence of + ...
... , copies preceding charcaters other than + ...
... , appends content following the + sequence, wraps it in a div tag with a class attribute containing the first word of said content ...
... , where the end of 'content' is marked by a sequence of whitespace, line feed or carriage return; ...
... repeats that substitution effectively for every line and concats the results;
In the second substitution, deletes the unwanted class attributes (in this case, 'details').
Notes
The 'details' could have been singled out differently,eg. by counting the + characters.
There are other possible solutions, in particular those employing advanced features of regex engines like positive lookbehind.
Caveat
Consider the solution as it stands as a proof of concept:
The specification of the context for substitutions in the regex crucially depends on the context in which the regex will be applied, ie. the set of potential strings against it will have to match.
The pattern that you have provided is a small set of samples. Literally taken, the pattern would precisely be the set of strings you've provided. The generalisations expressed by the regex are my own assumptions on the actual nature of your pattern, of course.
Update
OP updated his question and requested a different replacement:
$string =~ s/[+]{3}((\w+).*)([^ \r\n]*|$)/hii/g;
$string =~ s/[+]{2}((\w+).*)([^ \r\n]*|$)/how/g;
$string =~ s/[+]{1}((\w+).*)([^ \r\n]*|$)/fine/g;
The order by decreasing length of + sequences in the pattern is significant.
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I have a big list of simple expressions (2Mb file). For example:
11.*;112.*;113.*;12.*;123.*
I need to remove unnecessary expressions and come up with this:
11.*;12.*
bash version would be appreciated. thanks in advance
Here is something in Perl that would work, provided the only wildcards in your pattern are of the form .*:
#!/usr/bin/perl
use strict;
use warnings;
my %terms;
{
local $/;
%terms = map {$_ => 1} split /;|\n/, <>;
}
foreach my $k1 (keys %terms)
{
foreach my $k2 (keys %terms)
{
if ($k1 ne $k2 and $k1 =~ /^$k2$/)
{
delete $terms{$k1};
last;
}
}
}
print join ';', keys %terms;
It accepts your file as a command line argument.
This works by comparing keys to each other. In each comparison, one key is treated as a string and the other key is evaluated as a regex. This takes advantage of the fact the .* matches anything--including the literal characters .*. Thus an expression that matches the literal string of another pattern will also match all strings that pattern would match.
It will work even if there are multiple .* terms in a single pattern. For example, it correctly determines that 1.*1.* matches everything that 11.* matches, deleting the latter.
However, this is kind of a hacky simplification and will not work if you introduce other regex patterns. There is no easy solution to this problem in general, because you would have to parse all patterns and figure out what each would match.