This is driving me nuts!
I read a txt file into a string called $filestring.
sysopen(handle, $filepath, O_RDONLY) or die "WHAT?";
local $/ = undef;
my $filestring = <handle>;
I made a pattern variable called $regex which is generated dynamically, but takes on the format:
(a)|(b)|(c)
I search the text for patterns separated by a space
while($filestring =~ m/($regex)\s($regex)/g){
print "Match: $1 $2\n";
#...more stuff
}
Most of the matches are valid, but for some reason I get a match like the following every once and a while:
Match: and
whereas a normal match should have two outputs like the following:
Match: , and
Does anyone know what might be causing this?
EDIT: it appears that the NULL character is being matched in the pattern.
Each of the alternatives in your regexp is a separate capture group. The whole regexp looks like:
((a)|(b)|(c))\s((a)|(b)|(c))
12 3 4 56 7 8
I've notated it with the capture group number for each piece of the regexp.
So if $filestring is b a, $1 will be b, $2 will be the empty strying because nothing matched (a).
To avoid this, you should use non-capturing groups for the alternatives:
((?:a)|(?:b)|(?:c))\s((?:a)|(?:b)|(?:c))
Related
I want to use a Perl regex to extract certain values from file names.
They have the following (valid) names:
testImrrFoo_Bar001_off
testImrrFooBar_bar000_m030
testImrrFooBar_bar231_p030
From the above I would like to extract the first 3 digits (always guaranteed to be 3), and the last part of the string, after the last _ (which is either off, or (m orp) followed by 3 digits). So the first thing I would be extracting are 3 digits, the second a string.
And I came out with the following method (I realise this might be not the most optimal/nicest one):
my $marker = '^testImrr[a-zA-z_]+\d{3}_(off|(m|p)\d{3})$';
if ($str =~ m/$marker/)
{
print "1=$1 2=$2";
}
Where only $1 has a valid result (namely the last bit of info I want), but $2 turns out empty. Any ideas on how to get those 3 digits in the middle?
You were almost there.
Just :
- capture the three digits by adding parenthesis around: (\d{3})
- don't capture m|p by adding ?: after the parenthesis before it ((?:m|p)), or by using [mp] instead:
^testImrr[a-zA-z_]+(\d{3})_(off|[mp]\d{3})$
And you'll get :
1=001 2=off
1=000 2=m030
1=231 2=p030
You can capture both at once, e.g with
if ($str =~ /(\d{3})_(off|(?:m|p)\d{3})$/ ) {
print "1=$1, 2=$2".$/;
}
You example has two capture groups as well (off|(m|p)\d{3} and m|p). In case of you first filename, for the second capture group nothing is catched due to matching the other branch. For non-capturing groups use (?:yourgroup).
There's really no need for regular expressions when a simple split and substr will suffice:
use strict;
use warnings;
while (<DATA>) {
chomp;
my #fields = split(/_/);
my $digits = substr($fields[1], -3);
print "1=$digits 2=$fields[2]\n";
}
__DATA__
testImrrFoo_Bar001_off
testImrrFooBar_bar000_m030
testImrrFooBar_bar231_p030
Output:
1=001 2=off
1=000 2=m030
1=231 2=p030
I want to use a regular expression to check a string to make sure 4 and 5 are in order. I thought I could do this by doing
'$string =~ m/.45./'
I think I am going wrong somewhere. I am very new to Perl. I would honestly like to put it in an array and search through it and find out that way, but I'm assuming there is a much easier way to do it with regex.
print "input please:\n";
$input = <STDIN>;
chop($input);
if ($input =~ m/45/ and $input =~ m/5./) {
print "works";
}
else {
print "nata";
}
EDIT: Added Info
I just want 4 and 5 in order, but if 5 comes before at all say 322195458900023 is the number then where 545 is a problem 5 always have to come right after 4.
Assuming you want to match any string that contains two digits where the first digit is smaller than the second:
There is an obscure feature called "postponed regular expressions". We can include code inside a regular expression with
(??{CODE})
and the value of that code is interpolated into the regex.
The special verb (*FAIL) makes sure that the match fails (in fact only the current branch). We can combine this into following one-liner:
perl -ne'print /(\d)(\d)(??{$1<$2 ? "" : "(*FAIL)"})/ ? "yes\n" :"no\n"'
It prints yes when the current line contains two digits where the first digit is smaller than the second digit, and no when this is not the case.
The regex explained:
m{
(\d) # match a number, save it in $1
(\d) # match another number, save it in $2
(??{ # start postponed regex
$1 < $2 # if $1 is smaller than $2
? "" # then return the empty string (i.e. succeed)
: "(*FAIL)" # else return the *FAIL verb
}) # close postponed regex
}x; # /x modifier so I could use spaces and comments
However, this is a bit advanced and masochistic; using an array is (1) far easier to understand, and (2) probably better anyway. But it is still possible using only regexes.
Edit
Here is a way to make sure that no 5 is followed by a 4:
/^(?:[^5]+|5(?=[^4]|$))*$/
This reads as: The string is composed from any number (zero or more) characters that are not a five, or a five that is followed by either a character that is not a four or the five is the end of the string.
This regex is also a possibility:
/^(?:[^45]+|45)*$/
it allows any characters in the string that are not 4 or 5, or the sequence 45. I.e., there are no single 4s or 5s allowed.
You just need to match all 5 and search fails, where preceded is not 4:
if( $str =~ /(?<!4)5/ ) {
#Fail
}
I am looking to do a 2-step regular expression look-up in Perl, I have text that looks like this:
here is some text 9337 more text AA 2214 and some 1190 more BB stuff 8790 words
I also have a hash with the following values:
%my_hash = ( 9337 => 'AA', 2214 => 'BB', 8790 => 'CC' );
Here's what I need to do:
Find a number
Look up the text code for the number using my_hash
Check if the text code appears within 50 characters of the identified number, and if true print the result
So the output I'm looking for is:
Found 9337, matches 'AA'
Found 2214, matches 'BB'
Found 1190, no matches
Found 8790, no matches
Here's what I have so far:
while ( $text =~ /(\d+)(.{1,50})/g ) {
$num = $1;
$text_after_num = $2;
$search_for = $my_hash{$num};
if ( $text_after_num =~ /($search_for)/ ) {
print "Found $num, matches $search_for\n";
}
else {
print "Found $num, no matches\n";
}
This sort of works, except that the only correct match is 9337; the code doesn't match 2214. I think the reason is that the regular expression match on 9337 is including 50 characters after the number for the second-step match, and then when the regex engine starts again it is starting from a point after the 2214. Is there an easy way to fix this? I think the \G modifier can help me here, but I don't quite see how.
Any suggestions or help would be great.
You have a problem with greediness. The 1,50 will consume as much as it can. Your regex should be /(\d+)(.+?)(?=($|\d))/
To explain, the question mark will make the multiple match non-greedy (it will stop as soon as the next pattern is matched - the next pattern gets precedence). The ?= is a lookahead operator to say "check if the next element is a digit. If so, match but do not consume." This allows the first digit to get picked up by the beginning of the regex and be put into the next matched pattern.
[EDIT]
I added an optional end value to the lookahead so that it wouldn't die on the last match.
Just use :
/\b\d+\b/g
Why match everything if you don't need to? You should use other functions to determine where the number is :
/(?=9337.{1,50}AA)/
This will fail if AA is further than 50 chars away from the end of 9337. Of course you will have to interpolate your variables to match your hashe's keys and values. This was just an example for your first key/value pair.
I have list (multiline text string) with same number of line (order of items may differ in many ways and numbers of line may be however):
Ardei
Mere
Pere
Ardei
Castraveti
I want to find 2 th occurrence of a match line that contain 'Ardei' and replace name of item with another name and, separately in another regex, find 1 st occurrence of 'Ardei' and replace name with something else (perl).
Let's say you want to replace the 2nd "Ardei" with "XYZ". You could do that like this (PCRE syntax):
^(?s)(.*?Ardei.*?)Ardei
and replace it with:
$1XYZ
The $1 contains everything that is captured in (.*?Ardei.*?) and the (?s) will cause the . to match really every character (also line break chars).
A little demo:
#!/usr/bin/perl -w
my $text = 'Ardei
Mere
Pere
Ardei
Castraveti
Ardei';
$text =~ s/^(?s)(.*?Ardei.*?)Ardei/$1XYZ/;
# or just: $text =~ s/^(.*?Ardei.*?)Ardei/$1XYZ/s;
print $text;
will print:
Ardei
Mere
Pere
XYZ
Castraveti
Ardei
Ardei[\W\w]*?(Ardei)
will match exactly the second "Ardei" by its \1, so you can use it to replace exactly the second instance.
The original string is like this:
checksession ok:6178 avg:479 avgnet:480 MaxTime:18081 fail1:19
The last part "fail1:19" may appear 0 or 1 time. And I tried to match the number after "fail1:", which is 19, using this:
($reg_suc, $reg_fail) = ($1, $2) if $line =~ /^checksession\s+ok:(\d+).*(fail1:(\d+))?/;
It doesn't work. The $2 variable is empty even if the "fail1:19" does exist. If I delete the "?", it can match only if the "fail1:19" part exists. The $2 variable will be "fail1:19". But if the "fail1:19" part doesn't exist, $1 and $2 neither match. This is incorrect.
How can I rewrite this pattern to capture the 2 number correctly? That means when the "fail1:19" part exist, two numbers will be recorded, and when it doesn't exit, only the number after "ok:" will be recorded.
First, the number in fail field would end in $3, as those variables are filled according to opening parentheses. Second, as codaddict shows, the .* construct in RE is hungry, so it will eat even the fail... part. Third, you can avoid numbered variables like this:
my $line = "checksession ok:6178 avg:479 avgnet:480 MaxTime:18081 fail1:19";
if(my ($reg_suc, $reg_fail, $addend)
= $line =~ /^checksession\s+ok:(\d+).*?(fail1:(\d+))?$/
) {
warn "$reg_suc\n$reg_fail\n$addend\n";
}
Try the regex:
^checksession\s+ok:(\d+).*?(fail1:(\d+))?$
Ideone Link
Changes made:
.* in the middle has been made
non-greedy and
$ (end anchor) has been added.
As a result of above changes .*? will try to consume as little as possible and the end anchor forces the regex to match till the end of the string, matching fail1:number if present.
I think this is one of the few cases where a split is actually more robust than a regex:
$bar[0]="checksession ok:6178 avg:479 avgnet:480 MaxTime:18081 fail1:19";
$bar[1]="checksession ok:6178 avg:479 avgnet:480 MaxTime:18081";
for $line (#bar){
(#fields) = split/ /,$line;
$reg_suc = $fields[1];
$reg_fail = $fields[5];
print "$reg_suc $reg_fail\n";
}
I try to avoid the non-greedy modifier. It often bites back. Kudos for suggesting split, but I'd go a step further:
my %rec = split /\s+|:/, ( $line =~ /^checksession (.*)/ )[0];
print "$rec{ok} $rec{fail1}\n";