After having read this similar question and having tried my code several times, I keep on getting the same undesired output.
Let's assume the string I'm searching is "I saw wilma yesterday".
The regex should capture each word followed by an 'a' and its optional 5 following characters or spaces.
The code I wrote is the following:
$_ = "I saw wilma yesterday";
if (#m = /(\w+)a(.{5,})?/g){
print "found " . #m . " matches\n";
foreach(#m){
print "\t\"$_\"\n";
}
}
However, I kept on getting the following output:
found 2 matches
"s"
"w wilma yesterday"
while I expected to get the following one:
found 3 matches:
"saw wil"
"wilma yest"
"yesterday"
until I found out that the return values inside #m were $1 and $2, as you can notice.
Now, since the /g flag is on, and I don't think the problem is about the regex, how could I get the desired output?
You can try this pattern that allows overlapped results:
(?=\b(\w+a.{1,5}))
or
(?=(?i)\b([a-z]+a.{0,5}))
example:
use strict;
my $str = "I saw wilma yesterday";
my #matches = ($str =~ /(?=\b([a-z]+a.{0,5}))/gi);
print join("\n", #matches),"\n";
more explanations:
You can't have overlapped results with a regex since when a character is "eaten" by the regex engine it can't be eaten a second time. The trick to avoid this constraint, is to use a lookahead (that is a tool that only checks, but not matches) which can run through the string several times, and put a capturing group inside.
For another example of this behaviour, you can try the example code without the word boundary (\b) to see the result.
Firstly you want to capture everything inside the expression, i.e.:
/(\w+a(?:.{5,})?)/
Next you want to start your search from one character past where the last expression's first character matched.
The pos() function allows you to specify where a /g regex starts its search from.
$s = "I saw wilma yesterday";
while ($s =~ /(\w+a(.{0,5}))/g){
print "\t\"$1\"\n";
pos($s) = pos($s) - length($2);
}
Gives you:
"saw wil"
"wilma yest"
"yesterday"
But I don't know why you should get day and not yesterday.
Related
I want to create extract the first two words from a sentence using a Perl function in PostgreSQL. In PostgreSQL, I can do this with:
text = "I am trying to make this work";
Select substring(text from '(^\w+-\w+|^\w+(\s+)?(!|,|\&|'')?(\s+)?\w+)');
It would return "I Am"
I tried to build a Perl function in Postgresql that does the same thing.
CREATE OR REPLACE FUNCTION extract_first_two (text)
RETURNS text AS
$$
my $my_text = $_[0];
my $temp;
$pattern = '^\w+-\w+|^\w+(\s+)?(!|,|\&|'')?(\s+)?\w+)';
my $regex = qr/$pattern/;
if ($my_text=~ $regex) {
$temp = $1;
}
return $temp;
$$ LANGUAGE plperl;
But I receive a syntax error near the regular expression. I am not sure what I am doing wrong.
Extracting words is none trivial even in English. Take the following contrived example using Locale::CLDR
use 'Locale::CLDR';
my $locale = Locale::CLDR->new('en');
my #words = $locale->split_words('adf543. 123.25');
#words now contains
adf543
.
123.25
Note that the full stop after adf543 is split into a separate word but the one between 123 and 25 is kept as part of the number 123.25 even though the '.' is the same character
If gets worse when you look at non English languages and much worse when you use non Latin scripts.
You need to precisely define what you think a word is otherwise the following French gets split incorrectly.
Je avais dit «Elle a dit «Il a dit «Ni» il ya trois secondes»»
The parentheses are mismatched in our regex pattern. It has three opening parentheses and four closing ones.
Also, you have two single quotes in the middle of a singly-quoted string, so
'^\w+-\w+|^\w+(\s+)?(!|,|\&|'')?(\s+)?\w+)'
is parsed as two separate strings
'^\w+-\w+|^\w+(\s+)?(!|,|\&|'')?(\s+)?\w+)'
and
'^\w+-\w+|^\w+(\s+)?(!|,|\&|'
')?(\s+)?\w+)'
But I can't suggest how to fix it as I don't understand your intention.
Did you mean a double quote perhaps? In which case (!|,|\&|")? can be written as [!,&"]?
Update
At a rough guess I think you want this
my $regex = qr{ ^ \w++ \s* [-!,&"]* \s* \w+ }x;
$temp = $1 if $my_text=~ /($regex)/;
but I can't be sure. If you describe what you're looking for in English then I can help you better. For instance, it's unclear why you don't have question marks, full stops, and semicolons in the list of intervening punctuation.
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.
My perl is getting rusty. It only prints "matched=" but $1 is blank!?!
EDIT 1: WHo the h#$! downvoted this? There are no wrong questions. If you dont like it, move on to next one!
$crazy="abcd\r\nallo\nXYZ\n\n\nQQQ";
if ($crazy =~ m/([.\n\r]+)/gsi) {
print "matched=", $1, "\n";
} else {
print "not matched!\n";
}
EDIT 2: This is the code fragment with updated regex, works great!
$crazy="abcd\r\nallo\nXYZ\n\n\nQQQ";
if ($crazy =~ m/([\s\S]+)/gsi) {
print "matched=", $1, "\n";
} else {
print "not matched!\n";
}
EDIT 3: Haha, i see perl police strikes yet again!!!
I don't know if this is your exact problem, but inside square brackets, '.' is just looking for a period. I didn't see a period in the input, so I wondered which you meant.
Aside from the period, the rest of the character class is looking for consecutive whitespace. And as you didn't use the multiline switch, you've got newlines being counted as whitespace (and any character), but no indication to scan beyond the first record separator. But because of the way that you print it out, it also gives some indication that you meant more than the literal period, as mentioned above.
Axeman is correct; your problem is that . in a character class doesn't do what you expect.
By default, . outside a character class (and not backslashed) matches any character but a newline. If you want to include newlines, you specify the /s flag (which you seem to already have) on your regex or put the . in a (?s:...) group:
my $crazy="abcd\r\nallo\nXYZ\n\n\nQQQ";
if ($crazy =~ m/((?s:.+))/) {
print "matched=", $1, "\n";
} else {
print "not matched!\n";
}
. in a character class is a literal period, not match anything. What you really want is /(.+)/s. The /g flag says to match multiple times, but you are using the regex in scalar context, so it will only match the first item. The /i flag makes the regex case insensitive, but there are no characters with case in your regex. The \s flag makes . match newlines, and it always matches "\r", so instead of [.\n\r], you can just use ..
However, /(.+)/s will match any string with one or more characters, so you would be better off with
my $crazy="abcd\r\nallo\nXYZ\n\n\nQQQ";
if (length $crazy) {
print "matched=$crazy\n";
} else {
print "not matched!\n";
}
It is possible you meant to do something like this:
#!/usr/bin/perl
use strict;
use warnings;
my $crazy = "abcd\r\nallo\nXYZ\n\n\nQQQ";
while ($crazy =~ /(.+)[\r\n]+/g) {
print "matched=$1\n";
}
But that would probably be better phrased:
#!/usr/bin/perl
use strict;
use warnings;
my $crazy = "abcd\r\nallo\nXYZ\n\n\nQQQ";
for my $part (split /[\r\n]+/, $crazy) {
print "matched=$part\n";
}
$1 contains white space, that's why you don't see it in a print like that, just add something after it/quote it.
Example:
perl -E "qq'abcd\r\nallo\nXYZ\n\n\nQQQ'=~/([.\n\r]+)/gsi;say 'got(',length($1),qq') >$1<';"
got(2) >
<
Updated for your comments:
To match everything you can simply use /(.+)/s
[.] (dot inside a character class) does not mean "match any character", it just means match the literal . character. So in an input string without any dots,
m/([.\n\r]+)/gsi
will just match strings of \n and \r characters.
With the /s modifier, you are already asking the regex engine to include newlines with . (match any character), so you could just write
m/(.+)/gsi
I'm trying to parse my maillog, which contains a number of lines which look similar to the following line:
Jun 6 17:52:06 host sendmail[30794]: p569q3sX030792: to=<person#recipient.com>, ctladdr=<apache#host.com> (48/48), delay=00:00:03, xdelay=00:00:03, mailer=esmtp, pri=121354, relay=gmail-smtp-in.l.google.com. [1.2.3.4], dsn=2.0.0, stat=Sent (OK 1307354043 x8si28599066ict.63)
The rules I'm trying to apply are:
The date is always the first 2 words
The email address always occurs between " to=person#recipient.com, " however the email address might be surrounded by <>
There are some lines in the log which do not relate to a recipient, so I'd like to ignore those lines entirely.
The following code works for either rule individually, however I'm having trouble combining them:
if($_ =~ 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 $1 being assigned the match.
The look behind (?<\ to=) and look ahead (?=,\ ) do not capture anything, so these regexes only capture your target string.
The original string is like this:
checksession ok:6178 avg:479 avgnet:480 MaxTime:18081 fail1:19
The last part "fail1:19" may appear 0 or 1 time. And I tried to match the number after "fail1:", which is 19, using this:
($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";