I have a regex:
($value) = $line =~ /\ABC(.+?)\#/;
For input, e.g.:
(32321213321) ABC 24432.232 #Junk
Which is meant to catch the number between FD and #.
When I run it through the command line, it returns a space. Through Padre, it returns a space + the number before #.
Is there something wrong with the regex?
In your regex, you have escaped the A. This then becomes an escape sequence, an assertion \A to match the beginning of the string. Another version of the same escape is ^ . And your string does not start there, so the regex cannot match. You have another redundant escape as well, before #. The regex you need is
/ABC(.+?)#/
You can use:
$line =~ /ABC *([0-9 ]+?) *#/;
OR better:
$line =~ /ABC *(\d+(?: \d+)*) *#/;
Related
I want to search a character and replace it with a string. First, I search for ':' and replace it with 'to'. Next I want to search '$' and replace it with 'END'. This is the code that I've tried. In below code, it work for the first character but not the second character. I tried to use backslash to escape the special character '$' but it still did not work. What else can I do?
$string = "[9:8],
if ($string =~ /^.*:+/){
$stringreplaced =~ s/:/to/g;
}
elsif ($string =~ /^.*\$+/){
$stringreplaced =~ s/\$/END/g;
}
First of all, the code you posted doesn't even compile, yet you say it actually ran. Only post code that you've run.
Second, you're matching against the wrong string. You're checking if $string contains the character, but you replace the characters in $stringreplaced. ALWAYS use use strict; use warnings;. This would have caught this error.
Third, you only check if the character (: or $) is on the first line. This is because . doesn't match line feeds without /s.
Finally, You only check if the string contains $ if it doesn't contain : because you used elsif.
The following is all you need:
$string =~ s/:/to/g;
$string =~ s/\$/END/g;
There are numerous questions on how to do a multiline regex in Perl. Most of them mention the s switch that makes a dot match a newline. However, I want to match an exact phrase (so, not a pattern) and I don't know where the newlines will be. So the question is: can you ignore newlines, instead of matching them with .?
MWE:
$pattern = "Match this exact phrase across newlines";
$text1 = "Match\nthis exact\nphrase across newlines";
$text2 = "Match this\nexact phra\nse across\nnewlines";
$text3 = "Keep any newlines\nMatch this exact\nphrase across newlines\noutside\nof the match";
$text1 =~ s/$pattern/replacement text/s;
$text2 =~ s/$pattern/replacement text/s;
$text3 =~ s/$pattern/replacement text/s;
print "$text1\n---\n$text2\n---\n$text3\n";
I can put dots in the pattern instead of spaces ("Match.this.exact.phrase") but that does not work for the second example. I can delete all newlines as preprocessing but I would like to keep newlines that are not part of the match (as in the third example).
Desired output:
replacement text
---
replacement text
---
Keep any newlines
replacement text
outside
of the match
Just replace the literal spaces with a character class that matches a space or a newline:
$pattern = "Match[ \n]this[ \n]exact[ \n]phrase[ \n]across[ \n]newlines";
Or, if you want to be more lenient, use \s or \s+ instead, since \s also matches newlines.
Most of the time, you are treating newlines as spaces. If that's all you wanted to do, all you'd need is
$text =~ s/\n/ /g;
$text =~ /\Q$text_to_find/ # or $text =~ /$regex_pattern_to_match/
Then there's the one time you want to ignore it. If that's all you wanted to do, all you'd need is
$text =~ s/\n//g;
$text =~ /\Q$text_to_find/ # or $text =~ /$regex_pattern_to_match/
Doing both is next to impossible if you have a regex pattern to match. But you seem to want to match literal text, so that opens up some possibilities.
( my $pattern = $text_to_find )
=~ s/(.)/ $1 eq " " ? "[ \\n]" : "\\n?" . quotemeta($1) /seg;
$pattern =~ s/^\\n\?//;
$text =~ /$pattern/
It sounds like you want to change your "exact" pattern to match newlines anywhere, and also to allow newlines instead of spaces. So change your pattern to do so:
$pattern = "Match this exact phrase across newlines";
$pattern =~ s/\S\K\B/\n?/g;
$pattern =~ s/ /[ \n]/g;
It certainly is ugly, but it works:
M\n?a\n?t\n?c\n?h\st\n?h\n?i\n?s\se\n?x\n?a\n?ct\sp\n?h\n?r\n?a\n?s\n?e\sa\n?c\n?r\n?o\n?s\n?s\sn\n?e\n?w\n?l\n?i\n?n\n?e\n?s
For every pair of letters inside a word, allow a newline between them with \n?. And replace each space in your regex with \s.
May not be usable, but it gets the job done ;)
Check it out at regex101.
I am trying to remove all words that contain two keys (in Perl).
For example, the string
garble variable10 variable1 vssx vddx xi_21_vssx vddx_garble_21 xi_blahvssx_grbl_2
Should become
garble variable10 variable1
To just remove the normal, unappended/prepended keys is easy:
$var =~ s/(vssx|vddx)/ /g;
However I cannot figure out how to get it to remove the entire xi_21_vssx part. I tried:
$var =~ s/\s.*(vssx|vddx).*\s/ /g
Which does not work correctly. I do not understand why... it seems like \s should match the space, then .* matches anything up to one of the patterns, then the pattern, then .* matches anything preceding the pattern until the next space.
I also tried replacing \s (whitespace) with \b (word boundary) but it also did it work. Another attempt:
$var =~ s/ .*(vssx|vddx).* / /g
$var =~ s/(\s.*vssx.*\s|\s.*vddx.*\s)/ /g
As well as a few other mungings.
Any pointers/help would be greatly appreciated.
-John
I think the regex will just be
$var =~ s/\S*(vssx|vddx)\S*/ /g;
You can use
\s*\S*(?:vssx|vddx)\S*\s*
The problem with your regex were:
The .* should have been non-greedy.
The .* in front of (vssx|vddx) mustn't match whitespace characters, so you have to use \S*.
Note that there's no way to properly preserve the space between words - i.e. a vssx b will become ab.
regex101 demo.
I am trying to remove all words that [...]
This type of problem lends itself well to grep, which can be used to find the elements in a list that match a condition. You can use split to convert your string to a list of words and then filter it like this:
use strict;
use warnings;
use 5.010;
my $string = 'garble variable10 variable1 vssx vddx xi_21_vssx vddx_garble_21 xi_blahvssx_grbl_2';
my #words = split ' ', $string;
my #filtered = grep { $_ !~ /(?:vssx|vddx)/ } #words;
say "#filtered";
Output:
garble variable10 variable1
Try this as the regex:
\b[\w]*(vssx|vddx)[\w]*\b
How can I find the first substring until I find the first digit?
Example:
my $string = 'AAAA_BBBB_12_13_14' ;
Result expected: 'AAAA_BBBB_'
Judging from the tags you want to use a regular expression. So let's build this up.
We want to match from the beginning of the string so we anchor with a ^ metacharacter at the beginning
We want to match anything but digits so we look at the character classes and find out this is \D
We want 1 or more of these so we use the + quantifier which means 1 or more of the previous part of the pattern.
This gives us the following regular expression:
^\D+
Which we can use in code like so:
my $string = 'AAAA_BBBB_12_13_14';
$string =~ /^\D+/;
my $result = $&;
Most people got half of the answer right, but they missed several key points.
You can only trust the match variables after a successful match. Don't use them unless you know you had a successful match.
The $&, $``, and$'` have well known performance penalties across all regexes in your program.
You need to anchor the match to the beginning of the string. Since Perl now has user-settable default match flags, you want to stay away from the ^ beginning of line anchor. The \A beginning of string anchor won't change what it does even with default flags.
This would work:
my $substring = $string =~ m/\A(\D+)/ ? $1 : undef;
If you really wanted to use something like $&, use Perl 5.10's per-match version instead. The /p switch provides non-global-perfomance-sucking versions:
my $substring = $string =~ m/\A\D+/p ? ${^MATCH} : undef;
If you're worried about what might be in \D, you can specify the character class yourself instead of using the shortcut:
my $substring = $string =~ m/\A[^0-9]+/p ? ${^MATCH} : undef;
I don't particularly like the conditional operator here, so I would probably use the match in list context:
my( $substring ) = $string =~ m/\A([^0-9]+)/;
If there must be a number in the string (so, you don't match an entire string that has no digits, you can throw in a lookahead, which won't be part of the capture:
my( $substring ) = $string =~ m/\A([^0-9]+)(?=[0-9])/;
$str =~ /(\d)/; print $`;
This code print string, which stand before matching
perl -le '$string=q(AAAA_BBBB_12_13_14);$string=~m{(\D+)} and print $1'
AAAA_BBBB_
I'm new to both Perl and reg-ex's, and I'm trying to remove the first word in a string (or the first word in a line in a text file) , along with any whitespace that follows it.
For example, if my string is 'one two abd123words', I want to remove 'one '.
The code I was trying is: $line =~/(\S)$/i;
but this only gives me the last word.
If it makes any difference, the word i'm trying to remove is an input, and stored as $arg.
To remove the first word of each line use:
$line =~ s/^\S+\s*//;
EDIT for a explanation:
s/.../.../ # Substitute command.
^ # (Zero-width) Begin of line.
\S+ # Non-space characters.
\s* # Blank-space characters.
// # Substitute with nothing, so remove them.
You mean, like this? :
my $line = 'one two abd123words';
$line =~ s/^\s*\S+\s*//;
# now $line is 'two abd123words'
(That removes any initial whitespace, followed by a one or more non-whitespace characters, followed by any newly-initial whitespace.)
In one-liner form:
$ perl -pi.bak -e 's{^\s*\S+\s*}//' file.txt