I have a string "/project/pkt/sw/tool/xxx" and should be removed "sw/tool/xxx" from the original string.
Please let suggest me how to do it?
Input:
"/project/pkt/sw/tool/xxx";
Desired Output
"/project/pkt/"
Code
my $ str = "project/pkt/sw/tool/xxx";
$str =~ s|\w*/\w*/\w*$||;
print $str;
I am getting same original string here, please let me know how to remove last three words from the original string.
The following regex modifies $str to remove the last three words as defined in the question.
$str =~ s|\w*/\w*/\w*$||;
Related
I need a regex to match \' <---- literally backslash apostrophe.
my $line = '\'this';
$line =~ s/(\o{134})(\o{047})/\\\\'/g;
$line =~ s/\\'/\\\\'/g;
$line =~ s/[\\][']/\\\\'/g;
printf('%s',$line);
print "\n";
All I get out of this is
'this
When what I want is
\\'this
This occurs whether the string is declared using ' or ". This was a test script for tracking down a file parsing bug. I wanted to confirm that the regex was working as expected.
I don't know if when the backslash apostrophe is parsed by the regex it is not treated as 2 characters, but is instead treated as an escaped apostrophe.
Either way. what is the best way to match \' and print out \\'? I don't want to escape any other back slashes or apostrophes and I can't change the text I am parsing, just the way it is handled and outputted.
s/\\'/\\\\'/g
All three of your patterns match a backslash followed by a quote, the above being the simplest.
Your testing was in vain because your string doesn't contain any backslashes. Both string literals "\'this" (from earlier edit) and '\'this' (from later edit) produce the string 'this.
say "\'this"; # 'this
say '\'this'; # 'this
To produce the string \'this, you could use either of the following string literals (among others):
"\\'this"
'\\\'this'
say "\\'this"; # \'this
say '\\\'this'; # \'this
The answer is, of course
s/[\\][']/\\\\'/g
This will match
\'this
And substitute with this
\\'this
This was the only way I could get it to work.
Perl
Too much "regexing" in your snippet. Try:
my $line = '\'this';
$line =~ s/'/\\\\\'/g;
printf('%s',$line);
print "\n";
# \\'this
or... if you want another mode:
my $line = '\'this';
$line =~ s/'/\\'/g;
printf('%s',$line);
print "\n";
# \'this
an excerpt of a big binary file ($data) looks like this:
\n1ax943021C xxx\t2447\t5
\n1ax951605B yyy\t10400\t6
\n1ax919275 G2L zzz\t6845\t6
The first 25 characters contain an article number, filled with spaces. How can I convert all spaces between the article numbers and the next column into a \x09 ? Note the one or more spaces between different parts of the article number.
I tried a workaround, but that overwrites the article number with ".{25}xxx»"
$data =~ s/\n.{25}/\n.{25}xxx/g
Anyone able to help?
Thanks so much!
Gary
You can use unpack for fixed width data:
use strict;
use warnings;
use Data::Dumper;
$Data::Dumper::Useqq=1;
print Dumper $_ for map join("\t", unpack("A25A*")), <DATA>;
__DATA__
1ax943021C xxx 2447 5
1ax951605B yyy 10400 6
1ax919275 G2L zzz 6845 6
Output:
$VAR1 = "1ax943021C\txxx\t2447\t5";
$VAR1 = "1ax951605B\tyyy\t10400\t6";
$VAR1 = "1ax919275 G2L\tzzz\t6845\t6";
Note that Data::Dumper's Useqq option prints whitecharacters in their escaped form.
Basically what I do here is take each line, unpack it, using 2 strings of space padded text (which removes all excess space), join those strings back together with tab and print them. Note also that this preserves the space inside the last string.
I interpret the question as there being a 25 character wide field that should have its trailing spaces stripped and then delimited by a tab character before the next field. Spaces within the article number should otherwise be preserved (like "1ax919275 G2L").
The following construct should do the trick:
$data =~ s/^(.{25})/{$t=$1;$t=~s! *$!\t!;$t}/emg;
That matches 25 characters from the beginning of each line in the data, then evaluates an expression for each article number by stripping its trailing spaces and appending a tab character.
Have a try with:
$data =~ s/ +/\t/g;
Not sure exactly what you what - this will match the two columns and print them out - with all the original spaces. Let me know the desired output and I will fix it for you...
#!/usr/bin/perl -w
use strict;
my #file = ('\n1ax943021C xxx\t2447\t5', '\n1ax951605B yyy\t10400\t6',
'\n1ax919275 G2L zzz\t6845\t6');
foreach (#file) {
my ($match1, $match2) = ($_ =~ /(\\n.{25})(.*)/);
print "$match1'[insertsomethinghere]'$match2\n";
}
Output:
\n1ax943021C '[insertsomethinghere]'xxx\t2447\t5
\n1ax951605B '[insertsomethinghere]'yyy\t10400\t6
\n1ax919275 G2L '[insertsomethinghere]'zzz\t6845\t6
Trying to extract the alphanumeric characters from this string:
A_phase_I-II,_open-req_project_id_PX15RAD001
The problem is: the term PX15RAD001 can occur anywhere in the string.
Trying to extract the alpha-numeric part using the below expression. But this returns the entire string. I thought Alum was a valid keyword for alpha-numerics. Is that not the case?
(my $string = $line ) =~ s/\P{Alnum}//g;
print $string;
How can I extract the alphanumeric part of the afore mentioned string?
Thanks in advance.
-simak
At the end as per your input:
> echo "A_phase_I-II,_open-req_project_id_PX15RAD001"|perl -lne 'print $1 if(/id_([A-Z0-9]*)/)'
PX15RAD001
In the middle:
> echo "A_phase_I-II,_open-req_id_PX15RAD001_project" | perl -lne 'print $1 if(/id_([A-Z0-9]*)/)'
PX15RAD001
or in your terms:
$line=~m/id_([A-Z0-9]*)/g;
print $1;
Here are some testcases, produced with the comments of #Vijay s Answer:
my #line = (
'A_phase_I-II,_open-req_project_id_PX15RAD001',
'_PX15RAD001_A_phase_I-II,_open-req_project_id',
'A_pha3333se_I-II,_ope_PX15RAD001_n-req_project',
'A_phase_I-II,_PX15RAD001_open-req_projec123123123t_id',
'A_phase_I-II_PX15RAD001_roject_id'
);
foreach my $string ( #line ) {
$string =~ m{_([^_]{10})_?}g;
print $1 . "\n" if $1;
}
These kinds of questions are hard to answer because there is not enough information. What information we have is:
You say your target string is "alphanumeric", but the entire input string is alphanumeric, except for some punctuation, so that really doesn't tell us anything.
You say it is 12 characters long, but the sample you show is 10 characters long.
You seem to think that "alphanumeric" does not include underscore.
So, the reliable information I can sense from you is:
Target string is always delimited by underscore _
Target string is 10-12 characters, all alphanumeric except underscore.
The "reliable" solution based on this rather skimpy information is:
my $str = "A_phase_I-II,_open-req_project_id_PX15RAD001";
for my $field (split /_/, $str) {
if (length($field) <= 12 and
length($field) >= 10 and # field is 10-12 characters
$field !~ /\W/) { # and contains no non-alphanumerics
# do something
}
}
By splitting on underscore, we can easily isolate each field in the string and perform simpler tests on it, such as the ones above.
my code is as follows
my $string = $cells[71];
print $string;
this prints the string but where spaces should be there is a box with 01 10 in it. I opened it in Notepad++ and the box turned into a black GS (which i am assuming is group separator).
I looked online and it said to use:
s/[^[:print:]]+//g
but when i set the string to:
my $string =~s/[^[:print:]]+//g
and I run the program i get:
4294967295
How do i resolve this?
I did what HOBBS said and it worked... thanks :)
Is there anyway I could print an enter where each of these characters are ( the box with 1001)?
When doing a regex match, you need to be careful to write $var =~ /pattern/, not $var = ~ /pattern/. When you use the second one, you're doing /pattern/, which is a regex match against $_, returning a number in scalar context. Then you do ~, which takes the bitwise inverse of that number, then ($var =) you assign that result to $var. Not what you wanted at all.
You have to assign the variable first, then do the substitution:
my $string = $cells[71];
$string =~ s/[^[:print:]]+//g;
I have
print $str;
abcd*%1234$sdfsd..#d
The string would always have only one continuous stretch of numbers, like 1234 in this case. Rest all will be either alphabets or other special characters.
How can I extract the number (1234 in this case) and store it back in str?
This page suggests that I should use \d, but how?
If you don't want to modify the original string, you can extract the numbers by capturing them in the regex, using subpatterns. In list context, a regular expression returns the matches defined in the subpatterns.
my $str = 'abc 123 x456xy 789foo';
my ($first_num) = $str =~ /(\d+)/; # 123
my #all_nums = $str =~ /(\d+)/g; # (123, 456, 789)
$str =~ s/\D//g;
This removes all nondigit characters from the string. That's all that you need to do.
EDIT: if Unicode digits in other scripts may be present, a better solution is:
$str =~ s/[^0-9]//g;
If you wanted to do it the destructive way, this is the fastest way to do it.
$str =~ tr/0-9//cd;
translate all characters in the complement of 0-9 to nothing, delete them.
The one caveat to this approach, and Phillip Potter's, is that were there another group of digits further down the string, they would be concatenated with the first group of digits. So it's not clear that you would want to do this.
The surefire way to get one and only one group of digits is
( $str ) = $str =~ /(\d+)/;
The match, in a list context returns a list of captures. The parens around $str are simply to put the expression in a list context and assign the first capture to $str.
Personally, I would do it like this:
$s =~ /([0-9]+)/;
print $1;
$1 will contain the first group matched the given regular expression (the part in round brackets).