I am starting up with Perl and confused on how to render unicode characters given a hex string variable.
#!/usr/bin/perl
use warnings;
foreach my $i (0..10000) {
my $hex = sprintf("%X", $i);
print("unicode of $i is \x{$hex}\n");
}
print("\x{2620}\n");
print("\x{BEEF}\n");
Gives me the warning: Illegal hexadecimal digit '$' ignored at perl.pl line 9.
and no value prints for \x{$hex}
Both chr($num) and pack('W', $num) produce a string consisting of the single character with the specified value, just like "\x{XXXX}" does.
As such, you can use
print("unicode of $i is ".chr(hex($hex))."\n");
or just
print("unicode of $i is ".chr($i)."\n");
Note that your program makes no sense without
use open ':std', ':encoding(UTF-8)';
Yup. You can't do that. No variable interpolation allowed in the middle of a \x like that. You can use chr() to get that character though.
Randal's answer is correct. For more info, you might want to read perluniintro.
From there, you can find, for example:
At run-time you can use:
use charnames ();
my $hebrew_alef_from_name
= charnames::string_vianame("HEBREW LETTER ALEF");
my $hebrew_alef_from_code_point = charnames::string_vianame("U+05D0");
Related
I am trying to extract AAA and BBB from the output of the command "dspmq".
$dspmq <- this command gives output as -->
QMNAME(AAA) STATUS(Running)
QMNAME(BBB) STATUS(Running)
But it doesn't work with the below code.
perl -e 'use Data::Dumper qw(Dumper);my #qmgrlist = `dspmq`;$size = #qmgrlist;foreach my $i (#qmgrlist){my #temp1 = split /QMNAME\(/, $i;print #temp1;}'
AAA) STATUS(Running)
BBB) STATUS(Running)
I am able to truncate "QMNAME(" but unable to truncate those to the right of AAA and BBB. Basically I want to get the string between "QMNAME(" and the immediate ")". Please assist.
I think a regex approach is better than split() here, but you could use split() by splitting on parentheses and taking the second item in the returned list.
for (#qmgrlist) {
say +(split /[()]/)[0];
}
And a brief note on your use of command-line options to run this code. You can make it simpler if you a) pipe the output of qspmq into your code and b) use -n to process a record at a time.
$ perl -nE 'say +(split /[()]/)[1]' `dspmq`
There's also -M to load modules (e.g. -MData::Dumper), but you don't seem to be using Data::Dumper any more.
split isn't going to do what you need. I would just use a regular expression to match the sub-string you need
So change the loop from this
foreach my $i (#qmgrlist)
{
my #temp1 = split /QMNAME\(/, $i;
print #temp1;
}
to this
foreach my $i (#qmgrlist)
{
print "$1\n"
if /QMNAME\((.+?)\)/;
}
Try this perl one-liner:
dspmq | perl -lne 'print for m{ QMNAME [(] ( [^)]* ) [)] }x'
Here, dspmq STDOUT is fed using a pipe | into STDIN of the perl code, which has these flags:
-e tells Perl interpreter to look for the code inline rather than in a separate script file.
-n feeds the input line by line to the inline code (this way you do not need to store the output in an array - this matters for large outputs, not in your case).
-l strips the input record separator (newline on *NIX) before feeding it to the code, and appends it automatically after during print.
The print ... for ... m{... (...) ...} code prints every pattern captured in parentheses.
The captured pattern is [^)]*, which is maximum number (0 or more) chars that are not (^) listed in the character class, that is, that are not closing parens.
[(] ... [)] are literal parentheses escaped as character classes for readability. I prefer this to escaping like so: \( ... \).
QMNAME is used to make the programmer's intentions clear: you want the string that follows QMNAME in parens. I prefer this to using the field index, such as 1, which protects you against minor variation in output of your command used with different options, on different systems, etc.
Finally, the x regex modifier in m{...}x enables comments and whitespace to be ignored, and is preferred for readability.
RELATED:
Cutting the output of a dspmq command
Desired output can be achieved with following code
use strict;
use warnings;
use feature 'say';
map{ say $1 if /QMNAME\((.+?)\)/ } <DATA>;
__DATA__
QMNAME(AAA) STATUS(Running)
QMNAME(BBB) STATUS(Running)
output
AAA
BBB
and one liner (not tested - I am on Windows computer)
dspmq | perl -lne 'print $1 if /QMNAME\((.+?)\)/'
$name = "LPA";
$date= 39;
$file = "/pathtoFile/window/$name/cavin_l/'$name'_formula_24rw$date.csv";
I want this to set $file equal to:
/pathtoFile/window/LPA/cavin_l/LPA_formula_24rw39.csv
But it is setting $file to:
/pathtoFile/window/LPA/cavin_l/'39.csv
Could someone help me figure out the correct syntax for what I am trying to accomplish? I cannot seem to get it.
You fell victim to the archaic ' (quote) package separator.
$name'_formula_24 is interpreted as $name::_formula24, and since that variable is undefined, the empty string is interpolated.
You should change that to
$file = "/pathtoFile/window/$name/cavin_l/${name}_formula_24rw$date.csv"
Quoting from the man perldata:
There are two package separators in Perl: A double colon ("::") and
a single quote ("'"). Normal identifiers can start or end with a
double colon, and can contain several parts delimited by double
colons. Single quotes have similar rules, but with the exception
that they are not legal at the end of an identifier: That is,
"$'foo" and "$foo'bar" are legal, but "$foo'bar'" is not.
The double quotes "..." interpolate variables in the string. Then the problem is that in your attempt the variable names in the string are not cleanly delineated† and the interpreter ends up looking for
$name::_formula_24rw
That is, the variable _formula_24rw in the package $name; this is because a ' after a variable name (so ' after $name in your $file) is still interpreted as a package separator ::, for historical reasons. As there is no such variable you end up with ' and then the value for $date.
The most important takeaway: use warnings; would have caught this. Try to run
use warnings;
use strict;
my $name = "LPA";
my $date= 39;
my $file = "/pathtoFile/window/$name/cavin_l/'$name'_formula_24rw$date.csv";
and it prints
Use of uninitialized value $name::_formula_24rw in concatenation (.) or string at ...
Please always have use warnings; and use strict; at the beginning of a program.
So either use ${...} where {} delimit the variable name
my $file = "/pathtoFile/window/$name/cavin_l/${name}_formula_24rw$date.csv";
or piece it together from separate strings
my $file = '/pathtoFile/window/'
. $name . '/cavin_l/' . $name . '_formula_24rw' . "$date.csv";
The . cannot be in a variable name so $date in that string is resolved correctly.†
† See identifier parsing in perldata for details on how variable names are parsed in code.
One way is to use curly braces around the variable name like this:
$file = "/pathtoFile/window/$name/cavin_l/${name}_formula_24rw$date.csv";
This syntax is described in perldoc perldata
As in some shells, you can enclose the variable name in braces to
disambiguate it from following alphanumerics (and underscores).
like most things in Perl there is many ways to do the same task.
You can use {} to define the boundaries of your variable name
my $file1="/pathtoFile/window/$name/cavin_l/${name}_formula_24rw$date.csv";
Alternatively you could use concatenation to join the string.
my $file2="/pathtoFile/window/$name/cavin_l/" . $name ."_formula_24rw" . $date . ".csv";
Both these methods produce the same result.
use strict;
use warnings;
my $name="LPA";
my $date="39";
my $file1="/pathtoFile/window/$name/cavin_l/${name}_formula_24rw$date.csv";
my $file2="/pathtoFile/window/$name/cavin_l/" . $name ."_formula_24rw" . $date . ".csv";
print "$file1\n$file2\n";
output
/pathtoFile/window/LPA/cavin_l/LPA_formula_24rw39.csv
/pathtoFile/window/LPA/cavin_l/LPA_formula_24rw39.csv
I want to remove n characters from each line using PERL.
For example, I have the following input:
catbathatxx (length 11; 11%3=2 characters) (Remove 2 characters from this line)
mansunsonx (length 10; 10%3=1 character) (Remove 1 character from this line)
#!/usr/bin/perl -w
open FH, "input.txt";
#array=<FH>;
foreach $tmp(#array)
{
$b=length($tmp)%3;
my $c=substr($tmp, 0, length($tmp)-$b);
print "$c\n";
}
I want to output the final string (after the characters have been removed).
However, this program is not giving the correct result. Can you please guide me on what the mistake is?
Thanks a lot. Please let me know if there are any doubts/clarifications.
I am assuming trailing whitespace is not significant.
#!/usr/bin/env perl
use strict; use warnings;
use constant MULTIPLE_OF => 3;
while (my $line = <DATA>) {
$line =~ s/\s+\z//;
next unless my $length = length $line;
my $chars_to_remove = $length % MULTIPLE_OF;
$line =~ s/.{$chars_to_remove}\z//;
print $line, "\n";
}
__DATA__
catbathatxx
mansunsonx
0123456789
012345678
The \K regex sequence makes this a lot clearer; it was introduced in Perl v5.10.0.
The code looks like this
use 5.10.0;
use warnings;
for (qw/ catbathatxx mansunsonx /) {
(my $s = $_) =~ s/^ (?:...)* \K .* //x;
say $s;
}
output
catbathat
mansunson
In general you would want to post the result you are getting. That being said...
Each line in the file has a \n (or \r\n on windows) on the end of it that you're not accounting for. You need to chomp() the line.
Edit to add: My perl is getting rusty from non-use but if memory serves me correct you can actually chomp() the entire array after reading the file: chomp(#array)
You should use chomp() on your array, like this:
#array=<FH>;
chomp(#array);
perl -plwe 'chomp; $c = length($_) % 3; chop while $c--' < /tmp/zock.txt
Look up the options in perlrun. Note that line endings are characters, too. Get them out of the way using chomp; re-add them on output using the -l option. Use chop to efficiently remove characters from the end of a string.
Reading your code, you are trying to print just the first 'nx3' characters for the largest value of n for each line.
The following code does this using a simple regular expression.
For each line, it first removes the line ending, then greedy matches
as many .{3} as it can (. matches any character, {3} asks for exactly 3 of them).
The memory requirement of this approach (compared with using an array the size of your file) is fixed. Not too important if your file is small compared with your free memory, but sometimes files are gigabytes, and sometimes memory is very small.
It's always worth using variable names that reflect the purpose of the variable, rather than things like $a or #array. In this case I used only one variable, which I called $line.
It's also good practice to close files as soon as you have finished with them.
#!/usr/bin/perl
use strict;
use warnings; # This will apply warnings even if you use command perl to run it
open FH, '<', 'input.txt'; # Use three part file open - single quote where no interpolation required.
for my $line (<FH>){
chomp($line);
$line =~ s/((.{3})*).*/$1\n/;
print $line;
}
close FH;
Is there an easy way, using a subroutine maybe, to print a string in Perl without escaping every special character?
This is what I want to do:
print DELIMITER <I don't care what is here> DELIMITER
So obviously it will great if I can put a string as a delimiter instead of special characters.
perldoc perlop, under "Quote and Quote-like Operators", contains everything you need.
While we usually think of quotes as literal values, in Perl they function as operators, providing various kinds of interpolating and pattern matching
capabilities. Perl provides customary quote characters for these behaviors, but also provides a way for you to choose your quote character for any of
them. In the following table, a "{}" represents any pair of delimiters you choose.
Customary Generic Meaning Interpolates
'' q{} Literal no
"" qq{} Literal yes
`` qx{} Command yes*
qw{} Word list no
// m{} Pattern match yes*
qr{} Pattern yes*
s{}{} Substitution yes*
tr{}{} Transliteration no (but see below)
<<EOF here-doc yes*
* unless the delimiter is ''.
$str = q(this is a "string");
print $str;
if you mean quotes and apostrophes with 'special characters'
You can use the __DATA__ directive which will treat all of the following lines as a file that can be accessed from the DATA handle:
while (<DATA>) {
print # or do something else with the lines
}
__DATA__
#!/usr/bin/perl -w
use Some::Module;
....
or you can use a heredoc:
my $string = <<'END'; #single quotes prevent any interpolation
#!/usr/bin/perl -b
use Some::Module;
....
END
The printing is not doing special things to the escapes, double quoted strings are doing it. You may want to try single quoted strings:
print 'this is \n', "\n";
In a single quoted string the only characters that must be escaped are single quotes and a backslash that occurs immediately before the end of the string (i.e. 'foo\\').
It is important to note that interpolation does not work with single quoted strings, so
print 'foo is $foo', "\n";
Will not print the contents of $foo.
You can pretty much use any character you want with q or qq. For example:
#!/usr/bin/perl
use utf8;
use strict; use warnings;
print q∞This is a test∞;
print qq☼\nThis is another test\n☼;
print q»But, what is the point?»;
print qq\nYou are just making life hard on yourself!\n;
print qq¿That last one is tricky\n¿;
You cannot use qq DELIMITER foo DELIMITER. However, you could use heredocs for a similar effect:
print <<DELIMITER
...
DELIMETER
;
or
print <<'DELIMETER'
...
DELIMETER
;
but your source code would be really ugly.
If you want to print a string literally and you have Perl 5.10 or later then
say 'This is a string with "quotes"' ;
will print the string with a newline.. The importaning thing is to use single quotes ' ' rather than double ones " "
I have a string which holds a decimal value in it and I need to convert that string into a floating point variable. So an example of the string I have is "5.45" and I want a floating point equivalent so I can add .1 to it. I have searched around the internet, but I only see how to convert a string to an integer.
You don't need to convert it at all:
% perl -e 'print "5.45" + 0.1;'
5.55
This is a simple solution:
Example 1
my $var1 = "123abc";
print $var1 + 0;
Result
123
Example 2
my $var2 = "abc123";
print $var2 + 0;
Result
0
Perl is a context-based language. It doesn't do its work according to the data you give it. Instead, it figures out how to treat the data based on the operators you use and the context in which you use them. If you do numbers sorts of things, you get numbers:
# numeric addition with strings:
my $sum = '5.45' + '0.01'; # 5.46
If you do strings sorts of things, you get strings:
# string replication with numbers:
my $string = ( 45/2 ) x 4; # "22.522.522.522.5"
Perl mostly figures out what to do and it's mostly right. Another way of saying the same thing is that Perl cares more about the verbs than it does the nouns.
Are you trying to do something and it isn't working?
Google lead me here while searching on the same question phill asked (sorting floats) so I figured it would be worth posting the answer despite the thread being kind of old. I'm new to perl and am still getting my head wrapped around it but brian d foy's statement "Perl cares more about the verbs than it does the nouns." above really hits the nail on the head. You don't need to convert the strings to floats before applying the sort. You need to tell the sort to sort the values as numbers and not strings.
i.e.
my #foo = ('1.2', '3.4', '2.1', '4.6');
my #foo_sort = sort {$a <=> $b} #foo;
See http://perldoc.perl.org/functions/sort.html for more details on sort
As I understand it int() is not intended as a 'cast' function for designating data type it's simply being (ab)used here to define the context as an arithmetic one. I've (ab)used (0+$val) in the past to ensure that $val is treated as a number.
$var += 0
probably what you want. Be warned however, if $var is string could not be converted to numeric, you'll get the error, and $var will be reset to 0:
my $var = 'abc123';
print "var = $var\n";
$var += 0;
print "var = $var\n";
logs
var = abc123
Argument "abc123" isn't numeric in addition (+) at test.pl line 7.
var = 0
Perl really only has three types: scalars, arrays, and hashes. And even that distinction is arguable. ;) The way each variable is treated depends on what you do with it:
% perl -e "print 5.4 . 3.4;"
5.43.4
% perl -e "print '5.4' + '3.4';"
8.8
In comparisons it makes a difference if a scalar is a number of a string. And it is not always decidable. I can report a case where perl retrieved a float in "scientific" notation and used that same a few lines below in a comparison:
use strict;
....
next unless $line =~ /and your result is:\s*(.*)/;
my $val = $1;
if ($val < 0.001) {
print "this is small\n";
}
And here $val was not interpreted as numeric for e.g. "2e-77" retrieved from $line. Adding 0 (or 0.0 for good ole C programmers) helped.
Perl is weakly typed and context based. Many scalars can be treated both as strings and numbers, depending on the operators you use.
$a = 7*6; $b = 7x6; print "$a $b\n";
You get 42 777777.
There is a subtle difference, however. When you read numeric data from a text file into a data structure, and then view it with Data::Dumper, you'll notice that your numbers are quoted. Perl treats them internally as strings.
Read:$my_hash{$1} = $2 if /(.+)=(.+)\n/;.
Dump:'foo' => '42'
If you want unquoted numbers in the dump:
Read:$my_hash{$1} = $2+0 if /(.+)=(.+)\n/;.
Dump:'foo' => 42
After $2+0 Perl notices that you've treated $2 as a number, because you used a numeric operator.
I noticed this whilst trying to compare two hashes with Data::Dumper.