I am having a lot of trouble doing a simple search and replace. I tried the solution offered in
How do I remove white space in a Perl string?
but was unable to print this.
Here is my sample code:
#!/usr/bin/perl
use strict;
my $hello = "hello world";
print "$hello\n"; #this should print out >> hello world
#now i am trying to print out helloworld (space removed)
my $hello_nospaces = $hello =~ s/\s//g;
#my $hello_nospaces = $hello =~ s/hello world/helloworld/g;
#my $hello_nospaces = $hello =~ s/\s+//g;
print "$hello_nospaces\n"
#am getting a blank response when i run this.
I tried a couple of different ways, but I was unable to do this.
My end result is to automate some aspects of moving files around in a linux environment, but sometimes the files have spaces in the name, so I want to remove the space from the variable.
You're almost there; you're just confused about operator precedence. The code you want to use is:
(my $hello_nospaces = $hello) =~ s/\s//g;
First, this assigns the value of the variable $hello to the variable $hello_nospaces. Then it performs the substitution operation on $hello_nospaces, as if you said
my $hello_nospaces = $hello;
$hello_nospaces =~ s/\s//g;
Because the bind operator =~ has higher precedence than the assignment operator =, the way you wrote it
my $hello_nospaces = $hello =~ s/\s//g;
first performs the substitution on $hello and then assigns the result of the substitution operation (which is 1 in this case) to the variable $hello_nospaces.
As of 5.14, Perl provides a non-destructive s/// option:
Non-destructive substitution
The substitution (s///) and transliteration (y///) operators now support an /r option that copies the input variable, carries out the substitution on the copy, and returns the result. The original remains unmodified.
my $old = "cat";
my $new = $old =~ s/cat/dog/r;
# $old is "cat" and $new is "dog"
This is particularly useful with map. See perlop for more examples.
So:
my $hello_nospaces = $hello =~ s/\s//gr;
should do what you want.
You just need to add parentheses so Perl's parser can understand what you want it to do.
my $hello = "hello world";
print "$hello\n";
to
(my $hello_nospaces = $hello) =~ s/\s//g;
print "$hello_nospaces\n";
## prints
## hello world
## helloworld
Split this line:
my $hello_nospaces = $hello =~ s/\s//g;
Into those two:
my $hello_nospaces = $hello;
$hello_nospaces =~ s/\s//g;
From the official Perl Regex Tutorial:
If there is a match, s/// returns the number of substitutions made; otherwise it returns false.
Related
For example,
$test = "abc";
$test =~ s/b//g;
normally test will be ac. How do I manipulate the string "abc" so that it will go through this test and still return abc.
I am trying to do a command injection so I can't change $test =~ s/b//g; but can change the input.
Hopefully this makes sense.
You want to use the /r modifier. it's a new'ish feature so older versions will not run it
$test = "abc";
my $str= $test =~ s/b//gr;
http://perldoc.perl.org/perlre.html#Modifiers
There is no string you can pass to s/b//g that will result in a string that contains a b.
However, if you have access to $test and not just it's value, you could set its pos such that the substitution operator won't find anything.
$ perl -e'
my $test = "abc";
pos($test) = length($test);
$test =~ s/b//g;
CORE::say($test);
'
ac
I have a conf file which has the format of variable="value" where values may have special characters as well. An example line is:
LINE_D="(L#'id' == 'log') AND L#'id' IS NULL"
I have another file F which should replace values based on this conf file. For example, if there is line in F
PRINT '$LINE_D'
it should be replaced by
PRINT '(L#'id' == 'log') AND L#'id' IS NULL'
How can I a program in shell script which takes conf and F and generate the values in F replaced.
Thanks
Your definition of what's required leaves lots of gaps, so you'll probably need to tweak this script. It is a cut-down version of a more complex script originally designed to process makefiles. That means there is probably material you could remove from here without causing trouble, though I've gotten rid of most of the extraneous processing.
#!usr/bin/env perl
#
# Note: this script can take input from stdin or from one or more files.
# For example, either of the following will work:
# cat config file | setmacro
# setmacro file
use strict;
use warnings;
use Getopt::Std;
# Usage:
# -b -- omit blank lines
# -c -- omit comments
# -d -- debug mode (verbose)
# -e -- omit the environment
my %opt;
my %MACROS;
my $input_line;
die "Usage: $0 [-bcde] [file ...]" unless getopts('bcde', \%opt);
# Copy environment into hash for MAKE macros
%MACROS = %ENV unless $opt{e};
my $rx_macro = qr/\${?([A-Za-z]\w*)}?/; # Matches $PQR} but ideally shouldn't
# For each line in each file specified on the command line (or stdin by default)
while ($input_line = <>)
{
chomp $input_line;
do_line($input_line);
}
# Expand macros in given value
sub macro_expand
{
my($value) = #_;
print "-->> macro_expand: $value\n" if $opt{d};
while ($value =~ $rx_macro)
{
print "Found macro = $1\n" if $opt{d};
my($env) = $MACROS{$1};
$env = "" unless defined $env;
$value = $` . $env . $';
}
print "<<-- macro_expand: $value\n" if $opt{d};
return($value);
}
# routine to recognize macros
sub do_line
{
my($line) = #_;
if ($line =~ /^\s*$/o)
{
# Blank line
print "$line\n" unless $opt{b};
}
elsif ($line =~ /^\s*#/o)
{
# Comment line
print "$line\n" unless $opt{c};
}
elsif ($line =~ /^\s*([A-Za-z]\w*)\s*=\s*(.*)\s*$/o)
{
# Macro definition
print "Macro: $line\n" if $opt{d};
my $lhs = $1;
my $rhs = $2;
$rhs = $1 if $rhs =~ m/^"(.*)"$/;
$MACROS{$lhs} = ${rhs};
print "##M: $lhs = <<$MACROS{$lhs}>>\n" if $opt{d};
}
else
{
print "Expand: $line\n" if $opt{d};
$line = macro_expand($line);
print "$line\n";
}
}
Given a configuration file, cfg, containing:
LINE_D="(L#'id' == 'log') AND L#'id' IS NULL"
and another file, F, containing:
PRINT '$LINE_D'
PRINT '${LINE_D}'
the output of perl setmacro.pl cfg F is:
PRINT '(L#'id' == 'log') AND L#'id' IS NULL'
PRINT '(L#'id' == 'log') AND L#'id' IS NULL'
This matches the required output, but gives me the heebie-jeebies with its multiple single quotes. However, the customer is always right!
(I think I got rid of the residual Perl 4-isms; the base script still had a few remnants left over, and some comments about how Perl 5.001 handles things differently. It does use $` and $' which is generally not a good idea. However it works, so fixing that is an exercise for the reader. The regex variable is not now necessary; it was when it was also recognizing make macro notations — $(macro) as well as ${macro}.)
The shell script will be passed a string of arguments. The position of the key/value I am looking to parse out may change over time, i.e. it may come before or after another key at any time so parsing between two keys wouldn't be an option.
I am looking to parse the domain key out of a string like this:
maxpark 0 maxsub n domain sample.foo maxlst n max_defer_fail_percentage user oli force no_cache_update 0 maxpop n maxaddon 0 locale en contactemail
The key would be "domain" the value would be "sample.foo". The domain key could have more than one '.' in it so I would need to grab the entire domain key.
I am not the best with regular expressions but I imagine using 'sed' is what I'm going to need to do.
I am accessing this full string using $*, if I could simply reference the key by accessing $DOMAIN that would be great, but since my only option is to access based on position, $3, and the position could change, that isn't an option
Solved the problem using PERL.
#!/usr/bin/perl -w
use strict;
my %OPTS = #ARGV;
open(FILE, "</var/named/$OPTS{'domain'}.db") || die "File not found";
my #lines = <FILE>;
close(FILE);
my #newlines;
foreach(#lines) {
$_ =~ s/$LOCAL_IP/$PUBLIC_IP/g;
push(#newlines,$_);
}
open(FILE, ">/var/named/$OPTS{'domain'}.db") || die "File not found";
print FILE #newlines;
close(FILE);
If you do have perl, just use this one-liner from your shell script.
domain=$( echo $* | perl -ne '/domain\s([^\s]+)\s/ and print "$1"' )
Or if you'd rather just do it with sed:
domain=$( echo $* | sed 's/.*\<domain \([^ ]\+\).*/\1/' )
In my perl script I want to have both versions of $config directory:
my $config='$home/client/config';
and
my $config_resolved="$home/client/config";
But I want to get $config_resolved from $config, i.e. something like this:
my $config_resolved=resolve_vars($config);
How can I do such thing in perl?
From the Perl FAQ (which every Perl programmer should read at least once):
How can I expand variables in text strings?
(contributed by brian d foy)
If you can avoid it, don't, or if you can
use a templating system, such as Text::Template or Template Toolkit,
do that instead. You might even be able to get the job done with
sprintf or printf:
my $string = sprintf 'Say hello to %s and %s', $foo, $bar;
However, for the one-off simple case where I don't want to pull out a
full templating system, I'll use a string that has two Perl scalar
variables in it. In this example, I want to expand $foo and $bar to
their variable's values:
my $foo = 'Fred';
my $bar = 'Barney';
$string = 'Say hello to $foo and $bar';
One way I can do this involves the substitution operator and a double /e flag. The
first /e evaluates $1 on the replacement side and turns it into $foo. The
second /e starts with $foo and replaces it with its value. $foo,
then, turns into 'Fred', and that's finally what's left in the string:
$string =~ s/(\$\w+)/$1/eeg; # 'Say hello to Fred and Barney'
The /e will also silently ignore violations of strict, replacing undefined
variable names with the empty string. Since I'm using the /e flag
(twice even!), I have all of the same security problems I have with
eval in its string form. If there's something odd in $foo, perhaps
something like #{[ system "rm -rf /" ]}, then I could get myself in
trouble.
To get around the security problem, I could also pull the
values from a hash instead of evaluating variable names. Using a
single /e, I can check the hash to ensure the value exists, and if it
doesn't, I can replace the missing value with a marker, in this case
??? to signal that I missed something:
my $string = 'This has $foo and $bar';
my %Replacements = (
foo => 'Fred',
);
# $string =~ s/\$(\w+)/$Replacements{$1}/g;
$string =~ s/\$(\w+)/
exists $Replacements{$1} ? $Replacements{$1} : '???'
/eg;
print $string;
I use eval for this.
So, you must replace all scalars (their names) with their values.
$config = 'stringone';
$boo = '$config/any/string';
$boo =~ s/(\$\w+)/eval($1)/eg;
print $boo;
Because you are using my to declare it as private variable, you might as well use a /ee modifier. This can find variables declared to be in local scope:
$boo =~ s/(\$\w+)/$1/eeg;
This is most tidily and safely done by the double-eval modifier on s///.
In the program below, the first /e evaluates the string $1 to get $home, while the second evaluates $home to get the variable's value HOME.
use strict;
my $home = 'HOME';
my $config = '$home/client/config';
my $config_resolved = resolve_vars($config);
print $config_resolved, "\n";
sub resolve_vars {
(my $str = shift) =~ s/(\$\w+)/$1/eeg;
return $str;
}
output
HOME/client/config
I need to convert a name in the format Parisi, Kenneth into the format kparisi.
Does anyone know how to do this in Perl?
Here is some sample data that is abnormal:
Zelleb, Charles F.,,IV
Eilt, John,, IV
Wods, Charles R.,,III
Welkt, Craig P.,,Jr.
These specific names should end up as czelleb, jeilt, cwoods, cwelkt, etc.
I have one more condition that is ruining my name builder
O'Neil, Paulso far, Vinko Vrsalovic's answer is working the best when weird/corrupt names are in the mix, but this example above would come out as "pneil"... id be damned below judas if i cant get that o between the p and the n
vinko#parrot:~$ cat genlogname.pl
use strict;
use warnings;
my #list;
push #list, "Zelleb, Charles F.,,IV";
push #list, "Eilt, John,, IV";
push #list, "Woods, Charles R.,,III";
push #list, "Welkt, Craig P.,,Jr.";
for my $name (#list) {
print gen_logname($name)."\n";
}
sub gen_logname {
my $n = shift;
#Filter out unneeded characters
$n =~ s/['-]//g;
#This regex will grab the lastname a comma, optionally a space (the
#optional space is my addition) and the first char of the name,
#which seems to satisfy your condition
$n =~ m/(\w+), ?(.)/;
return lc($2.$1);
}
vinko#parrot:~$ perl genlogname.pl
czelleb
jeilt
cwoods
cwelkt
I would start by filtering the abnormal data so you only have regular names. Then something like this should do the trick
$t = "Parisi, Kenneth";
$t =~ s/(.+),\s*(.).*/\l$2\l$1/;
Try:
$name =~ s/(\w+),\s(\w)/$2$1/;
$name = lc $name;
\w here matches an alphanumerical character. If you want to be more specific, you could also use [a-z] instead, and pass the i flag (case insensitive):
$name =~ s/([a-z]+)\s([a-z])/$2$1/i;
Here's a one line solution, assuming you store all the names in a file called "names" (one per line) and you will do duplicated name detection somehow later.
cat names | perl -e 'while(<>) {/^\s*(\S*)?,\s*(\S)/; print lc "$2$1\n";}' | sed s/\'//g
It looks like your input data is comma-separated. To me, the clearest way to do this would be split into components, and then generate the login names from that:
while (<>) {
chomp;
my ($last, $first) = split /,/, lc $_;
$last =~ s/[^a-z]//g; # strip out nonletters
$first =~ s/[^a-z]//g; # strip out nonletters
my $logname = substr($first, 0, 1) . $last;
print $logname, "\n";
}
$rowfetch =~ s/['-]//g; #All chars inside the [ ] will be filtered out.
$rowfetch =~ m/(\w+), ?(.)/;
$rowfetch = lc($2.$1);
this is how I ended up using Vinko Vrsalovic's solution... its inside a while loop that goes through a sql query result ... thanks again vinko
This should do what you need
use strict;
use warnings;
use 5.010;
while ( <DATA> ) {
say abbreviate($_);
}
sub abbreviate {
for ( #_ ) {
s/[-']+//g;
tr/A-Z/a-z/;
tr/a-z/ /c;
return "$2$1" if /([a-z]+)\s+([a-z])/;
}
}
__DATA__
Zelleb, Charles F.,,IV
Eilt, John,, IV
Woods, Charles R.,,III
Welkt, Craig P.,,Jr.
O'Neil, Paul
output
czelleb
jeilt
cwoods
cwelkt
poneil