I have this simple command:
printf TEST | perl -nle 'print lc'
Which prints:
test
I want:
test
...without the newline. I tried perl's printf but that removes all newlines, and I'd like to keep existing one's in place. Plus, that wouldn't work for my second example that doesn't even use print in it:
printf "BOB'S BIG BOY" | perl -ple 's/([^\s.,-]+)/\u\L$1/g'
Which prints:
Bob's Big Boy
...with that annoying newline as well. I'm hoping for a magical switch like --no-newline but I'm guessing it's something more involved.
EDIT: I've changed my use of echo in the examples to printf to clarify the problem. A few commenters were correct in stating that my problem wouldn't actually be fixed as it was written.
You simply have to remove the -l switch, see perldoc perlrun
-l[octnum]
enables automatic line-ending processing. It has two separate
effects. First, it automatically chomps $/ (the input record
separator) when used with -n or -p. Second, it assigns $\ (the output
record separator) to have the value of octnum so that any print
statements will have that separator added back on. If octnum is
omitted, sets $\ to the current value of $/.
Related
It works as intended:
perl -ne "print uc" /etc/passwd
But following isn't (it just prints in original case":
perl -pe uc /etc/passwd
I don't understand what's wrong with it.
thanks.
You're doing different things. So it's not surprising that you get different results.
In the first example, you take the value of $_, pass it to uc and print the results (which is an upper case version of the original text).
In the second example, you take the value of $_, pass it to uc and print the value in $_. But you've done nothing to update $_ so you get the unaltered value. The fix (as you've already noted in a comment) is to update $_ with the value that is returned by uc.
perl -pe '$_ = uc' /etc/passwd
I have the following Perl 'one-liner' script (found it online, so not mine):
perl -lsne '
/$today.* \[([0-9.]+)\]:.+dovecot_(?:login|plain):([^\s]+).* for (.*)/
and $sender{$2}{r}+=scalar (split / /,$3)
and $sender{$2}{i}{$1}=1;
END {
foreach $sender(keys %sender){
printf"Recip=%05d Hosts=%03d Auth=%s\n",
$sender{$sender}{r},
scalar (keys %{$sender{$sender}{i}}),
$sender;
}
}
' -- -today=$(date +%F) /var/log/exim_mainlog | sort
I've been trying to understand its innards, because I would like to modify it to re-use its functionality.
Some questions I got:
What does the flag -lsne does? (From what I know, it's got to be, at least, 3 different flags in one)
Where does $sender gets its value from?
What about that (?:login|plain) segment, are they 'variables'? (I get that's ReGex, I'm just not familiarized with it)
What I'm trying to achieve:
Get the number of emails sent by each user in a SMTP relay periodically (cron job)
If there's an irregular number of emails (say, 500 in a 1-hour timespan), do something (like shutting of the service, or send a notification)
Why I'm trying to achieve this:
Lately, someone has been using my SMTP server to send spam, so I would like to monitor email activity so they stop abusing the SMTP relay resources. (Security-related suggestions are always welcomed, but out of topic for this question. Trying to focus on the script for now)
What I'm NOT trying to achieve:
To get the script done by third-parties. (Just try and point me in the right direction, maybe an example)
So, any suggestions, guidance,and friendly comments are welcomed. I understand this may be an out-of-topic question, yet I've been struggling with this for almost a week and my background with Perl is null.
Thanks in advance.
What does the flag -lsne does? (From what I know, it's got to be, at least, 3 different flags in one)
-l causes lines of input read in to be auto-chomped, and lines of
output printed out to have "\n" auto-appended
-s enables switch
parsing. This is what creates the variable $today, because a
command-line switch of --today=$(date +%F) was passed.
-n surrounds the entire "one-liner" in a while (<>) { ... } loop.
Effectively reading every line from standard input and running the
body of the one liner on that line
-e is the switch that tells
perl to execute the following code from the command line, rather
than running a file containing Perl code
Where does $sender gets its value from?
I suspect you are confusing $sender with %sender. The code uses $sender{$2}{r} without explicitly mentioning %sender. This is a function of Perl called "auto-vivification". Basically, because we used $sender{$2}{r}, perl automatically created a variable %sender, and added a key whose name is whatever is in $2, and set the value of that key in %sender to be a reference to a new hash. It then set that new hash to have a key 'r' and a value of scalar (split / /,$3)
What about that (?:login|plain) segment, are they 'variables'? (I get that's ReGex, I'm just not familiarized with it)
It's saying that this portion of the regular expression will match either 'login' or 'plain'. The ?: at the beginning tells Perl that these parentheses are used only for clustering, not capturing. In other words, the result of this portion of the pattern match will not be stored in the $1, $2, $3, etc variables.
-MO=Deparse is your friend for understanding one-liners (and one liners that wrap into five lines on your terminal):
$ perl -MO=Deparse -lsne '/$today.* \[([0-9.]+)\]:.+dovecot_( ...
BEGIN { $/ = "\n"; $\ = "\n"; }
LINE:
while ( defined($_ = <ARGV>) ) {
chomp $_;
$sender{$2}{'i'}{$1} = 1 if
/$today.* \[([0-9.]+)\]:.+dovecot_(?:login|plain):([^\s]+).* for (.*)/
and $sender{$2}{'r'} += scalar split(/ /, $3, 0);
sub END {
foreach $sender (keys %sender) {
printf "Recip=%05d Hosts=%03d Auth=%s\n",
$sender{$sender}{'r'},
scalar keys %{$sender{$sender}{'i'};}, $sender;
}
}
}
-e syntax OK
[newlines and indentation added for clarity]
What does the flag -lsne does? (From what I know, it's got to be, at least, 3 different flags in one)
You can access a summary of the available perl command line options by running '~$ perl -h' in the terminal. Below are filtered out the specific command line options you were asking about.
~$ perl -h|perl -ne 'print if /^\s+(-l|-s|-n|-e)/'
-e program one line of program (several -e's allowed, omit programfile)
-l[octal] enable line ending processing, specifies line terminator
-n assume "while (<>) { ... }" loop around program
-s enable rudimentary parsing for switches after programfile
Two examples of the '-s' command line option in use.
~$ perl -se 'print "Todays date is $today\n"' -- -today=`date +%F`
Todays date is 2016-10-17
~$ perl -se 'print "The sky is $color.\n"' -- -color='blue'
The sky is blue.
For detailed explanations of those command line options read the online documentation below.
http://perldoc.perl.org/perlrun.html
Or run the command below from your terminal.
~$ perldoc perlrun
Unrelated to the questions of the OP, I'm aware that this is not a complete answer (added as much as I was able to at the moment), so if this post/answer violates any SO rules, the moderators should remove it. Thx.
I am trying to run the following command in perl script :
#!/usr/bin/perl
my $cmd3 =`sed ':cycle s/^\(\([^,]*,\)\{0,13\}[^,|]*\)|[^,]*/\1/;t cycle' file1 >file2`;
system($cmd3);
but is not producing any output nor any error.
Although when I am running the command from command line it is working perfectly and gives desired output. Can you guys please help what I am doing wrong here ?
Thanks
system() doesn't return the output, just the exit status.
To see the output, print $cmd3.
my $cmd3 = `sed ':cycle s/^\(\([^,]*,\)\{0,13\}[^,|]*\)|[^,]*/\1/;t cycle' file1 >file2`;
print "$cmd3\n";
Edit:
If you want to check for exceptional return values, use CPAN module IPC::System::Simple:
use IPC::System::Simple qw(capture);
my $result = capture("any-command");
Running sed from inside Perl is just insane.
#!/usr/bin/perl
open (F, '<', "file1") or die "$O: Could not open file1: $!\n";
while (<F>) {
1 while s/^(([^,]*,){0,13}[^,|]*)\|[^,]*/$1/;
print;
}
Notice how Perl differs from your sed regex dialect in that grouping parentheses and alternation are unescaped, whereas a literal round parenthesis or pipe symbol needs to be backslash-escaped (or otherwise made into a literal, such as by putting it in a character class). Also, the right-hand side of the substitution prefers $1 (you will get a warning if you use warnings and have \1 in the substitution; technically, at this level, they are equivalent).
man perlrun has a snippet explaining how to implement the -i option inside a script if you really need that, but it's rather cumbersome. (Search for the first occurrence of "LINE:" which is part of the code you want.)
However, if you want to modify file1 in-place, and you pass it to your Perl script as its sole command-line argument, you can simply say $^I = 1; (or with use English; you can say $INPLACE_EDIT = 1;). See man perlvar.
By the way, the comment that your code "isn't producing any output" isn't entirely correct. It does what you are asking it to; but you are apparently asking for the wrong things.
Quoting a command in backticks executes that command. So
my $cmd3 = `sed ... file1 >file2`;
runs the sed command in a subshell, there and then, with input from file1, and redirected into file2. Because of the redirection, the output from this pipeline is nothing, i.e. an empty string "", which is assigned to $cmd3, which you then completely superfluously attempt to pass to system.
Maybe you wanted to put the sed command in regular quotes instead of backticks (so that the sed command line would be the value of $cmd3, which it then makes sense to pass to system). But because of the redirection, it would still not produce any visible output; it would create file2 containing the (possibly partially substituted) text from file1.
I was looking at this Perl one-liner
perl -n -e 'print "$. - $_"' file
and it says that this one liner gets converted to this:
LINE:
while (<>) {
print "$. - $_"
}
Which is fine, Ijust don't know what LINE: is. It doesn't seem like a filehandle, and if it is a variable, it does not have a $sign in front of it.
My guess is that it is something like #F: an idiom that is just used in Perl one liners. Is LINE just something that Perl uses in one-liners from the command line?
It's a label. They provide a way to mark a place in your code. Using labels is not idiomatic as they are rarely needed. Labels can be used with certain commands, namely next, last, redo and goto.
A label is a bareword followed by a colon, such as LINE:
More information can be found in perldoc perlsyn
I have a file, someFile, like this:
$cat someFile
hdisk1 active
hdisk2 active
I use this shell script to check:
$cat a.sh
#!/usr/bin/ksh
for d in 1 2
do
grep -q "hdisk$d" someFile && echo "$d : ok"
done
I am trying to convert it to Perl:
$cat b.sh
#!/usr/bin/ksh
export d
for d in 1 2
do
cat someFile | perl -lane 'BEGIN{$d=$ENV{'d'};} print "$d: OK" if /hdisk$d\s+/'
done
I export the variable d in the shell script and get the value using %ENV in Perl. Is there a better way of passing this value to the Perl one-liner?
You can enable rudimentary command line argument with the "s" switch. A variable gets defined for each argument starting with a dash. The -- tells where your command line arguments start.
for d in 1 2 ; do
cat someFile | perl -slane ' print "$someParameter: OK" if /hdisk$someParameter\s+/' -- -someParameter=$d;
done
See: perlrun
Sometimes breaking the Perl enclosure is a good trick for these one-liners:
for d in 1 2 ; do cat kk2 | perl -lne ' print "'"${d}"': OK" if /hdisk'"${d}"'\s+/';done
Pass it on the command line, and it will be available in #ARGV:
for d in 1 2
do
perl -lne 'BEGIN {$d=shift} print "$d: OK" if /hdisk$d\s+/' $d someFile
done
Note that the shift operator in this context removes the first element of #ARGV, which is $d in this case.
Combining some of the earlier suggestions and adding my own sugar to it, I'd do it this way:
perl -se '/hdisk([$d])/ && print "$1: ok\n" for <>' -- -d='[value]' [file]
[value] can be a number (i.e. 2), a range (i.e. 2-4), a list of different numbers (i.e. 2|3|4) (or almost anything else, that's a valid pattern) or even a bash variable containing one of those, example:
d='2-3'
perl -se '/hdisk([$d])/ && print "$1: ok\n" for <>' -- -d=$d someFile
and [file] is your filename (that is, someFile).
If you are having trouble writing a one-liner, maybe it is a bit hard for one line (just my opinion). I would agree with #FM's suggestion and do the whole thing in Perl. Read the whole file in and then test it:
use strict;
local $/ = '' ; # Read in the whole file
my $file = <> ;
for my $d ( 1 .. 2 )
{
print "$d: OK\n" if $file =~ /hdisk$d\s+/
}
You could do it looping, but that would be longer. Of course it somewhat depends on the size of the file.
Note that all the Perl examples so far will print a message for each match - can you be sure there are no duplicates?
My solution is a little different. I came to your question with a Google search the title of your question, but I'm trying to execute something different. Here it is in case it helps someone:
FYI, I was using tcsh on Solaris.
I had the following one-liner:
perl -e 'use POSIX qw(strftime); print strftime("%Y-%m-%d", localtime(time()-3600*24*2));'
which outputs the value:
2013-05-06
I was trying to place this into a shell script so I could create a file with a date in the filename, of X numbers of days in the past. I tried:
set dateVariable=`perl -e 'use POSIX qw(strftime); print strftime("%Y-%m-%d", localtime(time()-3600*24*$numberOfDaysPrior));'`
But this didn't work due to variable substitution. I had to mess around with the quoting, to get it to interpret it properly. I tried enclosing the whole lot in double quotes, but this made the Perl command not syntactically correct, as it messed with the double quotes around date format. I finished up with:
set dateVariable=`perl -e "use POSIX qw(strftime); print strftime('%Y-%m-%d', localtime(time()-3600*24*$numberOfDaysPrior));"`
Which worked great for me, without having to resort to any fancy variable exporting.
I realise this doesn't exactly answer your specific question, but it answered the title and might help someone else!
That looks good, but I'd use:
for d in $(seq 1 2); do perl -nle 'print "hdisk$ENV{d} OK" if $_ =~ /hdisk$ENV{d}/' someFile; done
It's already written on the top in one long paragraph but I am also writing for lazy developers who don't read those lines.
Double quotes and single quote has big different meaning for the bash.
So please take care
Doesn't WORK perl '$VAR' $FILEPATH
WORKS perl "$VAR" $FILEPATH