What does reading from <> do in Perl? For example, what will the following do?
print for(<>);
The so-called diamond operator (<>) reads line-by-line (in scalar context) from STDIN or the filename(s) specified as command-line arguments.
From perldoc perlop:
The null filehandle <> is special: it can be used to emulate the
behavior of sed and awk. Input from <> comes either from standard
input, or from each file listed on the command line. Here's how it
works: the first time <> is evaluated, the #ARGV array is checked,
and if it is empty, $ARGV[0] is set to "-", which when opened
gives you standard input. The #ARGV array is then processed as a
list of filenames.
In list context, <> returns all lines, with each line stored as an element in the list.
This means that print for <>; will do the same thing as print while <>;, albeit with more memory.
You've found the single most magical piece of Perl. Well, I'm sure there's more magical things, but this little idiom makes it very easy to write programs intended for shell pipeline use and file-operation use.
When run without any arguments, <> will read lines one-at-a-time from standard input.
When run with arguments, it'll treat the arguments as filenames and read lines one-at-a-time from the named files in turn.
A short demo:
$ cat > print.pl
#!/usr/bin/perl -w
print for(<>);
$ chmod 755 print.pl
$ echo hello world | ./print.pl
hello world
$ ./print.pl print.pl
#!/usr/bin/perl -w
print for(<>);
$ ./print.pl print.pl print.pl
#!/usr/bin/perl -w
print for(<>);
#!/usr/bin/perl -w
print for(<>);
$
I typed in the program by hand there; hit ^D when you've typed it in completely.
It reads from standard input, one line at a time, and stores it to $_. print then prints out $_ by default since it is not given an argument. This program reads from standard input and echoes to standard output until it reaches EOF.
Related
I have mulitple lines
QQQQl123
hsdhjhksd
QQQQl234
ajkdkjsdh
QQQQl564
i want to print all matching QQQQl[0-9]+
like
QQQQl123
QQQQl234
QQQQl564
how to do this using perl
I tried:
$ perl -0777pe '/QQQQl[0-9]+/' filename
it shows nothing
perl -we 'while(<>){ next unless $_=~/QQQQl[0-9]+/; print $_; }' < filename
perl -ne 'print if /QQQQl[0-9]+/' filename
Or, if, for some reason, you insist on using -0777, you could do
perl -0777nE 'say for /QQQQl[0-9]+/g' filename
(or print "$_\n" instead of say)
Your code doesn't work because /QQQQl[0-9]+/ returns true because $_ indeed contains that pattern, but you never asked Perl to do anything based on that return value.
-n is preferable to -p in that case, since you don't want to print every line but only some (-p automatically prints every line, and there is very little you can do about it).
Probably a very simple question:
I have a perl file sed.perl that takes as an input a string, makes some substitutions there and prints it on the standard output.
#!/usr/bin/perl
use warnings;
use strict;
use diagnostics;
use feature 'say';
#use Cwd;
my ($text) = #ARGV;
$text =~ s/\.\)\n/'\.'\)\n/;
print $text;
I want to feed the script with a string output from a terminal pipeline. Let's say in this way:
cat input.txt | perl sed.perl
but this doesn't work: Use of uninitialized value $text in substitution (s///) at
Using a score symbol doesn't works either:
cat input.txt | perl sed.perl -
#ARGV doesn't do what you think it does. It's literally the arguments passed to perl.
E.g. :
myscript.pl some arg
#ARGV will host 'some', 'arg'.
What you want is the STDIN file handle.
e.g.
#!/usr/bin/perl
use strict;
use warnings;
while ( <STDIN> ) {
s/something/somethingelse/g;
print;
}
Now what this is doing is reading STDIN line by line. Your pattern includes \n. Do you actually need it? It looks like you're 'just' using it as a line anchor, and so you could use:
s/\.\)$/'\.'\)/g;
$ is the regex for "end of line" - see perlre for more.
However, as noted in the comments by reinierpost - there's another thing that's probably useful to know - perl has the "diamond operator" <> which does two things:
If filenames are specified to the script, opens them and reads from them.
If no arguments specified, reads STDIN.
So you could do:
while ( <> ) {
s/something/somethingelse/g;
print;
}
And then your script can either be invoked by:
cat input.txt | ./yourscript.pl
Or:
./yourscript.pl input.txt
And you'll have the same result.
You're trying to read the text as if it were an argument, when in fact with a pipeline the information from input.txt will be something you read from stdin. With a pipeline stdout from the left process is connected to stdin of the right process.
As others have mentioned, #ARGV contains arguments to your script, and you'll want to use STDIN instead.
The shortest solution seems to be:
#/usr/bin/env perl
use strict;
while(<>) {
print s/\.\)$/'.')/r;
}
Note that you could also achieve the results of your example using a one-liner as follows:
cat input.txt | perl -pe 's/your/substitution/'
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.
What's the use of <> in Perl. How to use it ?
If we simply write
<>;
and
while(<>)
what is that the program doing in both cases?
The answers above are all correct, but it might come across more plainly if you understand general UNIX command line usage. It is very common to want a command to work on multiple files. E.g.
ls -l *.c
The command line shell (bash et al) turns this into:
ls -l a.c b.c c.c ...
in other words, ls never see '*.c' unless the pattern doesn't match. Try this at a command prompt (not perl):
echo *
you'll notice that you do not get an *.
So, if the shell is handing you a bunch of file names, and you'd like to go through each one's data in turn, perl's <> operator gives you a nice way of doing that...it puts the next line of the next file (or stdin if no files are named) into $_ (the default scalar).
Here is a poor man's grep:
while(<>) {
print if m/pattern/;
}
Running this script:
./t.pl *
would print out all of the lines of all of the files that match the given pattern.
cat /etc/passwd | ./t.pl
would use cat to generate some lines of text that would then be checked for the pattern by the loop in perl.
So you see, while(<>) gets you a very standard UNIX command line behavior...process all of the files I give you, or process the thing I piped to you.
<>;
is a short way of writing
readline();
or if you add in the default argument,
readline(*ARGV);
readline is an operator that reads a line from the specified file handle. Reading from the special file handle ARGV will read from STDIN if #ARGV is empty or from the concatenation of the files named by #ARGV if it's not.
As for
while (<>)
It's a syntax error. If you had
while (<>) { ... }
it get rewritten to
while (defined($_ = <>)) { ... }
And as previously explained, that means the same as
while (defined($_ = readline(*ARGV))) { ... }
That means it will read lines from (previously explained) ARGV until there are no more lines to read.
It is called the diamond operator and feeds data from either stdin if ARGV is empty or each line from the files named in ARGV. This webpage http://docstore.mik.ua/orelly/perl/learn/ch06_02.htm explains it very well.
In many cases of programming with syntactical sugar like this, Deparse of O is helpful to find out what's happening:
$ perl -MO=Deparse -e 'while(<>){print 42}'
while (defined($_ = <ARGV>)) {
print 42;
}
-e syntax OK
Quoting perldoc perlop:
The null filehandle <> is special: it can be used to emulate the
behavior of sed and awk, and any other Unix filter program that takes
a list of filenames, doing the same to each line of input from all of
them. Input from <> comes either from standard input, or from each
file listed on the command line.
it takes the STDIN standard input:
> cat temp.pl
#!/usr/bin/perl
use strict;
use warnings;
my $count=<>;
print "$count"."\n";
>
below is the execution:
> temp.pl
3
3
>
so as soon as you execute the script it will wait till the user gives some input.
after 3 is given as input,it stores that value in $count and it prints the value in the next statement.
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