For input abc, the code
perl -ne 'print "XYZ$_"'
prints XYZabc, but after switching the order of $_ and XYZ, i.e.
perl -ne 'print "$_XYZ"'
it prints nothing. Why?
XYZ can be part of a variable name, so $_XYZ is a variable name, rather than $_ followed by a literal XYZ.
You can split the string up:
perl -ne 'print $_ . "XYZ"'
Perl identifiers may contain any letters, digits, or underscore, so you are asking perl to print the value of the variable $_XYZ, which doesn't exist
You may surround the name of the variable with braces { ... } to separate it from any surrounding characters, like so
perl -ne 'print "${_}XYZ"'
Related
I have a file with the following
firsttext=cat secondtext=dog thirdtext=mouse
and I want it to return this string:
"firsttext=cat" "secondtext=dog" "thirdtext=mouse"
I yave tried this one-liner but it gives me an error.
cat oneline | perl -ne 'print \"$_ \" '
Can't find string terminator '"' anywhere before EOF at -e line 1.
I don't understand the error.Why can't it just add the quotation marks?
Also, if I have a variable in this string, I want it to be interpolated like:
firsttext=${animal} secondtext=${othervar} thirdtext=mouse
Which should output
"firsttext=cat" "secondtext=dog" "thirdtext=mouse"
perl -lne '#f = map qq/"$_"/, split; print "#f";' oneline
What you want is this:
cat oneline | perl -ne 'print join " ", map { qq["$_"] } split'
The -ne option only splits on lines, it won't split on arbitrary whitespace without other options set.
I ran into this problem while trying to print single quotes in a Perl one-liner. I eventually figured out you have to escape them with '\''. Here's some code to illustrate my question.
Let's start with printing a text file.
perl -ne 'chomp; print "$_\n"' shortlist.txt
red
orange
yellow
green
blue
Now let's print the name of the file instead for each line.
perl -ne 'chomp; print "$ARGV\n"' shortlist.txt
shortlist.txt
shortlist.txt
shortlist.txt
shortlist.txt
shortlist.txt
Then we can add single quotes around each line.
perl -ne 'chomp; print "'$_'\n"' shortlist.txt
shortlist.txt
shortlist.txt
shortlist.txt
shortlist.txt
shortlist.txt
Wait that didn't work. Let's try again.
perl -ne 'chomp; print "'\''$_'\''\n"' shortlist.txt
'red'
'orange'
'yellow'
'green'
'blue'
So I got it working now. But I'm still confused on why '$_' evaluates to the program name. Maybe this is something easy but can someone explain or link to some documentation?
edit: I'm running Perl 5.8.8 on Red Hat 5
To your shell, 'chomp; print "'$_'\n"' results in a string that's the concatenation of
chomp; print " (the first sequence inside single quotes),
the value of its variable $_, and
\n" (the second sequence inside single quotes).
In bash, $_ "... expands to the last argument to the previous command, after expansion. ...". Since this happens to be shortlist.txt, the following is passed to perl:
chomp; print "shortlist.txt\n"
For example,
$ echo foo
foo
$ echo 'chomp; print "'$_'\n"'
chomp; print "foo\n"
Note that the above mechanism shouldn't be used to pass values to a Perl one-liner. You shouldn't be generating Perl code from the shell. See How can I process options using Perl in -n or -p mode? for how to provide arguments to a one-liner.
You use single quotes in one-liners to protect your Perl code from being evaluated by the shell. In this command:
perl -ne 'chomp; print "'$_'\n"' shortlist.txt
you close the single quotes before $_, so the shell expands $_ to the last argument to the previous command. In your case, this happened to be the name of your input file, but the output would be different if you ran a different command first:
$ echo foo
$ perl -ne 'chomp; print "'$_'\n"' shortlist.txt
foo
foo
foo
foo
foo
I try to avoid quotes in one liners for just this reason. I use generalized quoting when I can:
% perl -ne 'chomp; print qq($_\n)'
Although I can avoid even that with the -l switch to get the newline for free:
% perl -nle 'chomp; print $_'
If I don't understand a one-liner, I use -MO=Deparse to see what Perl thinks it is. The first two are what you expect:
% perl -MO=Deparse -ne 'chomp; print "$_\n"' shortlist.txt
LINE: while (defined($_ = <ARGV>)) {
chomp $_;
print "$_\n";
}
-e syntax OK
% perl -MO=Deparse -ne 'chomp; print "$ARGV\n"' shortlist.txt
LINE: while (defined($_ = <ARGV>)) {
chomp $_;
print "$ARGV\n";
}
-e syntax OK
You see something funny in the one where you saw the problem. The variable has disappeared before perl ever saw it and there's a constant string in its place:
% perl -MO=Deparse -ne 'chomp; print "'$_'\n"' shortlist.txt
LINE: while (defined($_ = <ARGV>)) {
chomp $_;
print "shortlist.txt\n";
}
-e syntax OK
Your fix is curious too because Deparse puts the variable name in braces to separate it from the old package specifier ':
% perl -MO=Deparse -ne 'chomp; print "'\''$_'\''\n"' shortlist.txt
LINE: while (defined($_ = <ARGV>)) {
chomp $_;
print "'${_}'\n";
}
-e syntax OK
How to add a blank line after every grep result?
For example, grep -o "xyz" may give something like -
file1:xyz
file2:xyz
file2:xyz2
file3:xyz
I want the output to be like this -
file1:xyz
file2:xyz
file2:xyz2
file3:xyz
I would like to do something like
grep "xyz" | perl (code to add a new line after every grep result)
This is the direct answer to your question:
grep 'xyz' | perl -pe 's/$/\n/'
But this is better:
perl -ne 'print "$_\n" if /xyz/'
EDIT
Ok, after your edit, you want (almost) this:
grep 'xyz' * | perl -pe 'print "\n" if /^([^:]+):/ && ! $seen{$1}++'
If you don’t like the blank line at the beginning, make it:
grep 'xyz' * | perl -pe 'print "\n" if /^([^:]+):/ && ! $seen{$1}++ && $. > 1'
NOTE: This won’t work right on filenames with colons in them. :)½
If you want to use perl, you could do something like
grep "xyz" | perl -p -e 's/(.*)/\1\n/g'
If you want to use sed (where I seem to have gotten better results), you could do something like
grep "xyz" | sed 's/.*/\0\n/g'
This prints a newline after every single line of grep output:
grep "xyz" | perl -pe 'print "\n"'
This prints a newline in between results from different files. (Answering the question as I read it.)
grep 'xyx' * | perl -pe '/(.*?):/; if ($f ne $1) {print "\n"; $f=$1}'
Use a state machine to determine when to print a blank line:
#!/usr/bin/env perl
use strict;
use warnings;
# state variable to determine when to print a blank line
my $prev_file = '';
# change DATA to the appropriate input file handle
while( my $line = <DATA> ){
# did the state change?
if( my ( $file ) = $line =~ m{ \A ([^:]*) \: .*? xyz }msx ){
# blank lines between states
print "\n" if $file ne $prev_file && length $prev_file;
# set the new state
$prev_file = $file;
}
# print every line
print $line;
}
__DATA__
file1:xyz
file2:xyz
file2:xyz2
file3:xyz
In exploring an alternative answer to sarathi's current file line number question, I wrote this one-liner with the expectation that it would print the first line of all files provided:
$ perl -ne 'print "$ARGV : $_" if __LINE__ == 1;' *txt
This did not work as expected; all lines were printed.
Running the one-liner through -MO=Deparse shows that the conditional is not present. I assume this is because it has been constant-folded at compile time:
$ perl -MO=Deparse -ne 'print "$ARGV : $_" if __LINE__ == 1;' *txt
LINE: while (defined($_ = <ARGV>)) {
print "$ARGV : $_";
}
-e syntax OK
But why?
Run under Perl 5.8.8.
__LINE__ corresponds to the line number in the Perl source, not in the input file.
__LINE__ is the source line number i.e., the program line number.
$. will give you the input file line number.
if you want to print all the first lines of all the files then you can try this:
perl -lne '$.=0 if eof;print $_ if ($.==1)' *.txt
I'm evaluating whether or not certain variables match expected values. The variables are set in memory by a certain program, the values of which can be access from the shell with a custom program.
I'm piping the output of the shell command to awk to get the specific field I want and then I want to run it through perl to see if it matches the expected value. For example,
ysgrp autostart | awk -F\: '{print $1}' | perl -e 'print {"True"} else {print "False"} if /on/'
However, I'm getting complaints from perl about compilation errors near "} else". How does one handle if/then/else logic in a perl one-liner?
You can't use an else condition in a postfix conditional. You can either use a ternary conditional operator like this:
perl -e 'print /on/ ? "True" : "False"'
Or use explicit blocks like this:
perl -e 'if ( /on/ ) { print "True" } else { print "False" }'
This part:
awk -F\: '{print $1}' | perl -e 'print {"True"} else {print "False"} if /on/'
can be handled in perl (if I remember awk correctly):
perl -F/:/ -lane 'print $F[0] =~ /on/ ? "True" : "False"'
Note the use of the -n switch, without which your perl one-liner will not work. Also note the -l switch, which adds a newline to your print, which is something I assume you want. Otherwise your output will be something like:
TrueTrueTrueFalseTrueFalse
You could do:
... | perl -ne 'print /on/ ? "True" : "False"'
but please don't! You'd be better off doing:
... | grep -qF on && echo True || echo False
One does not, except by using the ternary ?: operator; the foo if bar syntax does not support else.