I have this simple code from Perl Cookbook which prints all directories and files recursively:
use File::Find;
#ARGV = qw(.) unless #ARGV;
find sub { print $File::Find::name, -d && '/', "\n" }, #ARGV;
I do not understand the grammar of print $File::Find::name, -d. How is this to interpret? If -d tests if $File::Find::nameis a directory so -dis a parameter of the function print? Or does Perl explicitly interpret a standalone -d as if -d?
No, the -d is a stand alone statement, it tests $_. So it is in essence identical to
-d $_ && '/'
Which says "if file is a directory, return a slash character (to print)". The sub code block is used by the find function from File::Find, where $_ contains the file name of the current file.
The commas , separate a list of statements that return strings for the print statement:
print $File::Find::name, # print the files name
-d && '/', # if it is a dir, print /
"\n" # print a newline
In the documentation for -d (contained in perldoc for -X where all the file tests are listed) states:
If the argument is omitted, tests $_ ...
This applies to all file tests under -X.
The reason && can be used this way is that it has a higher precedence than the comma operator ,. This is documented in perldoc perlop
Related
I am taking arguments from the command line to test if they exist print out their names if they do. In this code directories are printed out if they exist. How do I exclude directories?
#!/usr/bin/perl
use strict;
use warnings;
foreach my $x (#ARGV) {
if (-e $x){
print "$x ";
else {
print "'$x' does not exist"
}
}
exit;
Use the file test operator -d to test if you have a directory.
In your code, you can skip directories by adding:
next if -d $x;
See perldoc -f -X for details on this and all other file test functions.
This question already has answers here:
Using the -d test operator in perl
(3 answers)
Closed 8 years ago.
What does the -d in the following piece of code:
foreach my $filename (#files) {
my $filepath = $dir.$filename;
next if -d $filepath;
function1();
}
This is a short form for
if (-d $filepath) {
next;
}
Where -d $filepath is a test if $filepath is a directory.
See http://perldoc.perl.org/functions/-X.html for a full list of file tests.
-d tests if $filepath is a directory.
All such file tests are documented at perldoc -X:
-X FILEHANDLE
-X EXPR
-X DIRHANDLE
-X
A file test, where X is one of the letters listed below. This unary operator takes one argument, either a filename, a filehandle, or a dirhandle, and tests the associated file to see if something is true about it. If the argument is omitted, tests $_, except for -t, which tests STDIN. Unless otherwise documented, it returns 1 for true and '' for false. If the file doesn't exist or can't be examined, it returns undef and sets $! (errno). Despite the funny names, precedence is the same as any other named unary operator. The operator may be any of:
...
-f File is a plain file.
-d File is a directory.
...
It checks for the directory...
A short example to check that
$somedir = "c:/windows";
if (-d $somedir) {
print "$somedir exists";
} else {
print "$somedir does not exist!";
}
Also check the docs for other such cases
-f File is a plain file.
-d File is a directory.
-l File is a symbolic link.
-p File is a named pipe (FIFO), or Filehandle is a pipe.
-S File is a socket.
-b File is a block special file.
-c File is a character special file.
-t Filehandle is opened to a tty.
Essentially, next if -d $filepath; means "if this file is a directory, run the next iteration of the loop", which effectively skips the call of function1 for that file. In short, it is a way of applying function1 only to files which are NOT directories.
Is possible somewhat write a perl script in a bash script as heredoc?
This is not working (example only)
#/bin/bash
perl <<EOF
while(<>) {
chomp;
print "xxx: $_\n";
}
EOF
Is here some nice way how to embed a perl script into a bash script? Want run perl script from an bash script and don't want put it into external file.
The problem here is that the script is being passed to perl on stdin, so trying to process stdin from the script doesn't work.
1. String literal
perl -e '
while(<>) {
chomp;
print "xxx: $_\n";
}
'
Using a string literal is the most direct way to write this, though it's not ideal if the Perl script contains single quotes itself.
2. Use perl -e
#/bin/bash
script=$(cat <<'EOF'
while(<>) {
chomp;
print "xxx: $_\n";
}
EOF
)
perl -e "$script"
If you pass the script to perl using perl -e then you won't have the stdin problem and you can use any characters you like in the script. It's a bit roundabout to do this, though. Heredocs yield input on stdin and we need strings. What to do? Oh, I know! This calls for $(cat <<HEREDOC).
Make sure to use <<'EOF' rather than just <<EOF to keep bash from doing variable interpolation inside the heredoc.
You could also write this without the $script variable, although it's getting awfully hairy now!
perl -e "$(cat <<'EOF'
while(<>) {
chomp;
print "xxx: $_\n";
}
EOF
)"
3. Process substitution
perl <(cat <<'EOF'
while(<>) {
chomp;
print "xxx: $_\n";
}
EOF
)
Along the lines of #2, you can use a bash feature called process substitution which lets you write <(cmd) in place of a file name. If you use this you don't need the -e since you're now passing perl a file name rather than a string.
You know I never thought of this.
The answer is "YES!" it does work. As others have mentioned, <STDIN> can't be used, but this worked fine:
$ perl <<'EOF'
print "This is a test\n";
for $i ( (1..3) ) {
print "The count is $i\n";
}
print "End of my program\n";
EOF
This is a test
The count is 1
The count is 2
The count is 3
End of my program
In Kornshell and in BASH, if you surround your end of here document string with single quotes, the here document isn't interpolated by the shell.
Only small corection of #John Kugelman's answer. You can eliminate the useless cat and use:
read -r -d '' perlscript <<'EOF'
while(<>) {
chomp;
print "xxx: $_\n";
}
EOF
perl -e "$perlscript"
Here's another way to use a PERL HEREDOC script within bash, and take full advantage it.
#!/bin/sh
#If you are not passing bash var's and single quote the HEREDOC tag
perl -le "$(cat <<'MYPL'
# Best to build your out vars rather than writing directly
# to the pipe until the end.
my $STDERRdata="", $STDOUTdata="";
while ($i=<STDIN>){ chomp $i;
$STDOUTdata .= "To stdout\n";
$STDERRdata .= "Write from within the heredoc\n";
MYPL
print $STDOUTdata; #Doing the pipe write at the end
warn $STDERRdata; #will save you a lot of frustration.
)" [optional args] <myInputFile 1>prints.txt 2>warns.txt
or
#!/bin/sh
set WRITEWHAT="bash vars"
#If you want to include your bash var's
#Escape the $'s that are not bash vars, and double quote the HEREDOC tag
perl -le "$(cat <<"MYPL"
my $STDERRdata="", $STDOUTdata="";
while (\$i=<STDIN>){ chomp \$i;
\$STDOUTdata .= "To stdout\n";
\$STDERRdata .= "Write $WRITEWHAT from within the heredoc\n";
MYPL
print \$STDOUTdata; #Doing the pipe write at the end
warn \$STDERRdata; #will save you a lot of frustration.
)" [optional args] <myInputFile 1>prints.txt 2>warns.txt
I am trying to do a dynamic search and replace with Perl on the command line with part of the replacement text being the output of a grep command within backticks. Is this possible to do on the command line, or will I need to write a script to do this?
Here is the command that I thought would do the trick. I thought that Perl would treat the backticks as a command substitution, but instead it just treats the backticks and the content within them as a string:
perl -p -i -e 's/example.xml/http:\/\/exampleURL.net\/`grep -ril "example_needle" *`\/example\/path/g' `grep -ril "example_needle" *`
UPDATE:
Thanks for the helpful answers. Yes, there was a typo in my original one-liner: the target file of grep is supposed to be *.
I wrote a small script based on Schewrn's example, but am having confusing results. Here is the script I wrote:
#!/usr/bin/env perl -p -i
my $URL_First = "http://examplesite.net/some/path/";
my $URL_Last = "/example/example.xml";
my #files = `grep -ril $URL_Last .`;
chomp #files;
foreach my $val (#files) {
#dir_names = split('/',$val);
if(#dir_names[1] ne $0) {
my $url = $URL_First . #dir_names[1] . $URL_Last;
open INPUT, "+<$val" or die $!;
seek INPUT,0,0;
while(<INPUT>) {
$_ =~ s{\Q$URL_Last}{$url}g;
print INPUT $_;
}
close INPUT;
}
}
Basically what I am trying to do is:
Find files that contain $URL_Last.
Replace $URL_Last with $URL_First plus the name of the directory that the matched file is in, plus $URL_Last.
Write the above change to the input file without modifying anything else in the input file.
After running my script, it completely garbled the HTML code in the input file and it cut off the first few characters of each line in the file. This is strange, because I know for sure that $URL_Last only occurs once in each file, so it should only be matched once and replaced once. Is this being caused by a misuse of the seek function?
You should use another delimiter for s/// so that you don't need to escape slashes in the URL:
perl -p -i -e '
s#example.xml#http://exampleURL.net/`grep -ril "example_needle"`/example/path#g'
`grep -ril "example_needle" *`
Your grep command inside the regex will not be executed, as it is just a string, and backticks are not meta characters. Text inside a substitution will act as though it was inside a double quoted string. You'd need the /e flag to execute the shell command:
perl -p -i -e '
s#example.xml#
qq(http://exampleURL.net/) . `grep -ril "example_needle"` . qq(/example/path)
#ge'
`grep -ril "example_needle" *`
However, what exactly are you expecting that grep command to do? It lacks a target file. -l will print file names for matching files, and grep without a target file will use stdin, which I suspect will not work.
If it is a typo, and you meant to use the same grep as for your argument list, why not use #ARGV?
perl -p -i -e '
s#example.xml#http://exampleURL.net/#ARGV/example/path#g'
`grep -ril "example_needle" *`
This may or may not do what you expect, depending on whether you expect to have newlines in the string. I am not sure that argument list will be considered a list or a string.
It seems like what you're trying to do is...
Find a file in a tree which contains a given string.
Use that file to build a URL.
Replace something in a string with that URL.
You have three parts, and you could jam them together into one regex, but it's much easier to do it in three steps. You won't hate yourself in a week when you need to add to it.
The first step is to get the filenames.
# grep -r needs a directory to search, even if it's just the current one
my #files = `grep -ril $search .`;
# strip the newlines off the filenames
chomp #files;
Then you need to decide what to do if you get more than one file from grep. I'll leave that choice up to you, I'm just going to take the first one.
my $file = $files[0];
Then build the URL. Easy enough...
# Put it in a variable so it can be configured
my $Site_URL = "http://www.example.com/";
my $url = $Site_URL . $file;
To do anything more complicated, you'd use URI.
Now the search and replace is trivial.
# The \Q means meta-characters like . are ignored. Better than
# remembering to escape them all.
$whatever =~ s{\Qexample.xml}{$url}g;
You want to edit files using -p and -i. Fortunately we can emulate that functionality.
#!/usr/bin/env perl
use strict;
use warnings; # never do without these
my $Site_URL = "http://www.example.com/";
my $Search = "example-search";
my $To_Replace = "example.xml";
# Set $^I to edit files. With no argument, just show the output
# script.pl .bak # saves backup with ".bak" extension
$^I = shift;
my #files = `grep -ril $Search .`;
chomp #files;
my $file = $files[0];
my $url = $Site_URL . $file;
#ARGV = ($files[0]); # set the file up for editing
while (<>) {
s{\Q$To_Replace}{$url}g;
}
Everyone's answers were very helpful to my writing a script that wound up working for me. I actually found a bash script solution yesterday, but wanted to post a Perl answer in case anyone else finds this question through Google.
The script that #TLP posted at http://codepad.org/BFpIwVtz is an alternative way of doing this.
Here is what I ended up writing:
#!/usr/bin/perl
use Tie::File;
my $URL_First = 'http://example.com/foo/bar/';
my $Search = 'path/example.xml';
my $URL_Last = '/path/example.xml';
# This grep returns a list of files containing "path/example.xml"
my #files = `grep -ril $Search .`;
chomp #files;
foreach my $File_To_Edit (#files) {
# The output of $File_To_Edit looks like this: "./some_path/index.html"
# I only need the "some_path" part, so I'm going to split up the output and only use #output[1] ("some_path")
#output = split('/',$File_To_Edit);
# "some_path" is the parent directory of "index.html", so I'll call this "$Parent_Dir"
my $Parent_Dir = #output[1];
# Make sure that we don't edit the contents of this script by checking that $Parent_Dir doesn't equal our script's file name.
if($Parent_Dir ne $0) {
# The $File_To_Edit is "./some_path/index.html"
tie #lines, 'Tie::File', $File_To_Edit or die "Can't read file: $!\n";
foreach(#lines) {
# Finally replace "path/example.xml" with "http://example.com/foo/bar/some_path/path/example.xml" in the $File_To_Edit
s{$Search}{$URL_First$Parent_Dir$URL_Last}g;
}
untie #lines;
}
}
How can I list all available UNIX commands from Perl?
perl -MFile::Find -le 'find sub {print if -f and -x _}, split ":", $ENV{PATH}'
This code looks in each directory in your path (split ":", $ENV{PATH}) for files (-f) that are executable (-x), and prints the ones it finds. You may want to read about
File::Find
split
%ENV
filetest operators
An alternative that does not search subdirectories of the directories in the PATH is
perl -le '-f and -x _ and print for map { glob "$_/*" } split ":", $ENV{PATH}'