So I've done a research about the perl -pe command and I know that it takes records from a file and creates an output out of it in a form of another file. Now I'm a bit confused as to how this line of command works since it's a little modified so I can't really figure out what exactly is the role of perl pe in it. Here's the command:
cd /usr/kplushome/entities/Standalone/config/webaccess/WebaccessServer/etc
(PATH=/usr/ucb:$PATH; ./checkall.sh;) | perl -pe "s,^, ,g;"
Any idea how it works here?
What's even more confusing in the above statement is this part : "s,^, ,g;"
Any help would be much appreciated. Let me know if you guys need more info. Thank you!
It simply takes an expression given by the -e flag (in this case, s,^, ,g) and performs it on every line of the input, printing the modified line (i.e. the result of the expression) to the output.
The expression itself is something called a regular expression (or "regexp" or "regex") and is a field of learning in and of itself. Quick googles for "regular expression tutorial" and "getting started with regular expressions" turn up tons of results, so that might be a good place to start.
This expression, s,^, ,g, adds ten spaces to the start of the line, and as I said earlier, perl -p applies it to every line.
"s,^, ,g;"
s is use for substitution. syntax is s/somestring/replacement/.
In your command , is the delimiter instead of /.
g is for work globally, means replace all occurrence.
For example:
perl -p -i -e "s/oldstring/newstring/g" file.txt;
In file.txt all oldstring will replace with newstring.
i is for inplace file editing.
See these doc for information:
perlre
perlretut
perlop
Related
I have a fortran code with global comments, which start with a double exclamation mark (i.e., !!) and personal comments, which start with a single exclamation mark (i.e., !), and I just want to hide my personal comment lines (or substitute the line with another line, e.g., '! jw'). For example, the original code looks like this:
!! This is a global comment
Code..
Code..
! This is a personal comment
code... ! This is a personal comment
!! This is a global comment
code...
Then, I want to update the original code as:
!! This is a global comment
Code..
Code..
! jw
code... ! jw
!! This is a global comment
code...
I have tried to use "sed" and "awk", but I failed. So, would someone can help me? I prefer to use "sed" instead "awk" by the way.
Use Perl one-liner with negative lookbehind pattern:
perl -pe 's/(?<!!)!\s.*/! jw/' in_file > out_file
To change the file in-place:
perl -i.bak -pe 's/(?<!!)!\s.*/! jw/' in_file
To change multiple files in-place, for example ex*.f90 files:
perl -i.bak -pe 's/(?<!!)!\s.*/! jw/' ex*.f90
The Perl one-liner uses these command line flags:
-e : Tells Perl to look for code in-line, instead of in a file.
-p : Loop over the input one line at a time, assigning it to $_ by default. Add print $_ after each loop iteration.
-i.bak : Edit input files in-place (overwrite the input file). Before overwriting, save a backup copy of the original file by appending to its name the extension .bak.
(?<!!)! : Exclamation point that is not preceded by an exclamation point.
\s : Whitespace.
.* : Any character, repeated 0 or more times.
SEE ALSO:
perldoc perlrun: how to execute the Perl interpreter: command line switches
perldoc perlre: Perl regular expressions (regexes)
perldoc perlre: Perl regular expressions (regexes): Quantifiers; Character Classes and other Special Escapes; Assertions; Capture groups
perldoc perlre: Negative lookbehind
perldoc perlrequick: Perl regular expressions quick start
sed '/!!/!s/!.*/! jw/' file
/!!/! If the line does not contain !!, then
s/!.*/! jw/ substitute all following a exclamation mark with ! jw.
awk 'BEGIN{FS=OFS="!"}$2{$2=" jw"}1' file
BEGIN{FS=OFS="!"} Set the field separators to !.
$2{$2=" jw"} If the 2nd field is not empty, substitute it by jw.
1 Print the line.
If the line starts with ! then you could do something like
sed 's/^! /! jw/' mycode.fortran >newcodefile.fortran
I would put it into a new file then rename after. If you overwrite your file you could end up cause problems if anything goes wrong.
The s in the string to sed tells it to search, and replace.
the ^ means start of line, so if the comment is further in the line than the beginning this won't find that comment.
Then we search for a line that starts with ! followed by a space and replace with ! jw
If you just run it as:
sed 's/^! /! jw/' mycode.fortran
without redirecting the output to a file it will stream the output to your console so you can see if it's working. Then run it again output to a file with the redirect >, check the file then do your renaming. Don't get rid of your original code file until your completely sure it worked and didn't do anything you didn't want.
I came across this script which was not written by me, but because of an issue I need to know what it does.
What is the purpose of filtering the log file using this Perl one-liner?
cat log.txt | perl -pe 's/\e([^\[\]]|\[.*?[a-zA-Z]|\].*?\a)/ /g'
The log.txt file contains the output of a series of commands. I do not understand what is being filtered here, and why it might be useful.
It looks like the code should remove ANSI escape codes from the input, i.e codes to set colors, window title .... Since some of these code might cause harm it might be a security measure in case some kind of attack was able to include such escape codes into the log file. Since usually a log file does not contain any such escape codes this would also explain why you don't see any effect of this statement for normal log files.
For more information about this kind of attack see A Blast From the Past: Executing Code in Terminal Emulators via Escape Sequences.
BTW, while your question looks bad on the first view it is actually not. But you might try to improve questions by at least formatting it properly. Otherwise you risk that this questions gets down-voted fast.
First, the command line suffers from a useless use of cat. perl is fully capable of reading from a file name on the command line.
So,
$ perl -pe 's/\e([^\[\]]|\[.*?[a-zA-Z]|\].*?\a)/ /g' log.txt
would have done the same thing, but avoided spawning an extra process.
Now, -e is followed by a script for perl to execute. In this case, we have a single global substitution.
\e in a Perl regex pattern corresponds to the escape character, x1b.
The pattern following \e looks like the author wants to match ANSI escape sequences.
The -p option essentially wraps the script specified with -e in while loop, so the s/// is executed for each line of the input.
The pattern probably does the job for this simple purpose, but one might benefit from using Regexp::Common::ANSIescape as in:
$ perl -MRegexp::Common::ANSIescape=ANSIescape,no_defaults -pe 's/$RE{ANSIescape}/ /g' log.txt
Of course, if one uses a script like this very often, one might want to either use an alias, or even write a very short script that does this, as in:
#!/usr/bin/env perl
use strict;
use Regexp::Common 'ANSIescape', 'no_defaults';
while (<>) {
s/$RE{ANSIescape}/ /g;
print;
}
I've been using Codeigniter for my PHP project and I've been using their session class.
$this->session->userdata('variablename')
I've been having a lot of problems with this so i've decided to use PHP Native session.
$_SESSION['variablename']
This is what I've got so far
perl -p -i -e "s/$this->session->userdata('.*?$SOMEVAR.*?\')/$_SESSION['$1']/g" offer.php
But truth to be told I don't really know what I'm doing.
I would also like to do this on all php files in my project.
Help much appreciated.
The regex should be:
s/\$this->session->userdata\('(.?)'\)/$_SESSION['$1']/g
Issues with the version you posted are mostly with un-escaped characters--you can escape a $ or parenthesis by adding a \ prior to the character. For example, \$this will find the text "$this", while $this will search for the value of the $this variable.
For a more comprehensive look at escapes (and other quick tips), if you have $2, I highly recommend this cheat sheet.
Also, you don't need to use the .*?$SOMEVAR.*? construct you added in there...Perl will automatically capture the result found between the first pair of parentheses and store it in $1, the second set of parentheses gets $2, etc.
When shell quoting is getting complicated, the simplest thing to do is to just put the source into a file. You can still use it as a one-liner. I have used a negative lookahead assertion to make sure that it does not break for escaped single quotes inside the string.
# source file, regex.txt
s/\$this->session->userdata\('(.+?)(?!\\')'\)/\$_SESSION['$1']/g;
Usage:
perl -pi regex.txt yourfile.php
Note that you simply leave out the -e switch. Also note that -i requires a backup extension for Windows, e.g. -i.bak.
I am using following kind of script in my perl script and expecting entry at line 1. I am getting some error as below; any help?
plz ignore perl variable....
Error Messaage -
sed: -e expression #1, char 22: extra characters after command
# Empty file will not work for Sed line number 1
`"Security Concerns Report" > $outputFile`;
`sed '1i\
\
DATE :- $CDate \
Utility accounts with super access:- $LNumOfSupUserUtil \
Users not found in LDAP with super access: - $LNumOfSupUserNonLdap\
' $outputFile > $$`;
`mv $$ $outputFile`;
}
Your immediate problem is that the backslash character is interpreted by Perl inside the backtick operator, as is the dollar character. So your backslash-newline sequence turns into a newline in the shell command that is executed. If you replace these backslashes by \\, you'll go over this hurdle, but you'll still have a very brittle program.
Perl is calling a shell which calls sed. This requires an extra level of quoting for the shell which you are not performing. If your file names and data contain no special characters, you may be able to get away with this, until someone uses a date format containing a ' (among many things that would break your code).
Rather than fix this, it is a lot simpler to do everything in Perl. Everything sed and shells can do, Perl can do almost as easily or easier. It's not very clear from your question what you're trying to do. I'll focus on the sed call, but this may not be the best way to write your program.
If you really need to prepend some text to an existing file, there's a widely-used module on CPAN that already does this well. Use existing libraries in preference to reinventing the wheel. File::Slurp has a prepend_file method just for that. In the code below I use a here-document operator for the multiline string.
use File::Slurp; # at the top of the script with the other use directives
File::Slurp->prepend_file($outputFile, <<EOF);
DATE :- $CDate
Utility accounts with super access:- $LNumOfSupUserUtil
Users not found in LDAP with super access: - $LNumOfSupUserNonLdap
EOF
I saw from
How to use sed to replace only the first occurrence in a file?
How to do most of what I want:
sed -n '0,/.*\(something\).*/s//\1/p'
This finds the first match of something and extracts it (of course my real example is more complicated). I know sed has a 'quit' command which will make it stop early, but I don't know how to combine the 'q' with the above line to get my desired behavior.
I've tried replacing the 'p' with {p;q;} as I've seen in some examples, but that is clearly not working.
The "print and quit" trick works, if you also put the substitution command into the block (instead of only putting the print and quit there).
Try:
sed -n '/.*\(something\).*/{s//\1/p;q}'
Read this like: Match for the pattern given between the slashes. If this pattern is found, execute the actions specified in the block: Replace, print and exit. Typically, match- and substitution-patterns are equal, but this is not required, so the 's' command could also contain a different RE.
Actually, this is quite similar to what ghostdog74 answered for gawk.
My specific use case was to find out via 'git log' on a git-p4 imported tree which perforce branch was used for the last commit. git log (when called without -n will log every commit that every happened (hundreds of thousands for me)).
We can't know a-priori what value to give git for '-n.' After posting this, I found my solution which was:
git log | sed -n '0,/.*\[git-p4:.*\/\/depot\/blah\/\([^\/]*\)\/.*/s//\1/p; /\[git-p4/ q'
I'd still like to know how to do this with a non '/' separator, and without having to specify the 'git-p4' part twice, once for the extraction and once for the quit. There has got to be a way to combine both on the same line...
Sed usually has an option to specify more than one pattern to execute (IIRC, it's the -e option). That way, you can specify a second pattern that quits after the first line.
Another approach is to use sed to extract the first line (sed '1q'), then pipe that to a second sed command (what you show above).
use gawk
gawk '/MATCH/{
print "do something with "$0
exit
}' file
To "return something from first line and quit" you could also use grep:
grep --max-count 1 'something'
grep is supposed to be faster than sed and awk [1]. Even if not, this syntax is easier to remember (and to type: grep -m1 'something').
If you don't want to see the whole line but just the matching part, the --only-matching (-o) option might suffice.