Here's my current command:
sed 's/\./-/' file.txt > CLEANED.txt
What I'm trying to do is replace all periods in my file with a dash. Some lines have multiple periods and I need all of them replaced with a dash - but the command above seems to just replace the first one in each line.
What am I doing wrong for it to not replace all of the periods?
In perl, just add the /g modifier to your regex:
perl -pe 's/\./-/g' file.txt > CLEANED.txt
Explanation:
Switches:
-p: Creates a while(<>){...; print} loop for each “line” in your input file.
-e: Tells perl to execute the code on command line.
Code:
s/\./-/g: Replace all periods with dashes. Could also use the transliteration operator: y/./-/
Add /g for a global replacement, else it only affects the first occurrence.
Like so:
sed 's/\./-/g' file.txt > CLEANED.txt
Related
I want to rename files with 'sr' in their names, replacing 'sr' with 'SR'. This one succeeded：
ls | perl -e 'while(<>){chomp;if(/(.*)sr(.*)/){rename $_,$1."SR".$2}}'
But this one failed:
ls | perl -e "while(<>){chomp;if(/sr/){rename $_,$\`.'SR'.($')}}"
with this error message:
Not enough arguments for rename at -e line 1, near "rename ,"`
Execution of -e aborted due to compilation errors.
It seems that $_ has become an empty string, but I don't quite understand why. Thanks for any explanations.
Now quotes have been an interesting problem and this is my test:
ls | perl -e "while(<>){chomp;if(/sr/){print $_;print\"\n\";print $\`,$&,($');print \"\n\";print $_,$\`,$&,($');print\"\n\";print $_;print\"\n\"}}"
outputs this:
3sr
3sr
3sr
3sr
sr1
sr1
sr1
sr1
sr2
sr2
sr2
sr2
it seems that when using alone, $_ is not empty; but it become empty when using along with $`,$& and $'. According to the last line of each file, I guess $_ has temporarily changed when not using alone?
Besides, according to a1111exe's answer, I test this:
ls | perl -e "while(<>){chomp;if(/sr/){print \$_,$\`,$&,($');print \"\n\"}}"
and got this:
3sr3sr
sr1sr1
sr2sr2
First in linux we should use single quote instead of double quote.
And instead of ls command you can use perl inbuilt function glob
And to capture the pre and post match you can use the $POSTMATCH and $PREMATCH from English module
so your one liner should be
perl -MEnglish -e 'while(<*>){chomp;if(/sr/){rename $_,$PREMATCH."SR".$POSTMATCH}}'
EDITED
Single quote and double quote is not about Perl this is about shell.
Single quote
Enclosing characters in single quotes (') preserves the literal value of each character within the quotes. A single quote may not occur between single quotes, even when preceded by a backslash.
Double quote
Enclosing characters in double quotes (‘"’) preserves the literal value of all characters within the quotes, with the exception of ‘$’, ‘`’, ‘\’, and, when history expansion is enabled, ‘!’.
In shell script we are accessing the shell variable prefix with $, so while using $ inside the double quote it is looking for the shell variable not a Perl variable. For example you can run the following line in your terminal,
m=4; perl -e "print $m;"
Here
m=4; perl -e "print $m;"
^ ^
| Accessing shell variable
Assigning shell variable
Output is 4. Because m is shell variable you are accessing the shell variable inside your Perl script.
And in windows, we need to use double-quote instead of single quote
It seems that double quotes mess between your shell environment and Perl. You can certainly do what #mkHun suggested. One other way:
ls | perl -e 'while(<>){chomp;($new=$_)=~s/sr/SR/g;rename $_,$new}'
Also, if you escape the '$' sigil in '$_', your oneliner will work too:
ls | perl -e "while(<>){chomp;if(/sr/){rename \$_,$\`.'SR'.$'}}"
I still don't get why though.. But it really seems like bash/perl interpolation issue.
Here's what I use to match a string in a variable and delete the line where the match exists:
sed -i '/'"$domainAndSuffix.cfg"'/d' /etc/file
I'd like to know how to match a string in a variable, but if the match in the file has a . adjacent to it on the immediate left, then it will NOT delete this line and keep going through the file until it finds a match without a .
Sample file Contents:
happy.domain.com
pappy.domain.com
domain.com
String to match:
domain.com
Desired File Output:
happy.domain.com
pappy.domain.com
*Edit:
Actual File Contents:
cfg_file=/etc/nagios/objects/http_url/bob.ca.cfg
cfg_file=/etc/nagios/objects/http_url/therecord.com.cfg
cfg_file=/etc/nagios/objects/http_url/events.therecord.com.cfg
cfg_file=/etc/nagios/objects/http_url/read.therecord.com.cfg
cfg_file=/etc/nagios/objects/http_url/wheels.ca.cfg
cfg_file=/etc/nagios/objects/http_url/used-vehicle-search.autos.ca.msn.com.cfg
cfg_file=/etc/nagios/objects/http_url/womensweekendshow.com.cfg
cfg_file=/etc/nagios/objects/http_url/yorkregion.com.cfg
cfg_file=/etc/nagios/objects/http_url/yourclassifieds.ca.cfg
If the preceding substring is fixed, you can try the following:
PREFIX='cfg_file=\/etc\/nagios\/objects\/http_url\/'
DOMAIN='therecord.com'
sed -i "/^${PREFIX}${DOMAIN}/d" file
If it is not fixed, it would be nice to use a negative lookbehind, but sed can't do that. You can use ssed or GNU grep:
ssed -Ri '/(?<!\.)'"$DOMAIN"'.cfg/d' file
or
grep -vP '(?<!\.)'"$DOMAIN" > file1; mv file1 file
I have some output data from some Bash Shell commands. The output is delimited line by line with "\n" or "\0". I would like to know that is there any way to pipe the output into Perl and process the data line by line within Perl (just like piping the output to awk, but in my case it is in the Perl context.). I suppose the command may be something like this :
Bash Shell command | perl -e 'some perl commands' | another Bash Shell command
Suppose I want to substitute all ":" character to "#" character in a "line by line" basis (not a global substitution, I may use a condition, e.g. odd or even line, to determine whether the current line should have the substitution or not.), then how could I achieve this.
See perlrun.
perl -lpe's/:/#/g' # assumes \n as input record separator
perl -0 -lpe's/:/#/g' # assumes \0 as input record separator
perl -lne'if (0 == $. % 2) { s/:/#/g; print; }' # modify and print even lines
Yes, Perl may appear at any place in a pipeline, just like awk.
The command line switch -p (if you want automatic printing) or -n (if you don't want it) will do what you want. The line contents are in $_ so:
perl -pe's/\./\#/g'
would be a solution. Generally, you want to read up on the '<>' (diamond) operator which is the way to go for non-oneliners.
I'm trying to read a specific section of a line out of a file with Perl.
The file in question is of the following syntax.
# Sets $USER1$
$USER1$=/usr/....
# Sets $USER2$
#$USER2$=/usr/...
My oneliner is simple,
perl -ne 'm/^\$USER1\$\s*=\s*(\S*?)\s*$/m; print "$1";' /my/file
For some reason I'm getting the extraction for $1 repeated several times over, apparently once for every line in the file after my match occurs. What am I missing here?
You are executing print for every line of the file because print gets called for every line, whether the regex matches or not. Replace the first ; with an &&.
From perlre:
NOTE: Failed matches in Perl do not reset the match variables, which makes it easier to write code that tests for a series of more specific cases and remembers the best match.
Try this instead:
perl -ne 'print "$1" if m/^\$USER1\$\s*=\s*(\S*?)\s*$/m;' /my/file
$ cat test.txt
# Sets $USER1$
$USER1$=/usr/....
# Sets $USER2$
#$USER2$=/usr/...
$ perl -nle 'print if /^\$USER1/;' test.txt
$USER1$=/usr/....
Try this
perl -ne '/^.*1?=([\w\W].*)$/;print "$1";' file
I have a variable in a shell script,
var=1234_number
I want to replace all other than integer of $var .. how can I do it using a perl onliner?
You might be looking for something to edit the shell script, in which case, this might be sufficient:
perl -i.bak -e 's/\b(var=\d+).*/$1/' shellscript.sh
The '-i' overwrites the original file, saving a copy in shellscript.sh.bak; the substitute command finds assignments to 'var' (and not any longer name ending 'var') followed by an equals sign, some digits, and any non-digits, and leaves behind just the assignment of digits.
In the example, it gives:
var=1234
Note that the Perl regex is not foolproof - it will mangle this (dropping the closing brace).
: ${var=1234_number}
Dealing with all such possible variants is extremely fairly tricky:
echo $var=$other
OTOH, you might be looking to eliminate digits from a variable within a shell script, in which case:
var=$(echo $var | perl -e 's/\D//g')
You could also use 'sed' for the job:
var=$(echo $var | sed 's/[^0-9]//g')
No need to use anything but the shell for this
var=1234_abcd
var=${var%_*}
echo $var # => 1234
See 'Parameter Expansion' in the bash manual.