I've been using the following which is great, except that I inadvertently delete additional sections because they include "222", like "2224" and "2225" and "2226" when all I really want is "222"
$ awk -v RS= -v ORS='\n\n' '!/ID: 222/' file
{START-TAG
foo bar
ID: 111
foo bar
END-TAG}
{START-TAG
foo bar
ID: 333
foo bar
END-TAG}
otherwise:
$ awk '/{START-TAG/{f=1} f{rec=rec $0 ORS} /END-TAG}/{if (rec !~ /ID: 222/) print rec; rec=f=""}' file
{START-TAG
foo bar
ID: 111
foo bar
END-TAG}
{START-TAG
foo bar
ID: 333
foo bar
END-TAG}
Related
I want to replace every occurrence of a string in a file but in a way that the new text will be numbered. Example:
from:
foo bar a
foo bar b
foo bar c
to:
bar baz1 a
bar baz2 b
bar baz3 c
How to do this in command line?
One way with awk:
$ awk '{gsub(/foo bar/,"bar baz"++i)}1' file
bar baz1 a
bar baz2 b
bar baz3 c
perl -lape '#F[0,1] = ($F[1], "baz". ++$i); $_= "#F"' file
Using shell.
#! /usr/bin/bash
i=0
while read a b c
do
((i++))
echo "bar baz$i $c"
done < file
How can I do so? I have multiple e.g. foobar patterns in my file, how can I add after e.g. the 4th one some_text?
Does the following work for you?
sed ':a;$!{N;ba};s/\(foobar\)/\1\nsome_text/4' inputfile
For the input:
$ cat inputfile
line
foobar
foobar
foobar
line
foobar
line
foobar
line
foobar
line
This would generate
$ sed ':a;$!{N;ba};s/\(foobar\)/\1\nsome_text/4' inputfile
line
foobar
foobar
foobar
line
foobar
some_text
line
foobar
line
foobar
line
Use -e on FreeBSD:
sed -e :a -e '$!N' -e '$!ba' -e 's/\(foobar\)/\1\nsome_text/4' inputfile
an awk version:
awk '/foo/{x++} x==4{sub(/foo/,"&some_text")}1' file
example:
kent$ cat file
foo
foo
trash
foo
trash
foo
foo
foo
kent$ awk '/foo/{x++} x==4{sub(/foo/,"&some_text")}1' file
foo
foo
trash
foo
trash
foosome_text
foo
foo
my textfile looks like this
foo.en 14 :: xyz 1;foo bar 2;foofoo 5;bar 9
bar.es 18 :: foo bar 4;kjp bar 2;bar 6;barbar 8
Ignoring text before the :: delimiter, is there a one liner unix command (many pipes allowed) or one liner perl script that extract the text such that yields the output of unique words delimited by ; ?:
xyz
foo bar
foofoo
bar
kjp bar
barbar
i've tried looping through the textfile with a python script but i'm looking for a one-liner for the task.
ans = set()
for line in open(textfile):
ans.add(line.partition(" :: ")[1].split(";").split(" ")[:-1])
for a in ans:
print a
With Perl:
perl -nle 's/.*?::\s*//;!$s{$_}++ and print for split /\s*\d+;?/' input
Description:
s/.*?::\s*//; # delete up to the first '::'
This part:
!$s{$_}++ and print for split /\s*\d+;?/
can be rewritten like this:
foreach my $word (split /\s*\d+;?/) { # for split /\s*\d+;?/
if (not defined $seen{$word}}) { # !$s{$_}
print $word; # and print
}
$seen{$word}++; # $s{$_}++
}
Since the increment in !$s{$_}++ is a post increment, Perl first test for the false condition and then does the increment. An undefined hash value has the value 0. If the test fails, i.e., $s{$_} was previously incremented, then the and part is skipped due to short circuiting.
cat textfile | sed 's/.*:://g' | tr '[0-9]*;' '\n' | sort -u
Explanation:
sed 's/.*:://g' Take everything up to and including `::` and replace it with nothing
tr '[0-9];' '\n' Replace numbers and semicolon with newlines
sort -u Sort, and return unique instances
it does result in a sorted output, I believe...
You can try this:
$ awk -F ' :: ' '{print $2}' input.txt | grep -oP '[^0-9;]+' | sort -u
bar
barbar
foo bar
foofoo
kjp bar
xyz
If your phrases contains numbers, try this perl regex: '[^;]+?(?=\s+\d+(;|$))'
With only awk :
$ awk -F' :: ' '{
gsub(/[0-9]+/, "")
split($2, arr, /;/ )
for (a in arr) arr2[arr[a]]=""
}
END{
for (i in arr2) print i
}' textfile.txt
And a one-liner version :
awk -F' :: ' '{gsub(/[0-9]+/, "");split($2, arr, /;/ );for (a in arr) arr2[arr[a]]="";}END{for (i in arr2) print i}' textfile.txt
Basically I'm looking for the perl equivalent of gdb's "up" and "down" commands. If I break on subroutine bar, and I have a call stack that looks like this:
foo
\
baz
\
bar
I'd like to be able to (without returning from bar or baz) navigate up the foo frame and see what it was doing by manipulating variables as I ordinarily would using arguments to p or x.
Use the y command.
$ cat frames.pl
sub bar {
my $fnord = 42;
1
}
sub baz { bar }
sub foo {
my $fnord = 23;
baz
};
foo;
$ perl -d frames.pl
Loading DB routines from perl5db.pl version 1.37
Editor support available.
Enter h or 'h h' for help, or 'man perldebug' for more help.
main::(frames.pl:10): foo;
DB<1> c 3
main::bar(frames.pl:3): 1
DB<2> y 2 fnord
$fnord = 23
DB<3> T
. = main::bar() called from file 'frames.pl' line 5
. = main::baz() called from file 'frames.pl' line 8
. = main::foo() called from file 'frames.pl' line 10
it's possible to change N (for example second occurrence) in file using one-line sed/awk except such method?:
line_num=`awk '/WHAT_TO_CHANGE/ {c++; if (c>=2) {c=NR;exit}}END {print c}' INPUT_FILE` && sed "$line_num,$ s/WHAT_TO_CHANGE/REPLACE_TO/g" INPUT_FILE > OUTPUT_FILE
Thanks
To change the Nth occurence in a line you can use this:
$ echo foo bar foo bar foo bar foo bar | sed 's/foo/FOO/2'
foo bar FOO bar foo bar foo bar
So all you have to do is to create a "one-liner" of your text, e.g. using tr
tr '\n' ';'
do your replacement and then convert it back to a multiline again using
tr ';' '\n'
This awk solution assumes that WHAT_TO_CHANGE occurs only once per line. The following replaces the second 'one' with 'TWO':
awk -v n=2 '/one/ { if (++count == n) sub(/one/, "TWO"); } 1' file.txt