i have a file this format :
Word1 : "Word2" Word3
I want to remove "Word2"
Then file must be like this :
Word1 : Word3
How can i do this with sed?
Note : Word2 is changeable
If there is no extra stuff in the input_file, just substitute every quoted phrase with empty string:
sed 's/"[^"]*"//' input_file
If file.txt is your file, you can try this from command line:
sed -e "s_: \"[a-zA-Z0-9]*\" _: _g" -i file.txt
From the command line try
$ sed -i 's/\"Word2\"//g' /path/to/file.txt
If test is your file, try:
# cat test | sed -r 's/\".+\"\ //'
based on the given input. say, Word2 could be anything, but Word1 and Word3 are fixed. sed/awk oneliner could do the job 4 u:
kent$ echo 'Word1 : "Word2" Word3'|sed -r 's/(Word1 :) "[^"]*"( Word3)/\1\2/g'
Word1 : Word3
kent$ echo 'Word1 : "Word2" Word3'|awk -F"\"" '{print $1,$3}'
Word1 : Word3
Related
To delete/comment 3 lines befor a pattern (including the line with the pattern):
how can i achive it through sed command
Ref:
sed or awk: delete n lines following a pattern
the above ref blog help to achive the this with after a pattern match but i need to know before match
define host{
use xxx;
host_name pattern;
alias yyy;
address zzz;
}
the below sed command will comment the '#' after the pattern match for example
sed -e '/pattern/,+3 s/^/#/' file.cfg
define host{
use xxx;
#host_name pattern;
#alias yyy;
#address zzz;
#}
like this how can i do this for the before pattern?
can any one help me to resolve this
If tac is allowed :
tac|sed -e '/pattern/,+3 s/^/#/'|tac
If tac isn't allowed :
sed -e '1!G;h;$!d'|sed -e '/pattern/,+3 s/^/#/'|sed -e '1!G;h;$!d'
(source : http://sed.sourceforge.net/sed1line.txt)
Reverse the file, comment the 3 lines after, then re-reverse the file.
tac file | sed '/pattern/ {s/^/#/; N; N; s/\n/&#/g;}' | tac
#define host{
#use xxx;
#host_name pattern;
alias yyy;
address zzz;
}
Although I think awk is a little easier to read:
tac file | awk '/pattern/ {c=3} c-- > 0 {$0 = "#" $0} 1' | tac
This might work for you (GNU sed):
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/s/^/#/mg;P;D' file
Gather up 4 lines in the pattern space and if the last line contains pattern insert # at the beginning of each line in the pattern space.
To delete those 4 lines, use:
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/d;P;D' file
To delete the 3 lines before pattern but not the line containing pattern use:
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/s/.*\n//;P;D'
My file contains x number of lines, I would like to remove the string before and after the reference string at the beginning and end of each line.
The reference string and string to remove are separated by space.
The file contains :
test.user.passs
test.user.location
global.user
test.user.tel
global.pass
test.user.email string_err
#ttt...> test.user.car ->
test.user.address
è_ 788 test.user.housse
test.user.child
{kl78>&é} global.email
global.foo
test.user.foo
How to remove the string at the start of each line which contain "test" string and also the end of each line separated by space or tab with sed?
The desired result is :
test.user.passs
test.user.location
global.user
test.user.tel
global.pass
test.user.email
test.user.car
test.user.address
test.user.housse
test.user.child
{kl78>&é} global.email
global.foo
test.user.foo
I interpret your question as: find the first word that is "word characters and at least one dots"
Tcl:
echo '
set fh [open [lindex $argv 1] r]
while {[gets $fh line] != -1} {puts [regexp -inline {\w+(?:\.\w+)+} $line]}
' | tclsh - file
sed
sed -r 's/.*\<([[:alpha:]]+(\.[[:alpha:]]+)).*/\1/' file
perl
perl -nE '/(\w+(\.\w+)+)/ and say $1' file
using sed like
sed -r 's/^[^ ]+[ ]+([^ ]+)[ ]+[^ ]*/\1/' file
This might work for you (GNU sed):
sed -r 's/.*(test\S+).*/\1/' file
When I want to print an output like this
./myScript (prints some lines)
or
cat myFile
I want the output to show with linebreakers , for example each line will include not more than 100 chars.
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaffffff
vbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbf
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
there is something I can add to the command line to get this result ?
Thanks.
You can use sed if you want the line terminator as ,.
$ cat myfile
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaffffffvbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbfaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
$ sed -r 's/.{50}/&,\n/g' myfile
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa,
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaffffffvbbb,
bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbfaaaaaa,
aaaaaaaaaaaaaaaaaaaaaaaaaaaaa
fold is another utility but won't add a , at the end
$ fold -w50 myfile
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaffffffvbbb
bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbfaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaa
My text looks like this:
cat
catch
cat_mouse
catty
I want to replace "cat" with "dog".
When I do
sed "s/cat/dog/"
my result is:
dog
catch
cat_mouse
catty
How do I replace with sed if only part of the word matches?
There's a mistake :
You lack the g modifier
sed 's/cat/dog/g'
g
Apply the replacement to all matches to the regexp, not just the first.
See
http://www.gnu.org/software/sed/manual/html_node/The-_0022s_0022-Command.html
http://sed.sourceforge.net/sedfaq3.html#s3.1.3
If you want to replace only cat by dog only if part of the word matches :
$ perl -pe 's/cat(?=.)/dog/' file.txt
cat
dogch
dog_mouse
dogty
I use Positive Look Around, see http://www.perlmonks.org/?node_id=518444
If you really want sed :
sed '/^cat$/!s/cat/dog/' file.txt
bash-3.00$ cat t
cat
catch
cat_mouse
catty
To replace cat only if it is part of a string
bash-3.00$ sed 's/cat\([^$]\)/dog\1/' t
cat
dogch
dog_mouse
dogty
To replace all occurrences of cat:
bash-3.00$ sed 's/cat/dog/' t
dog
dogch
dog_mouse
dogty
awk solution for this
awk '{gsub("cat","dog",$0); print}' temp.txt
I want to print next line of matching word with sed.
I tried this command but it gives error :
sed -n '/<!\[CDATA\[\]\]>/ { N p}/' test.xml
what about grep -e -A 1 regex? It will print line below regex.
With sed, looking for pattern "dd", below works fine as you would:
sed -n '/dd/ {n;p}' file
For file content:
dd
aa
ss
aa
It prints:
aa
use awk
awk '/pattern/{getline;print}' file