how i can add Add text at the beginning of each line?
for example:- i have file contain:-
/var/lib/svn/repos/b1me/products/payone/generic/code/core
/var/lib/svn/repos/b1me/products/payone/generic/code/fees
/var/lib/svn/repos/b1me/products/payone/generic/code/2ds
i want it to become:-
svn+ssh://svn.xxx.com.jo/var/lib/svn/repos/b1me/products/payone/generic/code/core
svn+ssh://svn.xxx.com.jo/var/lib/svn/repos/b1me/products/payone/generic/code/fees
svn+ssh://svn.xxx.com.jo/var/lib/svn/repos/b1me/products/payone/generic/code/2ds
in other word i want to add "svn+ssh://svn.xxx.com.jo" at the beginning of each line of this file
One way to do this is to use awk.
awk '{ printf "svn+ssh://svn.xxx.com.jo"; print }' <filename>
If you want to modify the file in place, you can use sed with the -i switch.
sed -i -e 's_.*_svn+ssh://svn.xxx.com.jo&_' <filename>
Using sed:
printf "line1\nline2\n" | sed "s/^/new text /"
Using ex:
printf "line1\nline2\n" | ex -s +"%s/^/foo bar /e" +%p -cq! /dev/stdin
Using vim:
printf "line1\nline2\n" | vim - -es +"%s/^/foo bar /e" +%p -cq!
Using shell:
printf "line1\nline2\n" | while read line; do echo foo bar $line; done
ruby -pne 'sub(/^/,"svn+ssh://svn.xxx.com.jo")' file
Simple way:
sed -i 's_._svn+ssh://svn.xxx.com.jo_' <filename>
It can also be done with Perl:
perl -pe 's#^#svn+ssh://svn.xxx.com.jo#' input.file
Related
I've 95 files that looks like :
2019-10-29-18-00/dev/xx;512.00;0.4;/var/x/xx/xxx
2019-10-29-18-00/dev/xx;512.00;0.68;/xx
2019-10-29-18-00/dev/xx;512.00;1.84;/xx/xx/xx
2019-10-29-18-00/dev/xx;512.00;80.08;/opt/xx/x
2019-10-29-18-00/dev/xx;20480.00;83.44;/var/x/x
2019-10-29-18-00/dev/xx;3584.00;840.43;/var/xx/x
2019-10-30-00-00/dev/xx;2048.00;411.59;/
2019-10-30-00-00/dev/xx;7168.00;6168.09;/usr
2019-10-30-00-00/dev/xx;3072.00;1036.1;/var
2019-10-30-00-00/dev/xx;5120.00;348.72;/tmp
2019-10-30-00-00/dev/xx;20480.00;2033.19;/home
2019-10-30-12-00;/dev/xx;5120.00;348.72;/tmp
2019-10-30-12-00;/dev/hd1;20480.00;2037.62;/home
2019-10-30-12-00;/dev/xx;512.00;0.43;/xx
2019-10-30-12-00;/dev/xx;3584.00;794.39;/xx
2019-10-30-12-00;/dev/xx;512.00;0.4;/var/xx/xx/xx
2019-10-30-12-00;/dev/xx;512.00;0.68;/xx
2019-10-30-12-00;/dev/xx;512.00;1.84;/var/xx/xx
2019-10-30-12-00;/dev/xx;512.00;80.08;/opt/xx/x
2019-10-30-12-00;/dev/xx;20480.00;83.44;/var/xx/xx
2019-10-30-12-00;/dev/x;3584.00;840.43;/var/xx/xx
For some lines I've 2019-10-29-18-00/dev and for some other lines, I've 2019-10-30-12-00;/dev/
I want to add the ; before the /dev/ where it is missing, so for that I use this sed command :
sed 's/\/dev/\;\/dev/'
But How I can apply this command for each lines where the ; is missing ? I try this :
for i in $(cat /home/xxx/xxx/xxx/*.txt | grep -e "00/dev/")
do
sed 's/\/dev/\;\/dev/' $i > $i
done
But it doesn't work... Can you help me ?
Could you please try following with GNU awkif you are ok with it.
awk -i inplace '/00\/dev\//{gsub(/00\/dev\//,"/00;/dev/")} 1' *.txt
sed solution: Tested with GNU sed for few files and it worked fine.
sed -i.bak '/00\/dev/s/00\/dev/00\;\/dev/g' *.txt
This might work for you (GNU sed & parallel):
parallel -q sed -i 's#;*/dev#;/dev#' ::: *.txt
or if you prefer:
sed -i 's#;*/dev#;/dev#' *.txt
Ignore lines with ;/dev.
sed '/;\/dev/{p;d}; s^/dev^;/dev^'
The /;\/dev/ check if the line has ;/dev. If it has ;/dev do: p - print the current line and d - start from the beginning.
You can use any character with s command in sed. Also, there is no need in escaping \;, just ;.
How I can apply this command for each lines where the ; is missing ? I try this
Don't edit the same file redirecting to the same file $i > $i. Think about it. How can you re-write and read from the same file at the same time? You can't, the resulting file will be in most cases empty, as the > $i will "execute" first making the file empty, then sed $i will start running and it will read an empty file. Use a temporary file sed ... "$i" > temp.txt; mv temp.txt "$i" or use gnu extension -i sed option to edit in place.
What you want to do really is:
grep -l '00/dev/' /home/xxx/xxx/xxx/*.txt |
xargs -n1 sed -i '/;\/dev/{p;d}; s^/dev^;/dev^'
grep -l prints list of files that match the pattern, then xargs for each single one -n1 of the files executes sed which -i edits files in place.
grep for filtering can be eliminated in your case, we can accomplish the task with a single sed command:
for f in $(cat /home/xxx/xxx/xxx/*.txt)
do
[[ -f "$f" ]] && sed -Ei '/00\/dev/ s/([^;])(\/dev)/\1;\2/' "$f"
done
The easiest way would be to adjust your regex so that it's looking a bit wider than '/dev/', e.g.
sed -i -E 's|([0-9])/dev|\1;/dev|'
(note that I'm taking advantage of sed's flexible approach to delimiters on substitute. Also, -E changes the group syntax)
Alternatively, sed lets you filter which lines it handles:
sed -i '/[0-9]\/dev/ s/\/dev/;/dev/'
This uses the same substitution you already have but only applied on lines that match the filter regex
I'm trying to add a line at the beginning of a file, using
echo 'time/F:x1:x2' | cat - file.txt>newfile.txt
But this produces line breaks at each line in the new file (except for after the added 'time/F:x1:x2' line). Any ideas on how to avoid this?
Use -n to disable the trailing newline:
echo -n 'time/F:x1:x2' | cat - file.txt > newfile.txt
There are other ways, too:
sed '1s|^|time/F:x1:x2|' file.txt > newfile.txt
How about
{ echo 'time/F:x1:x2'; cat file.txt; } >newfile.txt
or
sed '1i\
time/F:x1:x2' file.txt > newfile.txt
Actually you don't even need the echo and pipe if you're using bash. Just use a herestring:
<<< 'time/F:x1:x2' cat - file.txt > newfile.txt
I am trying to extract text between pattern1 (fixed) and pattern2 (this can be p2-1/p2-2).
can you please tell me how to achieve this in a single command?
A file starts with start and ends with either end or close
File1:
======
junktest
data
start
stackoverflow
sed
close
File2:
======
data2
start
stackoverflow
end
I can extract text from File1 with
sed -n "/start/,/close/p"
And from File2 with
sed -n "/start/,/end/p"
I need a single sed command to achieve both..
something like:
sed -n "/start/, /close or end /p"
Both GNU sed and BSD sed:
sed -nE '/start/,/close|end/p' file
This awk looks better
awk '/start/,/end|close/' file
sed -n -E "/Word1/,/Word2-1/p" | sed -n -E "/Word1/,/Word2-2/p"
Easy with awk:
$ awk '/start/{p=1}p{print}/end|close/{p=0}' file
I have been trying quite a few ways with no luck. I have a file named test.txt that has some lorem ipsum and the text [staging: production] I simply want to add a few lines that I have saved in a variable in before it.
If you could explain where I have gone wrong with any of the below it would be much appreciated!
#!/bin/bash
test="lala\
kjdsh"
sed '/^#$/{N; /[staging: production]/ i \
<Location /cgis> \
</Location>\
}' ./test.txt
sed -i -e 's/\[staging\: production\]/\$test/g' ./test.txt
#sed -i 's/Lorem/beautiful/g' test.txt
#awk -v data=$test '{A[NR]=$0}/\[staging\: production\]/{ print data }' test.txt > testfile.txt
#read -a text <<<$(cat test.txt)
#echo ${#text[#]}
#for i in ${text[#]};
#do
# echo -n $i;
# sleep .2;
#done
#ed -s test.txt <<< $'/\[staging\: production\]/s/lalalala/g\nw'
#awk -v data=$test '/\(/\[staging\: production\]\)/ { print data }' test.txt > testfile.txt
# && mv testfile.txt test.txt
#sed -i -e '/\(\[staging\: production\]\)/r/$test\1/g' test.txt
#sed "/\(\[staging\: production\]\)/s//$test\1/g" test.txt
sed -i -e 's/\[staging\: production\]/\$test/g' ./test.txt
won't work because inside singe quotes BASH will not expand \$test.
Therefore you don't need to escape the $.
If you want to substitute with the contents of the variable $test do:
sed -i -e 's/\[staging: production\]/'$test'/g' ./test.txt
You also do not need to escape :
To insert before your pattern works for me this way:
sed -i -e '/\[staging: production\]/ i '$test'' ./test.txt
However to preserve the linebreak inside the variable I needed to define:
test="lala\nkjdsh"
Please note the \n to encode the linebreak.
Try it in perl, it seems to work fine:
perl -pe '{$rep="what\nnow"; s/(\[foo foo2\])/$rep$1/}' file
This might work for you (GNU sed):
test="lala\\
kjdsh"
sed '/\[staging: production\]/i\'"$test" test.txt
N.B. \\ in the variable and the variable is surrouded by "'s in the sed command.
Here's what output looks like, basically:
? RESTRequestParamObj.cpp
? plugins/dupfields2/_DupFields.cpp
? plugins/dupfields2/_DupFields.h
I need to get the filenames from second column and pass them to rm. There's AWK script that goes like awk '{print $2}' but I was wondering if there's another solution.
If you have spaces between the ? and the filename then:
cut -c9-
If they're tabs then:
cut -f2
Placed your output in file
$> cat ./text
? RESTRequestParamObj.cpp
? plugins/dupfields2/_DupFields.cpp
? plugins/dupfields2/_DupFields.h
Edit it with sed
$> cat ./text | sed -r -e 's/(\?[\ \t]*)(.*)/\2/g'
RESTRequestParamObj.cpp
plugins/dupfields2/_DupFields.cpp
plugins/dupfields2/_DupFields.h
Sed in here is matching 2 parts of line -
? with tabs or spaces
Other characters until the end f the line
And then it changes whole line only with second part.
This might work for you:
echo "? RESTRequestParamObj.cpp" | sed -e 's/^\S\+/rm /' | sh
or using GNU sed
echo "? RESTRequestParamObj.cpp"| sed -r 's/^\S+/rm /e'
bash only solution, assuming your output comes from stdin:
while read line; do echo ${line##* }; done
use cut/perl instead
cut -f2 -t'\t'|xargs rm -rf
<your output>|perl -ne '#cols = split /\t/; print $cols[1]'|xargs rm -rf