This question already has answers here:
sed unknown option to `s' in bash script [duplicate]
(4 answers)
Closed 2 years ago.
I am trying
grep searchterm myfile.csv | sed 's/replaceme/withthis/g'
and getting
unknown option to `s'
What am I doing wrong?
Edit:
As per the comments the code is actually correct. My full code resembled something like the following
grep searchterm myfile.csv | sed 's/replaceme/withthis/g'
# my comment
And it appears that for some reason my comment was being fed as input into sed. Very strange.
use the --expression option
grep searchterm myfile.csv | sed --expression='s/replaceme/withthis/g'
use "-e" to specify the sed-expression
cat input.txt | sed -e 's/foo/bar/g'
To make sed catch from stdin , instead of from a file, you should use -e.
Like this:
curl -k -u admin:admin https://$HOSTNAME:9070/api/tm/3.8/status/$HOSTNAME/statistics/traffic_ips/trafc_ip/ | sed -e 's/["{}]//g' |sed -e 's/[]]//g' |sed -e 's/[\[]//g' |awk 'BEGIN{FS=":"} {print $4}'
If you are trying to do an in-place update of text within a file, this is much easier to reason about in my mind.
grep -Rl text_to_find directory_to_search 2>/dev/null | while read line; do sed -i 's/text_to_find/replacement_text/g' $line; done
Open the file using vi myfile.csv
Press Escape
Type :%s/replaceme/withthis/
Type :wq and press Enter
Now you will have the new pattern in your file.
Related
This question already has an answer here:
shell variable in a grep regex
(1 answer)
Closed 4 years ago.
I am writing a bash shell script in which I want to use two shell commands sed and egrep.
My bash shall script read a text file q2.txt and then do some actions using egrep and sed.
The code is as given below.
#!/bin/bash
var=$(<q2.txt)
sed "s/^[ \t]*//" -i var
grep -v '^/\*.*\*/$' var
echo "$var"
I read the content of q2.txt in variable var. Then remove the tabs and spaces using sed s/^[ \t]*//" -i var and update my var.
Then execute grep -v '^/\*.*\*/$' var on my updated variable to select some lines with specific start and end.
But in the ouput, It seems like grep and sed are not applicable to var.
Output
sed: can't read var: No such file or directory
grep: var: No such file or directory
Here is a much simpler approach
cat q2.txt | sed "s/^[ \t]*//" | grep -v '^/\*.*\*/$'
The problem is more about shell variables, rather than sed/grep.
See:
https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html
http://wiki.bash-hackers.org/syntax/pe
Also better to use the Pipe (|) as the other answer has showed, so you don't have to store the intermediate result.
This question already has answers here:
Can grep show only words that match search pattern?
(15 answers)
Closed 8 years ago.
I have a bash file that has below line along with other lines.
var BUILD_VERSION = '2014.17.10_23';
I just want to extract 2014.17.10_23 and this value may change so something like grep for 2014* . However when I do that I get the whole line returned instead of the value 2014.17.10_23.
What would be the best way to achieve this?
Thanks
Using awk:
awk -F= '/BUILD_VERSION/{print $2}' input | tr -d "[' ;]"
And with sed:
sed -n "/BUILD_VERSION/s/.*'\([^']*\)'.*/\1/p" input
grep 'BUILD_VERSION' <your file> | sed -e 's/var BUILD_VERSION = //g'
Would get you '2014.17.10_23'; tweak the sed expression (or pipe it through a few more) to get rid of quotes.
It would be a 1 liner regex in Perl...
Here is another awk solution:
awk -F' = ' '/BUILD_VERSION/ {gsub(/\x27|;/,""); print $NF}'
You can use this awk
awk -F\' '/BUILD_VERSION/ {print $2}' file
2014.17.10_23
How to use sed to find lines with the word linux? As later display a first line 10 with the word linux?
EX.:
cat file | sed -e '/linux/!d' -e '10!d' ### I can not display the first 10 lines of the word linux
cat file | sed '/linux/!d' | sed '10!d' ### It is well
How to make it work with one sed?
cat file | sed -e '/linux/!d; ...?; 10!d'
...? - storing of the buffer linux? 10 later cut the lines?
Someone explain to me?
I would use awk:
awk '/linux/ && c<10 {print;c++} c==10 {exit}' file
This might work for you (GNU sed):
sed -nr '/linux/{p;G;/(.*\n){10}/q;h}' file
Print the line if it contains the required string. If the required number of lines has already been printed quit, otherwise store the line and previous lines in the hold space.
You could use perl:
perl -ne 'if (/linux/) {print; ++$a;}; last if $a==10' inputfile
Using GNU sed:
sed -rn "/linux/{p;x;s/^/P/;ta;:a;s/^P{10}$//;x;Tb;Q;:b}" filename
Thanks. You are great. All of the examples look very nice. Wooow :) It is a pity that I can not do that.
I have not seen for 'r' option in sed. I need to learn.
echo -e 'windows\nlinux\nwindows\nlinux\nlinux\nwindows' | sed -nr '/linux/{p;G;/(.*\n){2}/q;h}'
It works very well.
echo -e 'windows\nlinux\nwindows\nlinux\nlinux\nwindows' | sed -nr '/linux/{p;G;/(.*\n){2}/q;h}' | sed '2s/linux/debian/'
Can I ask you one more example? How to get a result at one sed?
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Use grep to report back only line numbers
I only want to see the line number. I don't need to see the remaining output.
Pipe your grep -n output, which normally looks something like:
11: stuff that matched
43: more stuff that matched
through sed to strip out the matching parts:
grep -n pattern file | sed -e 's/:.*//g'
11
43
grep -n or --line-number option will do this for you. You can find this information in the grep help file, which you can find by using grep --help or grep --help | less to read it more carefully. Also consider using the manual page: man grep
You could use awk too.
grep -n word file | awk -F: '{ print $1 }'
As #Barmar pointed out you could just use an awk one-liner as such:
awk '/regex/ { print NR }' file
Since you don't have awk you could also use cut:
grep -n word file | cut -d: -f1
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