Match specific lines in sed command - unix

How can I match a set of specific lines for a substitution command?
(incorrect):
sed -e'71,116,211s/[ ]+$//' ...
I want to strip trailing whitespace on lines 71, 116 and 211 only


You could try something like:
awk 'NR== 71 || NR == 116 || NR == 211 {sub(/ *$/,"",$0)}{print $0}'
or
sed '71s/ *$//;116s///;211s///'


sed '71bl;116bl;211bl;b;:l;s/[ ][ ]*$//' input
For any specified line, this script jumps to the label l. Other lines will jump to the end of the script with the bare branch.
And an awk solution:
awk -v k="71,116,221" 'BEGIN{split(k,a,",")}
(NR in a) { sub(/ *$/,"",$0) }1' input


This might work for you (GNU sed):
sed -r '/\s+$/!b;71s///;116s///;221s///' file
or perhaps:
sed -e '/ *$/!b' -e '71s///' -e '116s///' -e '221s///' file
or as has been said already:
sed -e '71ba' -e '116ba' -e '221ba' -e 'b' -e ':a' -e 's/ *$//' file

Related

Ignore specific patterns in line using sed

I'm trying to grep the word which starts with group keyword & ends with -wx in the given line. Also I need to ignore the below words.
Starts with default:group and ends with -wx
group::-wx
My Findings
echo "# file: /test/test123 # owner: own # group: acct user::r-- group::r-x mask::rwx other::r-x default:user::r-- default:user:an:--x default:group::r-x default:group:fin:-wx default:mask::rwx default:other::r-x" | grep -o "group:[^ ]*-wx" | sed '/group::-wx/d';'/default:[^ ]*:[^ ]*-wx/d'
Actual result
fin:-wx
Expected result
<null>

You already have a grep to select what you want, simply add grep statements to remove those you do not want.
Like so:
LINE="# file: /test/test123 # owner: own # group: acct user::r-- group::r-x mask::rwx other::r-x default:user::r-- default:user:an:--x default:group::r-x default:group:fin:-wx default:mask::rwx default:other::r-x"
echo $LINE | grep -o "group:[^ ]*-wx" \
| grep -vo "default:group:[^ ]*-wx" \
| grep -vo "group::-wx"
On my linux it returns nothing, which is what you expected. I do not have other test samples, but I think this is ok.

As you are first extracting the substring group:fin:-wx out of
default:group:fin:-wx with grep, the following sed filter
/default:[^ ]*:[^ ]*-wx/d no longer works.
A workaround is to change the order of filtering:
str="# file: /test/test123 # owner: own # group: acct user::r-- group::r-x mask::rwx other::r-x default:user::r-- default:user:an:--x default:group::r-x default:group:fin:-wx default:mask::rwx default:other::r-x"
echo "$str" | sed -e 's/default:group:[^ ]*-wx//' -e 's/group::-wx//' | grep -o 'group:[^ ]*-wx'
As an alternative, if your grep supports -P option, you can make use of positive lookbehind as:
echo "$str" | grep -Po '(?<= )group:[^ ]*-wx' | sed -e '/group::-wx/d' -e '/default:[^ ]*:[^ ]*-wx/d'
The pattern (?<= ) forces the pattern match preceded by a whitespace without
including it in the output.

using sed or awk to double quote comma separate and concatenate a list

I have the following list in a text file:
10.1.2.200
10.1.2.201
10.1.2.202
10.1.2.203
I want to encase in "double quotes", comma separate and join the values as one string.
Can this be done in sed or awk?
Expected output:
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203","10.1.2.204"

The easiest is something like this (in pseudo code):
Read a line;
Put the line in quotes;
Keep that quoted line in a stack or string;
At the end (or while constructing the string), join the lines together with a comma.
Depending on the language, that is fairly straightforward to do:
With awk:
$ awk 'BEGIN{OFS=","}{s=s ? s OFS "\"" $1 "\"" : "\"" $1 "\""} END{print s}' file
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"
Or, less 'wall of quotes' to define a quote character:
$ awk 'BEGIN{OFS=",";q="\""}{s=s ? s OFS q$1q : q$1q} END{print s}' file
With sed:
$ sed -E 's/^(.*)$/"\1"/' file | sed -e ':a' -e 'N' -e '$!ba' -e 's/\n/,/g'
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"
(With Perl and Ruby, with a join function, it is easiest to push the elements onto a stack and then join that.)
Perl:
$ perl -lne 'push #a, "\"$_\""; END{print join(",", #a)}' file
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"
Ruby:
$ ruby -ne 'BEGIN{#arr=[]}; #arr.push "\"#{$_.chomp}\""; END{puts #arr.join(",")}' file
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"

here is another alternative
sed 's/.*/"&"/' file | paste -sd,
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"

awk -F'\n' -v RS="\0" -v OFS='","' -v q='"' '{NF--}$0=q$0q' file
should work for given example.
Tested with gawk:
kent$ cat f
10.1.2.200
10.1.2.201
10.1.2.202
10.1.2.203
kent$ awk -F'\n' -v RS="\0" -v OFS='","' -v q='"' '{NF--}$0=q$0q' f
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"

$ awk '{o=o (NR>1?",":"") "\""$0"\""} END{print o}' file
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"

Using sed to enclose matches in double quotes

I'm trying to extract headers from emails and create a JSON fragment from them. I'm using sed to pull out the keys and values, but it's failing to put the trailing quote on each of the lines:
$ cat email1 | grep -i -e "^subject:" -e "^from:" -e "^to:" | \
sed -n 's/\^([^:]*\):[ ]*\(.*\)$/"\1":"\2"/gp'
"From":"Blah Blech <blah.blech#blahblech.com>
"To":"foo#bar.com
"Subject":"Yeah
I don't understand why the replacement pattern isn't working.

awk to the rescue!
$ awk -F": *" -vOFS=":" -vq="\"" 'tolower($0)~/^from|to|subject/
{print q$1q,q$2q}' email1
which combines cat or grep steps as well.

Stripping the carriage returns as #tripleee suggested fixed the issue with sed (using ctrl-v ctrl-m to capture the literal carriage return):
$ cat email1 | tr -d '^M' | grep -i -e "^subject:" -e "^from:" -e "^to:" | \
sed -n 's/^\([^:]*\):[ ]*\(.*\)$/"\1":"\2"/gp'
"From":"Blah Blech <blah.blech#blahblech.com>"
"To":"foo#bar.com"
"Subject":"Yeah"

to read a file in UNIX and manipulate

I have one requirement that i have to read the file and manipulate. I have to replace the single double quote into double double quote if it is found in any fields. fields are separated by |.
Please find below for better understanding.
Input:
1234567|9393874|"Hi"|"How are "you""
98647489|20370483|"i am "good""|"what about "you""
output :
1234567|9393874|"Hi"|"How are ""you"""
98647489|20370483|"i am ""good"""|"what about ""you"""

I would replace all the "edge" quotes with another character and then replace the "inner" ones:
sed -e 's/|"/|_/g' -e 's/"|/_|/g' -e 's/"$/_/' file | sed 's/"/""/g' | sed 's/_/"/g'
It returns:
1234567|9393874|"Hi"|"How are ""you"""
98647489|20370483|"i am ""good"""|"what about ""you"""
Step by step:
$ sed -e 's/|"/|_/g' -e 's/"|/_|/g' -e 's/"$/_/' a
1234567|9393874|_Hi_|_How are "you"_
98647489|20370483|_i am "good"_|_what about "you"_
$ sed -e 's/|"/|_/g' -e 's/"|/_|/g' -e 's/"$/_/' a | sed 's/"/""/g'
1234567|9393874|_Hi_|_How are ""you""_
98647489|20370483|_i am ""good""_|_what about ""you""_
$ sed -e 's/|"/|_/g' -e 's/"|/_|/g' -e 's/"$/_/' a | sed 's/"/""/g' | sed 's/_/"/g'
1234567|9393874|"Hi"|"How are ""you"""
98647489|20370483|"i am ""good"""|"what about ""you"""

Text replace on unix

I would like to replace these statement by the following, on a Unix system, does some one know how I can do that ?
/www/docs/syndrome.ms.fcm
by
$_SERVER['DOCUMENT_ROOT']

Yes, with sed :
sed -i "s#/www/docs/syndrome.ms.fcm#\$_SERVER['DOCUMENT_ROOT']#g" $(
grep -l "/www/docs/syndrome.ms.fcm" *files
)
If you don't have the -i switch :
for f in $(grep -l "/www/docs/syndrome.ms.fcm" *files); do
sed "s#/www/docs/syndrome.ms.fcm#\$_SERVER['DOCUMENT_ROOT']#g" "$f" > newfile &&
mv newfile "$f"
done

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