I figured out how to colorize column 3 in green like this:
green=$'\033[1;32m';off=$'\e[m';echo -e "num co1umn1 column2 column3\n=== === === ===\n1 this is me\n2 that is you"|column -t|sed "s/[^[:blank:]]\{1,\}/$green&$off/3";unset green off
CLI result
How do I need to alter my sed command to colorize the pattern 'is' only within column 3 so that the output becomes:
Wanted result
If you want to color the whole word is, you can use (with GNU sed):
sed "s/\bis\b/$green&$off/"
sed "s/\<is\>/$green&$off/"
Here, \b is a word boundary, \< is a leading word boundary and \> is a trailing word boundary.
Else, you can tell sed to start looking for matches from the third line:
sed "3,$ s/[^[:blank:]]\{1,\}/$green&$off/3"
Output:
One way to do this is to ignore the first two lines of the output in sed:
sed "1,2 ! s/[^[:blank:]]\{1,\}/$green&$off/3";
Using sed
$ ... | sed "/^[[:digit:]]/s/\(\([^ ]* \+\)\{2\}\)\([^ ]*\)/\1$green\3$off/"
Modifying the echo to cover a couple other instances of is ...
$ echo -e "num co1umn1 column2 column3\n=== === === ===\n1 is is me\n2 that isn't you" | column -t
num co1umn1 column2 column3
=== === === ===
1 is is me # only colorize the 2nd occurrence of "is"
2 that isn't you # don't colorize "isn't" in 3rd column
Extending OP's current sed solution:
sed -r "3,$ s/[^[:blank:]]{1,}/XXX&XXX/3; s/XXXisXXX/${green}is${off}/; s/XXX//g"
Where:
3,$ - apply following sed scripts against line numbers 3-to-EOF (ie, skip 1st 2 lines)
first we offset the 3rd column values with XXX bookends (choose a set of characters that you know won't show up anywhere in the data)
then colorize XXXisXXX (removing the XXXs at the same time)
then remove any remaining XXX (from 3rd column in other rows)
This generates:
Related
This picks all the text on single line after a pattern match, and converts it to camel case using non-alphanumeric as separator, remove the spaces at the beginning and at the end of the resulting string, (1) this don't replace if it has 2 consecutive non-alphanumeric chars, e.g "2, " in the below example, (2) is there a way to do everything using sed command instead of using grep, cut, sed and tr?
$ echo " hello
world
title: this is-the_test string with number 2, to-test CAMEL String
end! " | grep -o 'title:.*' | cut -f2 -d: | sed -r 's/([^[:alnum:]])([0-9a-zA-Z])/\U\2/g' | tr -d ' '
ThisIsTheTestStringWithNumber2,ToTestCAMELString
To answer your first question, change [^[:alnum:]] to [^[:alnum:]]+ to mach one ore more non-alnum chars.
You may combine all the commands into a GNU sed solution like
sed -En '/.*title: *(.*[[:alnum:]]).*/{s//\1/;s/([^[:alnum:]]+|^)([0-9a-zA-Z])/\U\2/gp}'
See the online demo
Details
-En - POSIX ERE syntax is on (E) and default line output supressed with n
/.*title: *(.*[[:alnum:]]).*/ - matches a line having title: capturing all after it up to the last alnum char into Group 1 and matching the rest of the line
{s//\1/;s/([^[:alnum:]]+|^)([0-9a-zA-Z])/\U\2/gp} - if the line is matched,
s//\1/ - remove all but Group 1 pattern (received above)
s/([^[:alnum:]]+|^)([0-9a-zA-Z])/\U\2/ - match and capture start of string or 1+ non-alnum chars into Group 1 (with ([^[:alnum:]]+|^)) and then capture an alnum char into Group 2 (with ([0-9a-zA-Z])) and replace with uppercased Group 2 contents (with \U\2).
I have files that need to be removed from comments and white space until keyword . Line number varies . Is it possible to limit multiple continued sed substitutions based on Keyword ?
This removes all comments and white spaces from file :
sed -i -e 's/#.*$//' -e 's/;.*$//' -e '/^$/d' file
For example something like this :
# string1
# string2
some string
; string3
; string4
####
<Keyword_Keep_this_line_and_comments_white_space_after_this>
# More comments that need to be here
; etc.
sed -i '1,/keyword/{/^[#;]/d;/^$/d;}' file
I would suggest using awk and setting a flag when you reach your keyword:
awk '/Keyword/ { stop = 1 } stop || !/^[[:blank:]]*([;#]|$)/' file
Set stop to true when the line contains Keyword. Do the default action (print the line) when stop is true or when the line doesn't match the regex. The regex matches lines whose first non-blank character is a semicolon or hash, or blank lines. It's slightly different to your condition but I think it does what you want.
The command prints to standard output so you should redirect to a new file and then overwrite the original to achieve an "in-place edit":
awk '...' input > tmp && mv tmp input
Use grep -n keyword to get the line number that contains the keyword.
Use sed -i -e '1,N s/#..., when N is the line number that contains the keyword, to only remove comments on the lines 1 to N.
For example, I have a file containing a line as below:
"abc":"def"
I need to insert 123 between "abc":" and def" so that the line will become: "abc":"123def".
As "abc" appears only once so I think I can just search it and do the insertion.
How to do this with bash script such as sed or awk?
AMD$ sed 's/"abc":"/&123/' File
"abc":"123def"
Match "abc":", then append this match with 123 (& will contain the matched string "abc":")
If you want to take care of space before and after :, you can use:
sed 's/"abc" *: *"/&123/'
For replacing all such patterns, use g with sed.
sed 's/"abc" *: *"/&123/g' File
sed:
$ sed -E 's/(:")(.*)/\1123\2/' <<<'"abc":"def"'
"abc":"123def"
(:") gets :" and put in captured group 1
(.*) gets the remaining portion and put in captured group 2
in the replacement, \1123\2 puts 123 between the groups
awk:
$ awk -F: 'sub(".", "&123", $2)' <<<'"abc":"def"'
"abc" "123def"
In the sub() function, the second ($2) field is being operated on, pattern is used as . (which would match "), and in the replacement the matched portion (&) is followed by 123.
echo '"abc":"def"'| awk '{sub(/def/,"123def")}1'
"abc":"123def"
this question originated from string pattaren-matching using awk , basically we are splitting a line of text in multiple groups based on a regex pattern, and then printing two groups only. Now the question is can we right align a group while printing through sed?
below is an example
$cat input.txt
it is line one
it is longggggggg one
itttttttttt is another one
now
$sed -e 's/\(.*\) \(.*\) \(.*\) \(.*\)/\1 \3/g' input.txt
it splits and prints group 1 and 3, but the output is
it line
it longggggggg
itttttttttt another
my question is can we do it through sed so that the output comes as
it line
it longggggggg
itttttttttt another
I did it with awk but I feel it can be done through sed, but I am not able to get how I am going to get the length of the second group and then pad correct number of spaces in between the groups, I am open to any suggestions to try out.
This might work for you (GNU sed):
sed -r 's/^(.*) .* (.*) .*$/\1 \2/;:a;s/^.{1,40}$/ &/;ta;s/^( *)(\S*)/\2\1/' file
or:
sed -r 's/^(.*) .* (.*) .*$/printf "%-20s%20s" \1 \2/e' file
You can use looping in sed to achieve what you want:
#!/bin/bash
echo 'aa bb cc dd
11 22 33333333 44
ONE TWO THREEEEEEEEE FOUR' | \
sed -e 's/\(.*\) \(.*\) \(.*\) \(.*\)/\1 \3/g' \
-e '/\([^ ]*\) \([^ ]*\)/ { :x ; s/^\(.\{1,19\}\) \(.\{1,19\}\)$/\1 \2/g ; tx }'
The two 19's control the width of your columns. The :x is a label which is looped to by tx whenever the preceding substitution succeeded. (You could add a p; before tx to "debug" it.
It most easy to use awk in this case...
You could too use a bash loop to calculate the number of space and run this command on the line covered :
while read; do
# ... calculate $SPACE ...
echo $REPLY|sed "s/\([^\ ]*\)\ *[^\ ]*\ *\([^\ ]*\)/\1$SPACES\2/g"
done < file
But I prefer use awk for do all that (or other advanced shell languages such as Perl, Python, PHP shell mode, ...)
TemplateSpace=" "
TemplateSize=${#TemplateSpace}
sed "
# split your group (based on word here but depend on your real need)
s/^ *\(\w\) \(\w\) \(\w\) \(\w\).*$/\1 \3/
# align
s/$/${TemplateSpace}/
s/^\(.\{${TemplateSize}\}\).*$/\1/
s/\(\w\) \(\w\)\( *\)/\1 \3\2/
"
or more simple for avoiding TemplateSize (and there are no dot in content)
TemplateSpace="............................................................."
and replace
s/^\(.\{${TemplateSize}\}.*$/\1/
by
s/^\(${TemplateSpace}\).*$/\1/
s/\./ /g
Del columns 2 and 4. Right justify resulting col 2 at line length of 23 chars.
sed -e '
s/[^ ]\+/ /4;
s/[^ ]\+//2;
s/^\(.\{23\}\).*$/\1/;
s/\(^[^ ]\+[ ]\+\)\([^ ]\+\)\([ ]\+\)/\1\3\2/;
'
or gnu sed with extended regex:
sed -r '
s/\W+\w+\W+(\w+)\W+\w+$/\1 /;
s/^(.{23}).*/\1/;
s/(+\W)(\w+)(\W+)$/\1\3\2/
'
This question is old, but I like to see it as a puzzle.
While I love the loop solution for its brevity, here is one without a loop or shell help.
sed -E "s/ \w+ (\w+) \w+$/ \1/;h;s/./ /g;s/$/# /;s/( *)#\1//;x;H;x;s/\n//;s/^( *)(\w+)/\2\1/"
or without extended regex
sed "s/ .* \(.*\) .*$/ \1/;h;s/./ /g;s/$/# /;s/\( *\)#\1//;x;H;x;s/\n//;s/^\( *\)\([^ ]*\)/\2\1/"
I have a string which i need to insert at a specific position in a file :
The file contains multiple semicolons(;) i need to insert the string just before the last ";"
Is this possible with SED ?
Please do post the explanation with the command as I am new to shell scripting
before :
adad;sfs;sdfsf;fsdfs
string = jjjjj
after
adad;sfs;sdfsf jjjjj;fsdfs
Thanks in advance
This might work for you:
echo 'adad;sfs;sdfsf;fsdfs'| sed 's/\(.*\);/\1 jjjjj;/'
adad;sfs;sdfsf jjjjj;fsdfs
The \(.*\) is greedy and swallows the whole line, the ; makes the regexp backtrack to the last ;. The \(.*\) make s a back reference \1. Put all together in the RHS of the s command means insert jjjjj before the last ;.
sed 's/\([^;]*\)\(;[^;]*;$\)/\1jjjjj\2/' filename
(substitute jjjjj with what you need to insert).
Example:
$ echo 'adad;sfs;sdfsf;fsdfs;' | sed 's/\([^;]*\)\(;[^;]*;$\)/\1jjjjj\2/'
adad;sfs;sdfsfjjjjj;fsdfs;
Explanation:
sed finds the following pattern: \([^;]*\)\(;[^;]*;$\). Escaped round brackets (\(, \)) form numbered groups so we can refer to them later as \1 and \2.
[^;]* is "everything but ;, repeated any number of times.
$ means end of the line.
Then it changes it to \1jjjjj\2.
\1 and \2 are groups matched in first and second round brackets.
For now, the shorter solution using sed : =)
sed -r 's#;([^;]+);$#; jjjjj;\1#' <<< 'adad;sfs;sdfsf;fsdfs;'
-r option stands for extented Regexp
# is the delimiter, the known / separator can be substituted to any other character
we match what's finishing by anything that's not a ; with the ; final one, $ mean end of the line
the last part from my explanation is captured with ()
finally, we substitute the matching part by adding "; jjjj" ans concatenate it with the captured part
Edit: POSIX version (more portable) :
echo 'adad;sfs;sdfsf;fsdfs;' | sed 's#;\([^;]\+\);$#; jjjjj;\1#'
echo 'adad;sfs;sdfsf;fsdfs;' | sed -r 's/(.*);(.*);/\1 jjjj;\2;/'
You don't need the negation of ; because sed is by default greedy, and will pick as much characters as it can.
sed -e 's/\(;[^;]*\)$/ jjjj\1/'
Inserts jjjj before the part where a semicolon is followed by any number of non-semicolons ([^;]*) at the end of the line $. \1 is called a backreference and contains the characters matched between \( and \).
UPDATE: Since the sample input has no longer a ";" at the end.
Something like this may work for you:
echo "adad;sfs;sdfsf;fsdfs"| awk 'BEGIN{FS=OFS=";"} {$(NF-1)=$(NF-1) " jjjjj"; print}'
OUTPUT:
adad;sfs;sdfsf jjjjj;fsdfs
Explanation: awk starts with setting FS (field separator) and OFS (output field separator) as semi colon ;. NF in awk stands for number of fields. $(NF-1) thus means last-1 field. In this awk command {$(NF-1)=$(NF-1) " jjjjj" I am just appending jjjjj to last-1 field.