I have a file which looks like below:
memory=500G
brand=HP
color=black
battery=5 hours
For every line, I want to remove everything after = and also the =.
Eventually, I want to get something like:
memory:brand:color:battery:
(All on one line with colons after every word)
Is there a one-line sed command that I can use?
sed -e ':a;N;$!ba;s/=.\+\n\?/:/mg' /my/file
Adapted from this fine answer.
To be frank, however, I'd find something like this more readable:
cut -d = -f 1 /my/file | tr \\n :
Here's one way using GNU awk:
awk -F= '{ printf "%s:", $1 } END { printf "\n" }' file.txt
Result:
memory:brand:color:battery:
If you don't want a colon after the last word, you can use GNU sed like this:
sed -n 's/=.*//; H; $ { g; s/\n//; s/\n/:/g; p }' file.txt
Result:
memory:brand:color:battery
This might work for you (GNU sed):
sed -i ':a;$!N;s/=[^\n]*\n\?/:/;ta' file
perl -F= -ane '{print $F[0].":"}' your_file
tested below:
> cat temp
abc=def,100,200,dasdas
dasd=dsfsf,2312,123,
adasa=sdffs,1312,1231212,adsdasdasd
qeweqw=das,13123,13,asdadasds
dsadsaa=asdd,12312,123
> perl -F= -ane '{print $F[0].":"}' temp
abc:dasd:adasa:qeweqw:dsadsaa:
My command is
First step:
sed 's/([a-z]+)(\=.*)/\1:/g' Filename |cat >a
cat a
memory:
brand:
color:
battery:
Second step:
sed -e 'N;s/\n//' a | sed -e 'N;s/\n//'
My output is
memory:brand:color:battery:
Related
I have lines that start like this: 2141058222 11/22/2017 and I want to append a ; at the end of the ten digit number like this: 2141058222; 11/22/2017.
I've tried sed with sed -i 's/^[0-9]\{10\}\\$/;&/g' which does nothing.
What am I missing?
Try this:
echo "2141058222 11/22/2017" | sed -r 's/^([0-9]{10})/&;/'
echo "2141058222 11/22/2017" | sed 's/ /; /'
Output:
2141058222; 11/22/2017
If the input is always in the format specified, GNU cut works, and might even be more efficient than sed:
cut -c -10,11- --output-delimiter ';' <<< "2141058222 11/22/2017"
Output:
2141058222; 11/22/2017
For an input file that'd be:
cut -c -10,11- --output-delimiter ';' file
I like to find some thing with sed in file that have many occurrence
File As below
"xyz": "somename_dsa", some other text, "xyz": "zcbr53", some other text, "xyz": "zms53",
Item needed
I need text after "xyz" :
Using gnu awk
awk -v RS='"xyz"' 'NR>1 {print $2}' file
"somename_dsa",
"zcbr53",
"zms53",
Or just the data
awk -v RS='"xyz"' -F\" 'NR>1 {print $2}' file
somename_dsa
zcbr53
zms53
This might work for you (GNU sed):
sed '/"xyz":\s*/!d;s//\n/g;s/^[^\n]*\n//' file
Delete any lines that don't have the required string. Replace the required string with a newline and chop off the first entry.
You may try this,
$ cat rr.txt
"xyz": "somename_dsa", some other text, "xyz": "zcbr53", some other text, "xyz": "zms53"
$ sed -r '/[^"]*"xyz": ([^,]*)/ s//\1 /g' rr.txt
"somename_dsa" "zcbr53" "zms53"
$ sed -r '/[^"]*"xyz": "([^"]*)"/ s//\1 /g' rr.txt
somename_dsa zcbr53 zms53
All the captured texts are separated by spaces.
I have a file inside that one line contains nested parenthesis, i want to display those words only.
Example:
(abc (defg) or hij(klmn)) and (opq(rstuv))
Expected Result:
defg
klmn
rstuv
I have tried with awk - awk -F "[(())]" '{ for (i=2; i<NF; i+=2) print $i}'
I have tried with sed - sed 's/.*(\([a-zA-Z0-9_]*\)).*/\1/'
Using perl global matching and lazy quantifiers:
#! /usr/bin/perl -n
use feature 'say';
while (/\((.*?\)[^(]*?)\)/g) {
$m=$1;
while ($m =~ /\((.*?)\)/g) {
say $1;
}
}
Output:
defg
klmn
rstuv
Maybe with grep?
$ echo "(abc (defg) or hij(klmn)) and (opq(rstuv))" | grep -o "([a-z]*)"
(defg)
(klmn)
(rstuv)
It catches the groups of ( + letters + ).
I tried to get rid of the paranthesis but could not. This is my approach:
grep -Po '(?<=()[a-z]*(?=))'
but it indicates that "grep: lookbehind assertion is not fixed length", as I guess it cannot decide up to which ) to look for.
This might work for you (GNU sed):
sed -r 's/\(([^()]*)\)/\n\1\n/;s/[^\n]*\n//;/[^()]/P;D' file
I'm trying to extract data/urls (in this case - someurl) from a file that contains them within some tag ie.
xyz>someurl>xyz
I don't mind using either awk or sed.
I think the best, easiest, way is with cut:
$ echo "xyz>someurl>xyz" | cut -d'>' -f2
someurl
With awk can be done like:
$ echo "xyz>someurl>xyz" | awk 'BEGIN { FS = ">" } ; { print $2 }'
someurl
And with sed is a little bit more tricky:
$ echo "xyz>someurl>xyz" | sed 's/\(.*\)>\(.*\)>\(.*\)/\2/g'
someurl
we get blocks of something1<something2<something3 and print the 2nd one.
grep was born to extract things:
kent$ echo "xyz>someurl>xyz"|grep -Po '>\K[^>]*(?=>)'
someurl
you could kill a fly with a bomb of course:
kent$ echo "xyz>someurl>xyz"|awk -F\> '$0=$2'
someurl
If your grep supports P option then you can use lookahead and lookbehind regular expression to identify the url.
$ echo "xyz>someurl>xyz" | grep -oP '(?<=xyz>).*(?=>xyz)'
someurl
This is just a sample to get you started not the final answer.
I have a file called data.txt.
I want to add the current date, or time, or both to the beginning or end of each line.
I have tried this:
awk -v v1=$var ' { printf("%s,%s\n", $0, v1) } ' data.txt > data.txt
I have tried this:
sed "s/$/,$var/" data.txt
Nothing works.
Can someone help me out here?
How about :
cat filename | sed "s/$/ `date`/"
The problem with this
awk -v v1=$var ' { printf("%s,%s\n", $0, v1) } ' data.txt > data.txt
is that the > redirection happens first, and the shell truncates the file. Only then does the shell exec awk, which then reads an empty file.
Choose one of these:
sed -i "s/\$/ $var/" data.txt
awk -v "date=$var" '{print $0, date}' data.txt > tmpfile && mv tmpfile data.txt
However, does your $var contain slashes (such as "10/04/2011 12:34") ? If yes, then choose a different delimiter for sed's s/// command: sed -i "s#\$# $var#" data.txt