Linux replace newline - linux

Input
aaaa
bbbb
cccc
Output
aaaa
< _jupijuice_ >
bbbb
< _jupijuice_ >
cccc
< _jupijuice_ >
In simple words, replace \n with \n< jupijuice >\n


awk '{print; print "< _jupijuice_ >"}' input.txt
Output:
aaaa
< _jupijuice_ >
bbbb
< _jupijuice_ >
cccc
< _jupijuice_ >


In regular expressions you can use $ to match end of line:
sed 's/$/\n< _jupijuice_ >/' input.txt


sed 'a\
< _jupijuice_ >' YourFile
append a line <...> after each line


This should do:
awk '$0=$0"\n< _jupijuice_ >"'

Related

How to read the file and remove the field of each line then add the extra row to each line

I do have two shell script , read the file(file1) and remove the first field of the file ( output result redirect to the file2)
, then second script again read above modified file (file2) then add extra field to each line (file3),
How to do this in single script without using two shell script .
#!/bin/bash
# loop on all .txt files
for i in File1.txt; do
# remove first column
cut -d' ' -f2- < $i > File2.txt
done
#!/bin/bash
filename='File2.txt'
while read line; do
# reading each line
echo "$RANDOM $line" >> File3.txt
done < $filename
File1.txt
Date Field2 Field3
20111 aaaa bbbb
33111 bbbb vvvv
44444 cccc gggg
File2.txt
Field2 Field3
aaaa bbbb
bbbb vvvv
cccc gggg
File3.txt
New Field2 Fileld3
1 aaaa bbbb
2 bbbb vvvv
1 cccc gggg

One idea:
new="New"
while read -r ignore rest_of_line
do
echo "$new $rest_of_line"
new=$RANDOM
done < file1.txt > file3.txt
This generates:
$ cat file3.txt
New Field2 Field3
29258 aaaa bbbb
31885 bbbb vvvv
15550 cccc gggg
NOTE: it's not clear (to me) what the input/output field delimiters are so for now I'm assuming any white space on input and a single space on output; should be (relatively) easy to modify per OP's requirement

How to replace rows starting with a certain character with emptly lines in text file with Linux/Unix?

I'm using Ubuntu and have a large text file where certain rows start with < character what I'd like to replace each of these to an empty row. So from this:
eeeeee
<
<
aaaa
bbbb
cccc
<
dddddd
<
ff
I want this:
eeee
aaaa
bbbb
cccc
dddddd
ff
(In case of multiple consecutive < rows only one empty rows is needed ideally)
How to perform this in command line?

This Perl one-liner should do what you're asking for:
perl -ne 'if (/^</) {print "\n" if !$f; $f=1} else {$f=0; print}' tmp.txt
Here it is in action:
# cat tmp.txt
eeeeee
<
<
aaaa
bbbb
cccc
<
dddddd
<
ff
# perl -ne 'if (/^</) {print "\n" if !$f; $f=1} else {$f=0; print}' tmp.txt
eeeeee
aaaa
bbbb
cccc
dddddd
ff
Commented code:
# Found a line starting with '<'
if (/^</) {
# Print a blank line if $f is false
print "\n" if !$f;
# Set $f to true so subsequent lines starting with '<' are ignored
$f=1;
} else {
# Not a line starting with '<'; reset $f to false
$f=0;
# Print the current line
print;
}

You could use sed. In this case the command would look something like:
sed '/</c\\' file.txt
This would find a line with a '<' character at the beginning, and replace the line with nothing. It would not however replace multiple empty rows with a single row.

Linux grep, how can I display lines that don't contain word 1 and word 2 but still display the lines that have both words in them

I need some help with displaying all lines that don't contain word1 or word2 but lines that contain both of them have to be shown.
Example:
aaaa bbbb cccc
bbbb bbbb bbbb
cccc cccc cccc
dddd dddd aaaa
if word1 = aaaa and word2 = bbbb then output should be:
aaaa bbbb cccc
cccc cccc cccc
Tried
grep -Ewv "word1/word2" file.txt
but this shows only lines that don't contain them, it doesn't show lines containing both
I need to do this with grep command, forgot to mention this

Grep version of both or none of each:
grep -v -P '((?=.*aaaa)(?!.*bbbb))|((?=.*bbbb)(?!.*aaaa))'
But please do not use grep in this case. Negative and positive look ahead can easily lead to Catastrophic Backtracking
GNU grep knows Perl compatible regular expression (PCRE) syntax (option -P). This thing is still called a "regular" expression, although it not regular anymore. Other people are more explicit and call backtracking irregular expressions.
How it works:
(?=.*aaaa) matches aaaa anywhere in the line, but does not move the cursor. After the match the next search starts at the beginning of the line.
(?!.*bbbb) matches when no bbbb is in the line and does not move the cursor either.
Both together matches lines, which include aaaa but do not include bbbb.
This is one of the cases you want to exclude, from your search results. The second behind the or condition (|) is the other one you want to exclude: any bbbb without a aaaa.
With the above, you have defined, what you do not want. Next use -v to invert the search to get what you want.

Bash version of both or none of each:
#! /bin/bash
word1=${1:-aaaa}
word2=${2:-bbbb}
while read -r line; do
if [[ $line =~ $word1 ]]; then
if [[ $line =~ $word2 ]]; then
printf "%s\n" "$line"
fi
else
if [[ $line =~ $word2 ]]; then
:
else
printf "%s\n" "$line"
fi
fi
done

In my opinion, the simplest way (even though possibly not the fastest) is to find separately the lines that contain neither word and the lines that contain both words, and to concatenate the results. For example (assuming file.txt is a text file in directory test, and I pass the input values as environment variables for generality - and we are only looking for full words, not word fragments):
[mathguy#localhost test]$ more file.txt
aaaa bbbb cccc
bbbb bbbb bbbb
cccc cccc cccc
dddd dddd aaaa
[mathguy#localhost test]$ word1=aaaa
[mathguy#localhost test]$ word2=bbbb
[mathguy#localhost test]$ ( grep "\b$word1\b" file.txt | grep "\b$word2\b" ; \
> grep -v "\b$word1\b" file.txt | grep -v "\b$word2\b" ) | cat
aaaa bbbb cccc
cccc cccc cccc

In Linux replace \n with \n\n from text (file)

Let say we have simple text
aaaa
bbbb
cccc
Output:
aaaa
bbbb
cccc
Any ideas

Why not just this awk
awk '{print $0"\n"}' file
aaaa
bbbb
cccc
Or this:
awk 1 ORS="\n\n" file
aaaa
bbbb
cccc
Or this:
awk '$0=$0"\n"' file
aaaa
bbbb
cccc

You could try the below awk command,
$ awk -v ORS="\n\n" '{print}' file
aaaa
bbbb
cccc

If you want to only have spaces between lines and not have an extra blank line in the end:
awk 'NR > 1 { printf ORS } 1' file
Or
awk 'NR > 1 { $0 = ORS $0 } 1' file

This might work for you (GNU sed):
sed G file

The pure bash soltion would be
while read line; do echo $line; echo; done < file;

Awk coping when column is occassionly empty

I have output from a foreach loop in the form:
ABC123603LP 44Bq AAAA
ABC123603P 3BU AAAA
ABC123603ZZP AAAA
ABC123604DP 3BU BBBB
ABC123604LP 44Bq BBBB
ABC123605AP 4q CCCC
ABC123605DP 33BGU CCCC
ABC123606AP 35Bjq DDDD
ABC123606DP 4B DDDD
From this I wish to print columns 1 and 2 to the terminal with
echo ... | awk '{print $1, $2}'
However the third row and others prints ABC123603ZZP AAAA as the second column is blank in this case. How do I get around this?

Check the number of field before you print:
$ awk 'BEGIN {OFS="\t"}{ if (NF==2) print $1; else print $1, $2}' file
ABC123603LP 44Bq
ABC123603P 3BU
ABC123603ZZP
ABC123604DP 3BU
ABC123604LP 44Bq
ABC123605AP 4q
ABC123605DP 33BGU
ABC123606AP 35Bjq
ABC123606DP 4B

You can use sed instead:
echo ... | sed 's/\(A[^ ]*[\t ]*\)\([^ \t]*[0-9][^ \t]*\)*.*/\1 \2/'

One technique to remove the last column is to make awk think there are 1 fewer fields:
awk -v OFS='\t' '{NF--; print}'

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