LinuxQuestions.org - [SOLVED] replace \n with space in csv file

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replace \n with space in csv file

Hi all, I have csv file with the some of the rows with "\n" values, and I wanted to replace "\n" with just empty space " ". How can i achieve it. I tried the below but doesnot work. can anyone help?

Code:

sed 's/\/n/ /g' "$input_file" > "$output_file"

my csv file looks like this

Code:

Name,Age,Location

Alice\nbhat,25,New York

Bob\n,30,Los Angeles

Charlie\nTC,22,Chicago

/n and \n are two different things. Which one do you want to replace?
You can use a different separator:

Code:

sed 's!/n! !g' infile > outfile

# or 

sed 's!\\n! !g' infile > outfile

nm - too slow typing ... :doh:

Quote:

Originally Posted by pan64 (Post 6452958)

/n and \n are two different things. Which one do you want to replace?
You can use a different separator:

Code:

sed 's!/n! !g' infile > outfile

# or 

sed 's!\\n! !g' infile > outfile

\n sorry

does it work for you? In that case you might want to mark the thread as solved.
And also if you want to say thanks just click on yes

Sample scrip using bash.exe at windows

Code:

#!/bin/bash

input_file="input_file.csv"

if [[ ! -f $input_file ]]

then

        echo Last line not ending with new line

        echo -ne "Name,Age,Location

Alice\nbhat,25,New York

Bob\n,30,Los Angeles

Charlie\nTC,22,Chicago" > $input_file

fi

LAST_MODIFIED_DDD_DD_MMM_YYYY_HH_MM_SS_PM_LOC=$(/usr/bin/date --date="$(/usr/bin/stat -c "%y" $input_file)" "+%a_%d_%b_%Y_%I_%M_%S_%p_%Z")

if [[ ! -f "$input_file".$LAST_MODIFIED_DDD_DD_MMM_YYYY_HH_MM_SS_PM_LOC ]]

then

        echo "cp -i $input_file "$input_file".$LAST_MODIFIED_DDD_DD_MMM_YYYY_HH_MM_SS_PM_LOC"

        /usr/bin/cp -i $input_file "$input_file".$LAST_MODIFIED_DDD_DD_MMM_YYYY_HH_MM_SS_PM_LOC

fi

output_file=output_file.csv

if [[ ! -f $output_file ]]

then

        #:n => n is the label name

        #we can use a-z for label names

        #N        take current line append this line to next line using then PATTERN(regex) N

        #$!bn        => $ representing last line. bn => go back to label again. $! excluding the last line go back to label n

        #Replace every new line with space on PATTERN

        /usr/bin/sed ':n;N;$!bn;s/\n/ /g' $input_file > $output_file

        LINE_COUNT=$(/usr/bin/wc -l $output_file | /usr/bin/awk '{ print $1}')

else

        LINE_COUNT=$(/usr/bin/wc -l $output_file | /usr/bin/awk '{ print $1}')

        if [[ 0 -ne $LINE_COUNT ]]

        then

                backup_file=backup_file.csv

                if [[ ! -f $backup_file ]]

                then

                        /usr/bin/sed ':n;N;$!bn;s/\n/ /g' $output_file > $backup_file

                        output_file=$backup_file

                else

                        echo "update this script to rename backup file name $backup_file inside this script to other non-existing file name"

                        read

                fi

        fi

fi

LINE_COUNT=$(/usr/bin/wc -l $output_file | /usr/bin/awk '{ print $1}')

echo "Number of lines at $output_file: $LINE_COUNT"

echo "====== cat $input_file last line not ending with new line ======"

/usr/bin/cat $input_file

#Since previous cat do not end with new line I am using echo

echo

echo "====== cat $output_file ======"

/usr/bin/cat $output_file