1

I have data in a file in this format as given below

"253539","","Company1 Name  
","0","1"
"255229","","Ramu ","0","1"
"253548","","Mater Name/Id 
","0","1"
"255229","","Ram Lakh","0","1"
"253619","","CHild Name/Ids 
","0","1"

Every line has \n appended to it.

How can I replace \n", to ", in shell script to get the output as 

"253539","","Company1 Name/Id  ","0","1"
"255229","","Ramu ","0","1"
"253548","","Mater Name/Id ","0","1"
"255229","","Ram Lakh","0","1"
"253619","","CHild Name/Id ","0","1"

Need to automate a process so have to write in shell script.

Please help.

that other guy
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Manish Verma
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4 Answers4

1

Not sure about doing this easily in shell, but I made a throw-away Perl script to do it:

perl -e 'foreach (<>) { chop; print; print "\n" if /"$/ }' < mydata.csv

It works by outputting a newline only if the line ends in ".

paddy
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0

Could do something like

#!/bin/bash

while read -r line; do
 [[ $line =~ '"'$ ]] && echo "$line" || echo -n "$line"
done < file

e.g. on your input file

> ./abovescript
"253539","","Company1 Name","0","1"
"255229","","Ramu ","0","1"
"253548","","Mater Name/Id","0","1"
"255229","","Ram Lakh","0","1"
"253619","","CHild Name/Ids","0","1"
Reinstate Monica Please
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0

With awk you can do:

$ awk '/"$/ {print a$0; a=""; next} {a=$0}' file
"253539","","Company1 Name  ","0","1"
"255229","","Ramu ","0","1"
"253548","","Mater Name/Id ","0","1"
"255229","","Ram Lakh","0","1"
"253619","","CHild Name/Ids ","0","1"

What it does is to store in a the line in case it does not end with ". If it does, it prints the stored line plus the current one.

fedorqui
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0

if @fedorqui's answer can be accepted, then here is the shorter code.

awk '{printf $0 (/"$/?RS:X)}' file
BMW
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