Linux/Unix bash basic script awk/sed

Question

I'm working on bash script.

var=$(ls -t1 | head -n1);
cat $var | sed 's/"//g' > latest.csv
cat latest.csv | sed -e 's/^\|$/"/g' -e 's/,/","/g' > from_epos.csv
echo "LATEST: $var";

Here's the whole script, it's meant to delete all quotation mark from current file and add new one, between each field.

INPUT:

"sku","item","price","qty"
5135,"ITEM1",1.79,5
5338,"ITEM2",1.39,5
5318,"ITEM3",1.09,5
5235,"ITEM4",1.09,5
9706,"ITEM5",1.99,5

OUTPUT:

"sku","item","price","qty"
"5135","ITEM1","1.79","5
"
"5338","ITEM2","1.39","5
"
"5318","ITEM3","1.09","5
"
"5235","ITEM4","1.09","5
"
"9706","ITEM5","1.09","5
"

My ideal output is:

"sku","item","price","qty"
"5135","ITEM1","1.79","5"
"5338","ITEM2","1.39","5"
"5318","ITEM3","1.09","5"
"5235","ITEM4","1.09","5"
"9706","ITEM5","1.99","5"

It seems like it's entering random character between line in current output like " and quotation mark is between CR and LF.

What's the problem and how to get it to my ideal vision?

Thanks,

Adam

Answer 1

awk 'BEGIN{FS=OFS=","}{gsub(/\"/,"");gsub(/[^,]+/,"\"&\"")}1' input

Answer 2

Solution using sed:

sed -e 's/"//g; s/,/","/g; s/^/"/; s/$/"/'

Long-piped-commented version:

sed -e 's/"//g' | # removes all quotations
sed -e 's/,/","/g' | # changes all colons to ","
sed -e 's/^/"/; s/$/"/' # puts quotations in the start and end of each line