how to get quotes in output in unix?

Question

abc.dat file contains lines

1
2
3

code is

tradeline=""
for line in $(cat abc.dat)
do
tradeline="$line,$tradeline"
done

sample output am getting is 1,2,3, but i want the output as "1","2","3",

can u help me plz

Answer 1

Escape the quotes with \:

tradeline="\"$line\",$tradeline"

Answer 2

Your code has a number of problems, the first being what you describe.

It will also put the numbers in reverse order and have a trailing comma, which is probably not what you want.

If that is what you want, just use the second solution.

Solution 1, right order and no trailing comma:

tradeline=""
sep=""
for line in $(cat abc.dat) ; do
    tradeline="${tradeline}${sep}\"${line}\""
    sep=","
done
echo ${tradeline}

Solution 2, right order with trailing comma:

tradeline=""
for line in $(cat abc.dat) ; do
    tradeline="${tradeline}\"${line}\","
done
echo ${tradeline}

Solution 3, much simpler, using tools meant for the job:

awk 'BEGIN {sep=""} {printf "%s\"%s\"",sep,$0;sep=","} END {print}' abc.dat

Solution 3 would be my preferred option since it'll be faster and be able to handle input lines with white space in them.

Answer 3

Note that using cat like that will mean that if you have several words on one line, you will still quote each word separately. Also, your code will generate '3,2,1' rather than '1,2,3'.

To get quotes around each line, you would need to use the read command:

tradeline=""
sep=""
cat abc.dat |
while read line
do
    tradeline="$tradeline$sep\"$line\""
    sep=","
done

This is quite similar to Pax's solution - except for the use of read to obtain each line.