How to Print output of shell script in different columns of csv file

Question

I have written a shell script and want to print output of 5 defined variables in csv file, I am using a condition, If condition 1 success then output should print in first 5 columns, else output should in next 5 columns, like below:

if condition 1 success

$DAY,$ModName,$Version,$END_TIME,$START_TIME

(should print in column number 1..5 of csv)

if condition 2 success

$DAY,$ModName,$Version,$END_TIME,$START_TIME

(should print in column number 6..10 of csv)

But using my code output always appends to next row

Below is my code:

if [ "$Version" = linux ] 
then 
echo "$DAY","$ModName","$Version","$END_TIME","$START_TIME" | awk -F "\"*,\"*" '{print $a ","}' >> output.csv;
else 
echo "$DAY","$ModName","$Version","$END_TIME","$START_TIME" | awk -F "\"*,\"*" '{print $b}' >> output.csv; 
fi

I tried n number of things apart from this code, but not able to find the solution.

I would appreciate your help :)

For instance: You are using awk variables a and b, but you never assign a value to them. — user1934428
– user1934428, Commented Jul 9, 2018 at 15:00
@user1934428, I tried doing awk -F "\",\"" '{print $1, $2, $3, $4, $5} , It works for 1st 5 values which satisfies with 1st condition, but applying awk -F "\",\"" '{print $6, $7, $8, $9, $10} is not printing values in column 6...10 . — Satish Shihani
– Satish Shihani, Commented Jul 9, 2018 at 16:38
@user1934428 , I know, this is where i am stuck with, do you have any suggestion for this, so that i can drive in right direction ? — Satish Shihani
– Satish Shihani, Commented Jul 11, 2018 at 17:15
He doesn't print in separate rows. Please update your posting by your complete program based on Walter's solution, so we can't investigate. — user1934428
– user1934428, Commented Jul 12, 2018 at 13:09

Walter A · Accepted Answer · 2018-07-09 19:37:43Z

1

{print $6, $7, $8, $9, $10} refers to input fields, not output.
When you want to start with 5 empty fields just printf them (avoiding a \n)

if [ "${Version}" != "linux" ]; then 
   printf "%s" ",,,,,"
fi
echo "${DAY},${ModName},${Version},${END_TIME},${START_TIME}"

(Next time please use lowercase variable names)

When a variable can have a ',', you might need to give values in double quotes.

answered Jul 9, 2018 at 19:37

Walter A

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1 Comment

Satish Shihani Over a year ago

your answer didn't help me out.

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