How to print the values of variables with incremented numbers using a loop in shell script?

Question

I have a Bash script where i have calculated many values and stored them in the variables which have a number for each row. For instance, i have a variable named as TC_5 which calculates the value of 5th row of the input csv file. I have calculated all the values and stored in variables which have the naming convention of TC_<Row_No> so that for 200 rows i have values stored in:

TC_1
TC_2
.
.
TC_200

Now, i want to write a loop which could print the values of all these variables together to an external file instead of manually printing them. For this, i am using the while loop as follows:

i=0
while [ "$i" != 201 ]
do
    echo "TC_$i" >> Out
    i=`expr $i + 1`
done

How can i modify the above code in such a way that the echo statement would print the variable TC_<RowNo> to the Out File?

Answer 1

Your current script stuck in an infinitive loop, because the condition [ "$i" != 201 ] was always true.

You must increase $i after each iteration and using eval to print the content of TC_<RowNo> variable (but it's not safe):

i=1
while [ "$i" -ne 201 ]
do
    eval printf '%s\\n' "\${TC_$i}"
    i=$((i+1))
done >> "Out"

Note that $i started at 1, the use of -ne for integer comparison and the redirection at the end of while loop.