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?

• MatthewRock: Why not use awk/sed?
• Costas: Do you hear about arrays?
• Ankit Vashistha: Hello Guys, could you please provide your suggestions as answers? I don't want to do it strictly with while loop. Any other option which is faster and better would be appreciated. Of course i have heard about arrays :)

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.

• Ankit Vashistha: Good point cuonglm. Actually i missed to put the increment statement in the above example. That is present in my main script. Thanks for your answer. I will try it and get back to you.

Answer 2

The preferred way to do this sort of thing in bash is to use an array:

TC[1]=something
TC[2]=somethingelse
...
TC[200]=somethingstillelse
i=1
while ((i <= 200)); do
 echo "${TC[i]}" >> Out
 ((i++))
done

But if you really want to embed the indexes in plain variable names, you can use an indirect variable reference:

TC_1=something
TC_2=somethingelse
...
TC_200=somethingstillelse
i=1
while ((i <= 200)); do
 varname=TC_$i # Need to store the variable name in a variable first
 echo "${!varname}" >> Out # The ! (and braces) trigger indirect expansion
 ((i++))
done

Note that neither of these features are available in basic POSIX shells like dash. Make sure your script is running under bash, not a generic shell.

Answer 3

I changed your while loop slightly and it worked for me:

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

Note the $ in front of TC_$i and the removal of the double quotes. (I included the modification cuonglm suggested for the starting value.)

(Using GNU bash in iTerm on a Mac running High Sierra.)

• G-Man Says 'Reinstate Monica': Really? That doesn't work for me. (Are you sure you're using bash?)
• J. Van Heuit: ~/CertificateTest/save 499> /bin/sh --version GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin17) Copyright (C) 2007 Free Software Foundation, Inc.
• Kusalananda: No, $TC_$i would expand to the concatenation of the values of the variables TC_ and i.
• J. Van Heuit: And yet, it really did work for me on GNU bash...