Bash script printing a variable value after a newline character

Question

I have bash script which calls another script(some_script). some_script expects some input from the user. I have used printf statement for this purpose. But the problem is the variable value is not being accepted by the target script. I think this is because '\\' is being taken as an escape character in the script

The statement somewhat looks like this

 printf 'yes\n$var1\n$var2\n$var3' | some_script

If i directly replace the variable with values it runs perfectly but i want the script to take the values from the variables. How do i achieve this?

Answer 1

There is a difference between " and ' . Try

printf "yes\n$var1\n$var2\n$var3" | some_script

because with ' the variables won't get substituted.

Answer 2

Yes, \\ is a character that has to be escaped.

Use \\\\n .

For more details we would need more details on how your script works.