echo a one-liner bash script with variables

Question

I would like to do that:

i="1"; echo -e '#!/usr/bin/env bash\nmyprogram -i "input_${i}.txt"'

and pipe it to a job scheduler.

However, this doesn't replace the variable i by its value. Instead, I obtain this:

#!/usr/bin/env bash
myprogram -i "input_${i}.txt"

I played a bit with option -e of echo and with single-/double-quote but could not make it work. For instance, I get this:

i="1"; echo -e "#!/usr/bin/env bash\nmyprogram -i \"input_${i}.txt\""
-bash: !/usr/bin/env: event not found

My bash version is 4.1.2.

Solution

Try this:

i="1"; echo -e '#!/usr/bin/env bash\nmyprogram -i '"\"input_${i}.txt\""

You can echo single- and double-quoted strings at the same time.