echo printing -e in bash

Question

I\'ve gotten this script I\'ve created in Bash, and one of the functions I\'m using is echo and I\'m using the -e flag for interpretations of

Answer 1

You wrote, "I've gotten this script I've created in Bash", but you haven't told us exactly what you mean by that.

UPDATE : The question has been updated. The script's first line is #!/bin/sh. Read on for an explanation and solution.

I can reproduce the problem on my system by including your code in a script starting with

#!/bin/sh

I can correct the problem by changing that first line to

#!/bin/bash

/bin/sh on my system happens to be a symlink to dash. The dash shell has echo as a builtin command, but it doesn't support the -e option. The #! line, g othe

There are numerous implementations of the echo command: most shells provide it as a builtin command (with various features depending on the shell), and there's likely to be an external command /bin/echo with its own subtly different behavior.

If you want consistent behavior for printing anything other than a simple line of text, I suggest using the printf command. See https://unix.stackexchange.com/questions/65803/why-is-printf-better-than-echo (cited by Josh Lee in a comment).

The #! line, known as a shebang, controls what shell is used to execute your script. The interactive shell you execute the script from is irrelevant. (It pretty much has to be that way; otherwise scripts would behave differently for different users). In the absence of a #! line, a script will (probably) be executed with /bin/sh, but there's not much point in not being explicit.