I am using following shebang in my script:
#!/bin/sh To check whether I haven't used any bash syntax, I checked my script with checkbashisms script and it says
possible bashism in my_script.sh line 3 (echo -e): echo -e "hello world" And I am running this script on embedded board with busybox shell.
How to resolve this?
POSIX allows echo -e as the default behavior. You may do
var="string with escape\nsequence" printf "%s\n" "$var" # not interpreting backslashes printf "%b\n" "$var" # interpreting backslashes This should pass the checkbashisms test
help echo -e enable interpretation of the following backslash escapes ...
If your code doesn't need to interpret backslash escapes then you can safely remove the -e flag.
Sometimes, people tend to use -e just for interpreting \n as newline. If that's the case, know that you can have multiline strings just by quoting them and writing as it is. An example:
var="Hi, I am a multiline string" But if you do need to interpret backslash escapes, then using printf will be your best bet.
Though I would recommend using printf whether you need backslash interpretation or not.
With printf:
printf "%s\n" "$var" # no backslash interpretation printf "$var" # backslash interpretation occurs 
echo is a shell builtin. While the man pages might correctly describe it on a GNU system, they are probably for /bin/echo which (in theory) could be different. On the version of UNIX I am running on, the shell built-in supports -e but /bin/echo does not, and the man pages do not have the line you show, however help echo does. There can be similar issues with printf, being both a shell built-in and a program in (on my system) /usr/bin.
-eflag or not? We cannot know that without seeing some code.-eoption is forechoto parse escape code (used for example for VT100 escape codes or embedded newlines etc.). If your string doesn't have any escape code, then you don't need it.-eoption is not in the POSIX standard forecho. It's less of a "Bashism" and more of a GNUism" I would say, especially since in Linux it's not only a Bash built-in command, but also part of GNU coreutils.command echo -e .... Thecommandcommand, which is part of the POSIX standard, causes theechofrom your PATH to be executed, instead of the builtinecho.