"echo -n" prints "-n"

Solution 1:

There are multiple versions of the echo command, with different behaviors. Apparently the shell used for your script uses a version that doesn't recognize -n.

The printf command has much more consistent behavior. echo is fine for simple things like echo hello, but I suggest using printf for anything more complicated.

What system are you on, and what shell does your script use?

Solution 2:

bash has a "built-in" command called "echo":

$ type echo
echo is a shell builtin

Additionally, there is an "echo" command that is a proper executable (that is, the shell forks and execs /bin/echo, as opposed to interpreting echo and executing it):

$ ls -l /bin/echo
-rwxr-xr-x 1 root root 22856 Jul 21  2011 /bin/echo

The behavior of either echo's with respect to \c and -n varies. Your best bet is to use printf, which is available on four different *NIX flavors that I looked at:

$ printf "a line without trailing linefeed"
$ printf "a line with trailing linefeed\n"

Solution 3:

Try with

echo -e "Some string...\c"

It works for me as expected (as I understood from your question).

Note that I got this information from the man page. The man page also notes the shell may have its own version of echo, and I am not sure if bash has its own version.

Solution 4:

To achieve this there are basically two methods which I frequently use:

1. Using the cursor escape character (\c) with echo -e

Example :

for i in {0..10..2}; do
  echo -e "$i \c"              
done
# 0 2 4 6 8 10

-e flag enables the Escape characters in the string.
\c brings the Cursor back to the current line.

2. Using the printf command

Example:

for ((i = 0; i < 5; ++i)); do
  printf "$i "
done
# 0 1 2 3 4

Solution 5:

If you use echo inside an if with other commands, like "read", it might ignore the setting and it will jump to a new line anyway.