"echo -n" prints "-n"

Question

I have a problem with echo in my script:

echo -n "Some string..."

prints

-n Some string...

and moves to the next line. In the console it's working correcly without newline:

Some string...

Answer 1

There are multiple versions of the echo command, with different behaviors. It's often built into the shell. If it isn't, echo will typically invoke /bin/echo or /usr/bin/echo, which itself can be from any of a number of implementations.

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?

Answer 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"

Answer 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.

Answer 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.

OR

2. Using the printf command

Example:

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

Answer 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.

Answer 6

Just for the most popular Linux distribution, Ubuntu and its Bash:

1. Check which shell are you using. Mostly the below works, else see this:

   echo $0

2. If above prints bash, then the below will work:

   printf "hello with no new line printed at end"

   Or

   echo -n "hello with no new line printed at end"

Answer 7

enable -n echo
echo -n "Some string..."

Answer 8

I believe right now your output prints as below

~ echo -e "String1\nString2"
String1
String2

You can use xargs to get multiline standard output into same line.

 ~ echo -e "String1\nString2" | xargs
String1 String2

 ~

Answer 9

Note that /usr/bin/echo and /bin/echo on AIX don't support any arguments, so neither -n nor -e work if using sh or KornShell (ksh) shells.

C shell and Bash have their own built-in echo which supports -n.
This is relevant, because a lot of shell scripts explicitly use sh or KornShell.

AIX does have /usr/bin/printf, so as suggested in some earlier answers,

$ printf "whatever"

is equivalent to echo -n "whatever" where -n is supported.

Answer 10

When you go and write your shell script, always use #!/usr/bin/env bash as the first line.

This shell doesn't omit or manipulate escape sequences.

Example:

echo "This is first \\n line"

prints

This is first \n line.

Answer 11

I had the same issue in IBM z/OS, so I used print instead of echo, and it worked.

print -n "Some string ...."

print — Return arguments from the shell