How to prevent grep from printing a trailing newline?

Use tr -d to delete characters in a string:

$ grep -c ' ' /etc/passwd | tr -d '\n'69$ grep -c ' ' /etc/passwd | tr -d '\n' | xxd 0000000: 3639                                     69$ 

You can pipe it through tr and translate the \n to a \0 character:

tr '\n' '\0'

I know this is old, and tr works just as well, but I happened across this question and noticed OP stated:I am executing in sh, not bash. So nesting it into echo -n "$(grep -c pattern)" doesn't work either.

This isn't grep or sh so much as how echo is being used. For future visitors, the only reason this didn't work is due to the double quotes around the substituted command. The following does, in fact, work even using sh.

echo -n $(grep -c pattern)

Examples:

$ ls /dev/sd? #example of formatted output/dev/sda  /dev/sdc  /dev/sde  /dev/sdg  /dev/sdi  /dev/sdk/dev/sdb  /dev/sdd  /dev/sdf  /dev/sdh  /dev/sdj$ echo $(ls /dev/sd?) #without -n, appends \n only at the end/dev/sda /dev/sdb /dev/sdc /dev/sdd /dev/sde /dev/sdf /dev/sdg /dev/sdh /dev/sdi /dev/sdj /dev/sdk$ echo -n $(ls /dev/sd?) #with -n, does not append the \n, but still strips the line breaks from the string/dev/sda /dev/sdb /dev/sdc /dev/sdd /dev/sde /dev/sdf /dev/sdg /dev/sdh /dev/sdi /dev/sdj /dev/sdk$ echo -n "$(ls /dev/sd?)" #output when double quotes are used/dev/sda/dev/sdb/dev/sdc/dev/sdd/dev/sde/dev/sdf/dev/sdg/dev/sdh/dev/sdi/dev/sdj/dev/sdk