How to avoid printing a newline when seq completes?

Assuming you just want to print 1234567890123456789, you can do it with:

$ printf "%s" $(seq 1 9) $(seq 0 9) 1234567890123456789$ 

That won't have a trailing newline at all though, so maybe you prefer:

$ printf "%s" $(seq 1 9) $(seq 0 9) $'\n' 1234567890123456789 $ 

A few simpler choices if you don't need to use seq:

$ perl -le 'print 1..9,0,1..9' 1234567890123456789 $ printf "%s" {1..9} {0..9} $'\n' 1234567890123456789 

Since you mentioned portability, I recommend you use the perl approach, or if you are likely to encounter systems without perl, and yet need the same command to run in shells including bash, sh, dash, tcsh etc, try Kamil's approach.

You can remove newlines from any stream with tr -d '\n'. In your case

(seq -s '' 1 9 ; echo -n 0; seq -s '' 1 9) | tr -d '\n' 

While other answers may concentrate on modifying your original approach, this is the way that should just remove newline characters regardless of what is before |.

I assume you don't want just that particular string, but strings of incrementing digits with variable lengths. For that, it may be useful to be able to change the width by changing one number, instead of having to build it from multiple calls to seq.

In Bash, you could use something like this (for a 19-long sequence):

for ((i=1; i <= 19; i++)); do printf "%d" "$(( i % 10 ))"; done; echo 

This should work in a standard shell (and works in at least dash and busybox, anyway):

i=1; while [ "$i" -le 19 ]; do printf "%d" "$(( i % 10 ))"; i=$((i+1)); done; echo 

Consider also:

printf '%d' $(seq -w 1 1 99 | cut -c2) 

We generate the numbers 01..99, zero-padded (-w), then strip off the tens-place. Those ones-place numbers are then sent to printf to be printed individually.

To get your desired output, use 19 instead:

$ printf '%d' $(seq -w 1 1 19 | cut -c2) 1234567890123456789 

When using tr to get rid of trailing new-line it's worth realising that it deletes all occurrences of the characters given in its -d option. Thus you can get it a bit more slim:

(seq 1 9; echo 0; seq 1 9) | tr -d '\n' 

— even echo doesn't need to have -n anymore.

And for completeness sake: BSD's version of seq when -s '' specified doesn't produce new-line on its exit.

To remove the trailing newline, just use a command execution:

 $ echo "test$(seq -s '' 9)no-newline" test12345679no-newline 

Or capture it in a variable:

 $ var=$(seq -s '' 9); echo "${var}0${var}" 1234567890123456789 

Or also:

 $ printf '%s' {{0..9},0,{0..9}}; echo 1234567890123456789 

I wonder why you don't just:

echo 1234567890123456789 

And what's the point of printing just 19 columns instead of 20?

If you want it repeated (by default: for the whole width of the terminal)

chdr(){ c=${1:-$COLUMNS}; while [ "$c" -gt 0 ]; do printf '%.*s' "$c" 1234567890; c=$((c-10)); done; echo; } chdr 17 12345678901234567 chdr 12345678901234567890123456789012345678901234567890123456789012345678901234567890 

Of course, that won't work in csh (huh).

FYI: seq isn't portable (there's jot on *BSD, but it's quite different).