Add spaces to variable in shell script

Question

I want create a yaml configuration file from a sh script. So I need to add multiple spaces to a variable like that:

SERVERS=""
for i in $(seq 1 5); do
    SERVERS="$SERVERS      -server`printf "%03d" $i`.$URL\n"
done

But the spaces are reduced to only one space. How do I ensure that the spaces aren't replaced?

Note that you aren't adding newlines to your string; you are adding two literal characters "\" and "n". — chepner
– chepner, Commented Feb 19, 2013 at 14:25
I realized this already. When I print the variable by echo "$SERVERS", the \n is interpreted as a newline, but when I store it in a file, it just dumps "\n". How do I make a real line break? — Teots
– Teots, Commented Feb 20, 2013 at 8:40

Gilles Quénot · Accepted Answer · 2013-02-19 17:59:07Z

2

I suspect that you forgot to use double quotes when you echo $SERVER =)

SERVERS=""
for i in $(seq 1 5); do
    SERVERS="$SERVERS      -server$(printf "%03d" $i).$URL\n"
done
echo "$SERVER"

Output :

      -server001.\n      -server002.\n      -server003.\n      -server004.\n(...)

That's the rules of the (not so) terrific word splitting

Another solution using perl :

perl -le 'for (1..5) { print "   -server$_"}' | tee -a file.yaml

Or using awk :

awk 'BEGIN{for (i=1; i<=5; i++) print "   -server"++c}'  |
    tee -a file.yaml

answered Feb 19, 2013 at 14:20

Gilles Quénot

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1 Comment

Thanks, I really just forgot to add double quotes.

chepner · Accepted Answer · 2013-02-19 14:28:12Z

0

What do you ultimately do with SERVERS? Rather than try to build up a multi-line string, something like this might be better:

for i in $(seq 1 5); do
    echo "     -server$(printf "%03d" $i).$URL"
done >> file

answered Feb 19, 2013 at 14:28

chepner

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