Sorting number in command line

Question

What I want is if I enter any number with space delimited, it will sort it in ascending order. I have this in my bash created file.

re='^[0-9]+$'
if ! [[ $1 =~ $re ]] ; then
   echo "error: Not a number" >&2; exit 1
else 
   printf "%s\n" $@ | sort -n 
fi

Basically what id does above is if the user enter a non numeric it will display an error else will sort the numbers.

So if I enter in command: $: sh sort.sh 12 0 13. This will order it out to

0

12

13

Now my problem is I don't want it in a new line, instead I want it in a space.

Aaron · Accepted Answer · 2018-04-15 18:37:49Z

1

sort is a line-based tool, so you can't remove the linefeeds before you call it. You can however replace them with spaces once the data is sorted, for example with tr :

re='^[0-9]+$'
if ! [[ $1 =~ $re ]] ; then
   echo "error: Not a number" >&2; exit 1
else 
   printf "%s\n" $@ | sort -n | tr '\n' ' '
fi

answered Apr 15, 2018 at 18:37

Aaron

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9 Comments

Wondering Coder Over a year ago

hi, when I tried that it showing an extra character "%". -1, 0 12 13 %.

Aaron Over a year ago

Weird, I tried it and it worked fine for me with 12 0 13

Wondering Coder Over a year ago

i run it like this: sh sort.sh 12 0 -1, 13. output was -1, 0 12 13 %

Aaron Over a year ago

With -1 I get "error: Not a number" due to your guard condition. The comma seems weird too and should make your regex reject it

Wondering Coder Over a year ago

im using mac os btw. Also I noticed if I put the comma beside the -1 it works. But on the other number it get the error not a number.

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