Padding trailing whitespaces in a string with another character

Question

I'd like to output hello world over 20 characters.

printf "%-20s :\n\n" 'hello world!!'

# Actual output
hello world!!        :

# Wanted output
hello world!!========:

However, I don't want to complete with spaces but with "=" instead. How do I do that?

Answer 1

filler='===================='
string='foo'

printf '%s\n' "$string${filler:${#string}}"

Gives

foo=================

${#string} is the length of the value $string, and ${filler:${#string}} is the substring of $filler from offset ${#string} onwards.

The total width of the output will be that of the maximum width of $filler or $string.

The filler string can, on systems that has jot, be created dynamically using

filler=$( jot -s '' -c 16 '=' '=' )

(for 16 = in a line). GNU systems may use seq:

filler=$( seq -s '=' 1 16 | tr -dc '=' )

Other systems may use Perl or some other faster way of creating the string dynamically.

Answer 2

printf "%.20s:\n\n" "$str========================="

where %.20s is the string truncating format

Answer 3

One way to do it:

printf "====================:\r%s\n\n" 'hello world!!'

Answer 4

Updated answer to be more general solution. see also my another answer below using only shell brace expansion and printf.

$ str='Hello World!'
$ sed -r ':loop; s/ (=*):$/\1=:/; t loop' <<< "$(printf '%-20s:\n' "$str" )"
Hello World!========:

How it works?

this (=*):$/ captures one space, one-or-more = that followed by a colon : in the end of its input; we make the set of = as a group match and \1 will be its back-reference.

With :loop we defined a label named loop and with t loop it will jump to that label when a s/ (=*):$/\1=:/ has done successful substitution;

In replacement part with \1=:, it will always increment the number of =s and back the colon itself to the end of string.

Answer 5

A Perl approach:

$ perl -le '$k="hello world!!"; while(length($k)<20){$k.="=";} print "$k\n"'
hello world!!=======

Or, better, @SatoKatsura pointed out in the comments:

perl -le '$k = "hello world!!"; print $k, "=" x (20-length $k), "\n"'

If you need to support UTF multi-byte characters, use:

PERL_UNICODE='AS' perl -le '$k = "hello world!!"; print $k, "=" x (20-length $k), "\n"'

Same idea in the shell:

v='hello world!!'; while [ ${#v} -lt 20 ]; do v="$v""="; done; printf '%s\n\n' "$v"

Answer 6

Another way is using only printf command and generate the character padding pattern first by Shell Brace Expansion (You can put end with a number ≥ formatting area you want to print in {1..end}) and get only every first character of it %.1s which is =s and then print only first 20 characters length area of that %.20s. This is kind of better way to having repeated characters/word instead of duplicating them.

printf '%.20s:\n' "$str$(printf '%.1s' ={1..20})"
Hello World!!=======:

---

Explanations:

Normally as Brace Expansion, shell expanding {1..20} as following if we print those.

printf '%s ' {1..20}
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 

So with adding an equal sign to it ={1..20}, shell will expand as following.

printf '%s ' ={1..20}
=1 =2 =3 =4 =5 =6 =7 =8 =9 =10 =11 =12 =13 =14 =15 =16 =17 =18 =19 =20 

And with printf '%.1s' which is actually means printf '%WIDE.LENGTH', we are printing only one LENGTH of those at above with default 1 WIDE. so will result =s only and 20 times repeated itself.

Now with printf '%.20s:\n' we are printing only the 20 length of $str and if length of $str<20, the rest will take from generated =s to fill with instead of spaces.