I’d like to output hello world over 20 characters.
printf "%-20s :nn" '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?
Answers:
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Method 1
filler='===================='
string='foo'
printf '%sn' "$string${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=$( seq -s '=' 1 16 | tr -dc '=' )
Other systems may use Perl or some other faster way of creating the string dynamically.
Method 2
printf "%.20s:nn" "$str========================="
where %.20s is the string truncating format
Method 3
One way to do it:
printf "====================:r%snn" 'hello world!!'
Method 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.
Method 5
A Perl approach:
$ perl -le '$k="hello world!!"; while(length($k)<20){$k.="=";} print "$kn"'
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 '%snn' "$v"
Method 6
Another way is using only printf command and generate the character padding pattern first by Shell Brace Expansion (You can put end with a number ≥ formatting area you want to print in {1..end}) and get only every first character of it %.1s which is =s and then print only first 20 characters length area of that %.20s. This is kind of better way to having repeated characters/word instead of duplicating them.
printf '%.20s:n' "$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.
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0