Show a string of certain length without truncating

Question

I am displaying a string of certain length in ruby. only 80 characters of that string can be displayed in one line. for example if string length is 82 then it will be shown

Answer 1

def wrap(str, max_line_len)
  str.scan /(?<=\A| ).{1,#{max_line_len}}(?= |\z)/
end

str = "Little Miss Muffet she sat on her tuffet, eating her curds and whey. Along " + 
      "came a spider who sat down beside her and frightened Miss Muffet away."

         1         2         3    
123456789012345678901234567890123

puts wrap(str, 31)
Little Miss Muffet she sat on
her tuffet, eating her curds
and whey. Along came a spider
who sat down beside her and
frightened Miss Muffet away.

puts wrap(str, 32)
Little Miss Muffet she sat on
her tuffet, eating her curds and
whey. Along came a spider who
sat down beside her and
frightened Miss Muffet away.

puts wrap(str, 33)
Little Miss Muffet she sat on her
tuffet, eating her curds and
whey. Along came a spider who sat
down beside her and frightened
Miss Muffet away.

See String#scan. The regular expression reads, "match between 1 and max_line_len characters, immediately preceded by the beginning of the string or a space and immediately followed by a space or the end of the string". (?<=\A| ) is a positive lookbehind and (?= |\z) is a positive lookahead.

Answer 2

A very quick and dirty, iterative version:

max_chars = 80
text = 'Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam erat, sed diam voluptua. At vero eos et accusam et justo duo dolores et ea rebum. Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit amet. Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam erat, sed diam voluptua. At vero eos et accusam et justo duo dolores et ea rebum. Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit amet.'

lines = []
words = text.split

while words.length > 0
  line = words.shift

  while words.first && (line.length + words.first.length + 1) <= max_chars
    line << " #{words.shift}" 
  end

  lines << line
end

lines.each { |line|  puts line }
#=> Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam nonumy eirmod
#   tempor invidunt ut labore et dolore magna aliquyam erat, sed diam voluptua. At
#   vero eos et accusam et justo duo dolores et ea rebum. Stet clita kasd gubergren,
#   no sea takimata sanctus est Lorem ipsum dolor sit amet. Lorem ipsum dolor sit
#   amet, consetetur sadipscing elitr, sed diam nonumy eirmod tempor invidunt ut
#   labore et dolore magna aliquyam erat, sed diam voluptua. At vero eos et accusam
#   et justo duo dolores et ea rebum. Stet clita kasd gubergren, no sea takimata
#   sanctus est Lorem ipsum dolor sit amet.

Answer 3

Just out of curiosity:

loop.inject([[], input]) do |(acc, src)|
  if m = src[/.{1,79}(\s|\z)/]
    [acc << m, $']
  else
    break acc << src
  end
end

Answer 4

Other option, not so clean, but...

The idea is to find the index of the spaces and substitute with \n the spaces at index close to the line length.

So, given the string str and the max_len:

delta = 0
(str + " ")
.each_char.with_index.with_object([]) { |(c, i), o| o << i if c == " "} # find the index of the spaces
.each_cons(2).with_object([]) do |(a, b), tmp| # select the index to be substituted
  if b > (tmp.size + 1) * max_len + delta
    tmp << a 
    delta = tmp.last - max_len * tmp.size + 1
  end
end.each { |i| str[i] = "\n" } # substitute

Now str has \n when line length is close to max_len. This alterates the original string.