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Last Edited at 07/28/07 11:38am

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Format strings like PRINT USING

/ Published in: Ruby

This is an extension to Ruby's String class. Just put this code in a file and include it. This basically takes a string and adds literals and padding to it. For example, you can format a phone number with optional area code like this:

"5445556747".using('(###) ###-####', '', true)
=> (544) 555-6747

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require 'strscan'

class String

# Returns the string formatted according to a pattern.
  #
  # The pattern consists of placeholders and literals. The string is placed in
  # the placeholders, leaving the literals as they are. The result may be
  # truncated or padded if there are more placeholders than strings.
  #
  # Placeholders are '#' or '&'. Each '#' is replaced by one character from
  # the string, or the filler character if the string has no characters left.
  # The '&' is replaced by any remaining characters, or left out of the result
  # if there are no remaining characters. There can only be one '&' in the
  # pattern. If there is no '&' and more characters than placeholders, the
  # remaining characters are discarded.
  # 
  # '#' or '&' may be replaced by other characters if they are needed as
  # literals.
  #
  # Examples:
  # "123456789".using('###-##-####')
  #    => "123-45-6789"
  # "12345".using('###-##-####')
  #    => "123-45"
  # "12345".using('###-##-####', nil)
  #    => "12345"
  # "12345".using('###-##-####', ' ')
  #    => "123-45-    "
  # "873555121276668".using ('(###) ###-#### ext &', '', true)
  #    => "(873) 555-1212 ext 76668"
  # "8735551212".using ('(###) ###-#### ext &', '', true)
  #    => "(873) 555-1212"
  # "5551212".using ('(###) ###-#### ext &', '', true)
  #    => "555-1212"
  # "KB5774X".using ('##-&-#')
  #    => "KB-5774-X"
  #
  # Parameters:
  # pattern -- The format string, see above.
  # fill    -- A string for padding. If the empty string, then the pattern is
  #            filled as much as possible, and the rest of the pattern is
  #            truncated. If nil, and the pattern cannot be filled exactly,
  #            the string is returned unchanged. If some other string, the
  #            pattern is filled as much as possible and the remainder is 
  #            padded with the string. Defaults to the empty string.
  # right   -- If true, the pattern is filled from right-to-left instead of
  #            from left-to-right, and truncated on the left instead of the
  #            right if needed. Default is false.
  # fixchar -- The single-character placeholder. Default is '#'.
  # remchar -- The remaining-character placeholder. Default is '&'.
  #
  def using(pattern, fill='', right=false, fixchar='#', remchar='&')

remCount = pattern.count(remchar)
    raise ArgumentError.new("Too many #{remchar}") if remCount > 1
    raise ArgumentError.new("#{fixchar} too long") if fixchar.length > 1
    raise ArgumentError.new("#{remchar} too long") if remchar.length > 1
    raise ArgumentError.new("#{fill} too long")    if fill.length > 1
    remaining = remCount != 0
    slots = pattern.count(fixchar)

# Return the string if it doesn't fit and we shouldn't even try,
    if fill.nil?
      return self if self.length < slots
      return self if self.length > slots and !remaining
    end

# Pad and clone the string if necessary.
    source =  if fill.nil? || fill.empty? then
                self
              elsif right then
                self.rjust(slots, fill)
              else
                self.ljust(slots, fill)
              end
    
    # Truncate the string if necessary.
    if source.length > slots && !remaining then
      source = right ? source[-source.length, source.length] :
                       source[0, source.length]
    end

# Truncate pattern if needed.
    if !fill.nil? && fill.empty? then
      
      if source.length < slots  # implies '&' can be ignored
        keepCount = source.length # Number of placeholders we are keeping
        leftmost, rightmost = 0, pattern.length - 1
        if right then
          # Look right-to-left until we find the last '#' to keep.
          # Loop starts at 1 because 0th placeholder is in the inject param.
          leftmost = (1...keepCount).inject(pattern.rindex(fixchar)) {
            |leftmost, n| pattern.rindex(fixchar, leftmost - 1) }
        else
          # Look left-to-right until we find the last '#' to keep.
          rightmost = (1...keepCount).inject(pattern.index(fixchar)) {
            |rightmost, n| pattern.index(fixchar, rightmost + 1) }
        end
        pattern = pattern[leftmost..rightmost]
        slots = pattern.count(fixchar)
      end
    
      # Trim empty '&' up to nearest placeholder. If a '&' at the end goes
      # empty, the literals between it and the nearest '#' are probably also
      # unnecessary.
      if source.length == slots then
        if pattern.match("^#{Regexp.escape(remchar)}") then
          pattern = pattern[pattern.index(fixchar) || 0 ... pattern.length]
        elsif pattern.match("#{Regexp.escape(remchar)}$") then
          pattern = pattern[0 ... (pattern.rindex(fixchar) + fixchar.length) || pattern.length]
        end
      end

end
      
    # Figure out how long the remainder will be when we get to it.
    remSize = source.length - slots
    if remSize < 0 then remSize = 0; end
    
    # Make the result.
    scanner = ::StringScanner.new(pattern)
    sourceIndex = 0
    result = ''
    fixRegexp = Regexp.new(Regexp.escape(fixchar))
    remRegexp = Regexp.new(Regexp.escape(remchar))
    while not scanner.eos?
      if scanner.scan(fixRegexp) then
        result += source[sourceIndex].chr
        sourceIndex += 1
      elsif scanner.scan(remRegexp) then
        result += source[sourceIndex, remSize]
        sourceIndex += remSize
      else
        result += scanner.getch
      end
    end
    
    result
  end
end

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