Programming Praxis

12 Responses to “Form Letters”

1. Programming Praxis – Form Letters « Bonsai Code said

   November 30, 2010 at 10:25 AM

   […] Praxis – Form Letters By Remco Niemeijer In today’s Programming Praxis exercise, we have to write a program to generate form letters. Let’s get […]
2. Remco Niemeijer said

   November 30, 2010 at 10:25 AM

   My Haskell solution (see http://bonsaicode.wordpress.com/2010/11/30/programming-praxis-form-letters/ for a version with comments):

   import Control.Applicative ((*>), (<*>), (<$>))
   import Text.CSV
   import Text.Parsec
   
   fillWith :: String -> [String] -> String
   fillWith text vars = either show concat $ parse form "" text where
       form = many $ escape <|> count 1 anyChar
       escape = char '$' *> (string "$" <|>
           ((vars !!) . read <$> option "0" (many1 digit)))
   
   formLetters :: FilePath -> FilePath -> IO [String]
   formLetters schema vars = either (return . show) . map . fillWith <$>
       readFile schema <*> parseCSVFromFile vars
   
3. Chris Teixeira said

   November 30, 2010 at 5:10 PM

   My (slightly golfed up) Ruby version:

   d = File.read(‘input.csv’).split(“\n”).map {|line| line.split(“,”)}
   s = File.read(‘letter.schema’)
   d.each {|r| puts s.scan(/\$\d+/).inject(s) {|a,m| a = a.gsub(m,r[m[/\d+/].to_i])}}
4. slabounty said

   November 30, 2010 at 5:16 PM

   A ruby version …

   require 'csv'
   
   form = File.read(ARGV[0])
   
   CSV.foreach(ARGV[1]) do |row|
       form_filled = form
       row.each_with_index do |v, i|
           form_filled = form_filled.gsub("$#{i}", v)
       end
       puts "#{form_filled}"
   end
   

   If you’re using ruby 1.8, then require fastercsv.
5. slabounty said

   November 30, 2010 at 7:11 PM

   OK, both mine and Chris’ have the same two issues in that they don’t work with > 10 elements and the $$ doesn’t work. This one should work (it’s uglier, but works as is often the case). It works backwards down the list so that $10 will be subbed out before $1 and it will leave a “$$” alone. At the end it changes the “$$” to a single “$”.

   require 'csv'
   
   form = File.read(ARGV[0])
   
   CSV.foreach(ARGV[1]) do |row|
       form_filled = form
       (row.size-1).downto(0) do |i|
           form_filled = form_filled.gsub(/([^$])\$#{i}/, "#{$1}#{row[i]}")
       end
       form_filled.gsub!(/\$\$/, "$")
       puts "#{form_filled}"
   end
   
   
6. Ian Tyndall said

   December 1, 2010 at 2:53 AM

   <?php
   function formLetters($schema_file,$csv_inputfile) {
       $s = file_get_contents($schema_file);
       $dcount = 0;
       if (($handle = fopen($csv_inputfile, "r")) !== FALSE) {
           while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
               if ($dcount == 0) $dcount = count($data);
               echo preg_replace(array('/\$(\d+)/e','/\$\$/'), array('$data[\\1]','$'), $s) . "\n";
           }
       }
   }
   formLetters('/dev/form1.schema','/dev/data.csv');
   ?>
   
7. Ian Tyndall said

   December 1, 2010 at 2:58 AM

   Left the $dcount variable in mine by accident, that would clean it up by two lines.
8. Axio said

   December 1, 2010 at 5:32 AM

   Works with “$$” and more than 10 fields.

   #!/usr/bin/env perl
   ($a,$b)=@ARGV;open($S,$a);$s=join/\n/,<$S>;open
   ($D,$b);while(chomp($_=<$D>)){@d=split/,/;$_=$s
   ;s/\$(\d+)/$d[$1]/g;s/\$\$/\$/g;print}
9. Graham said

   December 2, 2010 at 4:34 AM

   A bit longer than everyone else’s. My answer can deal with arbitrarily many elements, but handles only the subcase
   of the $$ problem where no other $n remain after a $$ in a line of the schema.
   I’ve included the imports, hashbang line, and the test at the end (copy-pasted schema and data from first page):

   #!/usr/bin/env python2.6
   
   import csv
   from string import digits
   
   def form_letter(schema, data):
       reader = csv.reader(open(data))
       s = open(schema)
       lines = s.readlines()
       s.close()
       for row in reader:
           output = ''
           for line in lines:
               x = 0
               while x != -1:
                   x = line.find("$", x)
                   i = x+1
                   while line[i] in digits:
                       i += 1
                   if i != x+1:    # swap out $n with data's row[n]
                       n = int(line[x+1:i])
                       line = line[:x] + row[n] + line[i:]
                   else:
                       break   # handles $$ case, as long as no other numbers are
                               # after it
               output = output + line
           print output + "\n"
       return
   
   if __name__ == "__main__":
       form_letter("schema.txt", "data.txt")
   
10. programmingpraxis said

    December 2, 2010 at 1:44 PM

    Graham: Read the input a character at a time instead of a line at a time, and you won’t have a problem with $n following $$ on the same schema line.
11. Mirko Grönroos said

    December 3, 2010 at 9:36 PM

    I came up with the following version, which can handle many elements but still fails when the schema contains “invalid” markup such as $$$$1 where the amount of $s is unbalanced.
    I also used some map/lambda foo to make it more interesting :)

    #!/usr/bin/env python2.6
    import csv, re
    
    def form_letter(schema, data):
        tmpl = open(schema).read()
        v_re = re.compile(r'(?<!$)\$\d+')
        order = map(lambda key: int(key[1:]), v_re.findall(tmpl))
        tmpl = v_re.sub('%s', tmpl).replace('$$', '$')
    
        reader = csv.reader(open(data))
        for row in reader:
            print tmpl % tuple(map(lambda key: row[key], order))
    
    if __name__ == '__main__':
        form_letter('schema.txt', 'data.csv')
    
12. John Doig said

    December 9, 2010 at 4:42 AM

    import string, csv

    def form_letter(letter, data):
    input = csv.reader(open(data))
    template = string.Template(re.sub(r”\$([0-9]+)”, r”$_\1″, open(letter).read()))
    for row in input:
    context = dict([(“_%s”%j,k) for j,k in enumerate(row)])
    print template.substitute(**context)