How to replace newline followed by space in Parsley PEG grammar

Question

all I'm learning Parsley by writing parser that parse vCard format. My problem is that some lines in vCard format may be split over several lines. Multiple-line representation using the RFC 822 “folding” technique. That is, wherever there may be linear white space a CRLF immediately followed by at least one LWSP-char. My solution is first parse raw text, there I can find newline followed by space and join that lines. However I can't get it to work: there is my latest try: The code returns errors there is last: TypeError: unsupported operand type(s) for +: 'bool' and 'str

import parsley
my_newline = "\n"
space = " "

def isLogicalLine(txt=""): #Function called from Parsley, just search for newline followed by space
    result = True
    if txt[:-2] == my_newline+space:
        result = False
    return result

tst_line2 = parsley.makeGrammar("""
CRLF = '\n' 
line = (~(CRLF) anything+:x ?(isLogicalLine(x))):first_part  -> first_part+ ' test '
""", {"isLogicalLine":isLogicalLine})
# ~(CRLF) = NOT CRLF
# ?(isLogicalLine(x)) = python expression that is true or false
# -> first_part+ ' test ' == modify result by Python express

test_string = "BEGIN:VCARD"+my_newline+"VERSION:2.1"+my_newline+\
              "N;CHARSET=UTF-8;ENCODING=8BIT:Bab"+my_newline+\
              "TEL;PREF;VOICE;ENCODING=8BIT:123456789"+my_newline+"END:VCARD"+\
              my_newline+" END:VCARD"

print(tst_line2(test_string).line())

score 0 · Answer 1 · answered Jun 24 '14 at 02:57

You can fix it by moving the test out of the binding, like so:

line = (~(CRLF) anything+:x):first_part ?(isLogicalLine(x)) -> first_part + ' test '

Not sure if it's a bug or not, but first_part was being bound to the result of the isLogicalLine test, so first_part + ' test ' was trying to add a bool and a string.

How to replace newline followed by space in Parsley PEG grammar

1 Answers1