python - Where is the circular reference? -
i'm writing python classes want encode json. when try jsonify objects, error mentioning 'circular reference'. think understand circular reference means, can't find examples in code.
the relationship (has a/is a) between objects
- signup has a
- registrant has a
- address
code (python):
class address: def __init__(self, address1, address2, city, state, zip): self.address1 = address1 self.address2 = address2 self.city = city self.state = state self.zip = zip class signup: def __init__(self, registrant, classid, date, time, paid, seatcost, notes, classname, seats, groupid, agentname, agentcompany): self.registrant = registrant self.classid = classid self.date = date self.time = time self.paid = paid self.seatcost = seatcost self.notes = notes self.classname = classname self.seats = seats self.groupid = groupid self.agentname = agentname self.agentcompany = agentcompany class registrant: def __init__(self, firstname, lastname, address, phone, email): self.firstname = firstname self.lastname = lastname self.address = address self.phone = phone self.email = email def scrape(br): signups = [] soup = libstuff.getsoup(br, 'http://thepaintmixer.com/admin/viewdailysignups.php') table = soup.find(id='calendar') rows = table.find_all('tr') rownumber = 0 row in rows: if rownumber == 0: rownumber = rownumber + 1 continue cells = row.find_all('td') cellnumber = 0 cell in cells: if cellnumber == 0: try: firstname = cell.contents[0] except indexerror: firstname = none elif cellnumber == 1: try: lastname = cell.contents[0] except indexerror: lastname = none elif cellnumber == 2: try: address1 = cell.contents[0] except indexerror: address1 = none elif cellnumber == 3: try: address2 = cell.contents[0] except indexerror: address2 = none elif cellnumber == 4: try: city = cell.contents[0] except indexerror: city = none elif cellnumber == 5: try: state = cell.contents[0] except indexerror: state = none elif cellnumber == 6: try: zip = cell.contents[0] except indexerror: zip = none elif cellnumber == 7: try: phone = cell.contents[0] except indexerror: phone = none elif cellnumber == 8: try: email = cell.contents[0] except indexerror: email = none elif cellnumber == 9: try: classid = cell.contents[0] except indexerror: classid = none elif cellnumber == 10: try: date = cell.contents[0] except indexerror: date = none elif cellnumber == 11: try: time = cell.contents[0] except indexerror: time = none elif cellnumber == 12: try: paid = cell.contents[0] except indexerror: paid = none elif cellnumber == 13: try: seatcost = cell.contents[0] except indexerror: seatcost = none elif cellnumber == 14: try: notes = cell.contents[0] except indexerror: notes = none elif cellnumber == 15: try: classname = cell.contents[0] except indexerror: classname = none elif cellnumber == 16: try: seats = cell.contents[0] except indexerror: seats = none elif cellnumber == 17: try: groupid = cell.contents[0] except indexerror: groupid = none elif cellnumber == 18: try: agentname = cell.contents[0] except indexerror: agentname = none elif cellnumber == 19: try: agentcompany = cell.contents[0] except indexerror: agentcompany = none cellnumber = cellnumber + 1 address = address(address1, address2, city, state, zip) registrant = registrant(firstname, lastname, address, phone, email) signup = signup(registrant, classid, date, time, paid, seatcost, notes, classname, seats, groupid, agentname, agentcompany) signups.append(signup) return signups #i call json.dumps() on returned list json.dumps(scrape(br), default=lambda o: o.__dict__) are constructors messed up? passing shouldn't?
the cause cell.contents[0] returning complex beautifulsoup object rather straight text. beautifulsoup objects know parents, siblings, parser class, attribute, , other objects might shared or circular.
the situation arises when <td> element contains inner html. common in tables (perhaps table entry bolded or italicized example).
a solution problem make sure use beautifulsoup's .text make sure you're getting text , not inner beautifulsoup elements:
columns = [col.text col in row.findall('td')] fwiw, here easy diagnostic technique see going on. modify default function in json.dumps() make output visible:
def view_dict(obj): print '--------------' print 'type:', obj.__class__ d = obj.__dict__ pprint.pprint(d) return d json.dumps(scrape(br), default=view_dict) the circular reference should pop right out. hope clears mystery (because otherwise code looks fine , not explicitly creating circular references).
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