odeits wrote:
I am looking to clean up this code... any help is much appreciated.
Note: It works just fine, I just think it could be done cleaner.
The result is a stack of dictionaries. the query returns up to
STACK_SIZE ads for a user. The check which i think is very ugly is
putting another contraint saying that all of the ni have to be the
same.
stack = []
rows = self.con.execute(adquerystring,(user,STACK_SIZE)).fetchall()
for row in rows:
ad = dict()
ad['ni'] = row['ni']
ad['adid'] = row['adid']
ad['rundateid'] = row['rundateid']
ad['rundate'] = row['rundate']
if row['city'] is None:
ad['city'] = 'None'
else:
ad['city'] = row['city']
if row['state'] is None:
ad['state'] = 'None'
else:
ad['state'] = row['state']
ad['status'] = row['status']
try:
if stack[0]['ni'] != ad['ni']:
break;
except IndexError:
pass
stack.append(ad)
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Random ideas:
def fetchsomedata(self, user):
query = 'some query'
return list(self._fetch(query, user, STACKSIZE))
def _fetchsomedata(self, query, *params):
keys = 'adid rundateid rundate status'.split()
specialkeys = 'city state'.split()
nikey = 'ni'
ni = None
for row in self.con.execute(query, params).fetchall():
nival = row[nikey]
if ni != nival:
break
if ni is None:
ni = nival
data = {nikey: nival}
for k in keys:
data[k] = row[k]
for k in specialkeys:
data[k] = 'None' if row[k] is None else row[k]
yield data
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