Help cleaning up some code
odeits at gmail.com
Sat Mar 7 06:52:30 CET 2009
Thanks a lot! that copy function is a really neat trick!
Yes the ni's are sorted, and yes the query SHOULD be the one to limit it to
just one set of ni's; however, i am having trouble implementing that in sql.
I am using sqlite3 atm and right now i have a very large select statment the
gets me what i want, i just dont know how to filter it again base on data
that is from the result itself. I THINK normally one would select into a
temporary table then filter it from there, but i am not sure.
On Fri, Mar 6, 2009 at 9:43 PM, andrew cooke <andrew at acooke.org> wrote:
> 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['ni'] != ad['ni']:
> > break;
> > except IndexError:
> > pass
> > stack.append(ad)
> NI = 'ni'
> def copy(src, dst, name, quote_none=False):
> value = src[name]
> dst[name] = 'None' if quote_none and value is None else value
> stack = 
> for row in self.con.execute(adquerystring,(user,STACK_SIZE)).fetchall():
> ad = dict()
> for name in (NI, 'adid', 'rundateid', 'rundate', 'status'):
> copy(row, ad, name)
> for name in ('city', 'state'):
> copy(row, ad, name, quote_none=True)
> if stack and stack[NI] != ad[NI]:
> but couldn't you change the query so that it only returns a single 'ni'?
> also, do you really want to break, or just skip the append? maybe you
> if not stack or stack['ni'] == ad['ni']:
> or perhaps the 'ni' are sorted (in which case what you are doing makes
> some sense).
> > --
> > http://mail.python.org/mailman/listinfo/python-list
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