Help cleaning up some code
Martin P. Hellwig
xng at xs4all.nl
Sat Mar 7 12:12:39 EST 2009
Martin P. Hellwig wrote:
<cut>
> def query_parser(QUERY, USER, STACK_SIZE):
> indexes = ['ni','adid','rundateid','rundate','city','state','status']
> empty = 'None'
>
> stack = []
> query_result = self.con.execute(QUERY,(USER,STACK_SIZE)).fetchall()
> ni = indexes[0]
>
> for row in query_result:
> temp_dict = dict()
>
> if len(stack) > 0:
> if stack[0][ni] != row[ni]:
> # Break if the current ni is not equal to the first
> break
>
> else:
> for item in indexes:
> try:
> temp_dict[item] = row[item]
> except:
> temp_dict[item] = empty
>
> stack.append(temp_dict)
>
> return(stack)
>
> hth
Hold my horses :-), I made logical errors in the break, else and for
item parts (thats what you get from rewriting instead of rethinking ;-) )
This should be more like it:
def query_parser(QUERY, USER, STACK_SIZE):
indexes = ['ni','adid','rundateid','rundate','city','state','status']
empty = 'None'
stack = []
query_result = self.con.execute(QUERY,(USER,STACK_SIZE)).fetchall()
ni = indexes[0]
for row in query_result:
temp_dict = dict()
ignore = False
if len(stack) > 0:
if stack[0][ni] != row[ni]:
ignore = True
if ignore == False:
for item in indexes:
try:
temp_dict[item] = row[item]
except:
temp_dict[item] = empty
stack.append(temp_dict)
return(stack)
Though it is likely I made another stupid mistake.
--
mph
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