Hello
I'm getting some unwanted result when SELECTing data from an SQLite
database:
======
sql = 'SELECT id FROM master'
rows=list(curso r.execute(sql))
for id in rows:
sql = 'SELECT COUNT(code) FROM companies WHERE code="%s"' % id[0]
result = list(cursor.exe cute(sql))
print "Code=%s, number=%s" % (id[0],result[0])
======
Code=0111Z, number=(47,)
======
I expected to see "number=47" . Why does Python return "(47,)"?
Thank you.
Nov 11 '08
12 1598
John Machin wrote:
On Nov 11, 10:47 pm, Steve Holden <st...@holdenwe b.comwrote:
>Chris Rebert wrote:
>>On Tue, Nov 11, 2008 at 12:56 AM, Gilles Ganault <nos...@nospam. comwrote: On Mon, 10 Nov 2008 20:02:39 -0600, Andrew <alif...@gmail. comwrote: sql = 'SELECT id FROM master' rows=list(c ursor.execute(s ql)) for id in rows: sql = 'SELECT COUNT(code) FROM companies WHERE code="%s"' % id[0] result = list(cursor.exe cute(sql)) print "Code=%s, number=%s" % (id[0],result[0][0]) Using liberal "term rewriting", consider the following rough equivalenci es in the code: id[0] <==rows[INDEX_HERE][0] <==list(cursor. execute(sql))[INDEX_HERE][0] result[0][0] <==list(cursor. execute(sql))[0][0] Note that in both cases, the list is sliced twice; the for-loop just conceals the `[INDEX_HERE]` implicit slicing that is caused by iterating over the list.
You might also want to consider saving some time by using a SQL solution (assuming SQLite supports it, which it should) (untested):
cursor.execute (""" SELECT master.id, count(companies .code) FROM master JOIN companies ON master.id = companies.code GROUP BY companies.code" "")
Shouldn't it be GROUP BY master.id? I would have thought that SQL
would be sad about a non-aggregate (master.id) that's in the SELECT
list but not also in the GROUP BY list.
Well, I did say "untested". But in SQL Server, for example, any field
argument to COUNT() must be an aggregated column. So it may depend on
the SQL implementation. I should really have said
GROUP BY master.id, companies.code
which is the kind of stupidity SQL's brainless implementations force one
to resort to.
>for id, count in cursor.fetchall (): print "Code=%s, number=%s" % (id, count)
I'd like to think it makes the Python a bit more readable too ...
Agreed. result[0][0] is an abomination.
Though one I am sure we have all used at times. The original code wasn't
too bad for a beginner.
regards
Steve
--
Steve Holden +1 571 484 6266 +1 800 494 3119
Holden Web LLC http://www.holdenweb.com/
Steve Holden wrote in news:mailman.38 04.1226412496.3 487.python- li**@python.org in comp.lang.pytho n:
>Shouldn't it be GROUP BY master.id? I would have thought that SQL would be sad about a non-aggregate (master.id) that's in the SELECT list but not also in the GROUP BY list.
Well, I did say "untested". But in SQL Server, for example, any field
argument to COUNT() must be an aggregated column. So it may depend on
the SQL implementation. I should really have said
You must mean an "SQL Server" other than the Microsofts one, as:
select count( aid ) as "count"
from table_1
group by aid
count
-----------
8
8
8
8
8
8
8
8
(8 row(s) affected)
and:
select count( aid ) as "count"
from table_1
count
-----------
64
(1 row(s) affected)
Like it should.
Rob.
-- http://www.victim-prime.dsl.pipex.com/
Rob Williscroft wrote:
Steve Holden wrote in news:mailman.38 04.1226412496.3 487.python- li**@python.org in comp.lang.pytho n:
>>Shouldn't it be GROUP BY master.id? I would have thought that SQL would be sad about a non-aggregate (master.id) that's in the SELECT list but not also in the GROUP BY list.
Well, I did say "untested". But in SQL Server, for example, any field argument to COUNT() must be an aggregated column. So it may depend on the SQL implementation. I should really have said
You must mean an "SQL Server" other than the Microsofts one, as:
select count( aid ) as "count"
from table_1
group by aid
count
-----------
8
8
8
8
8
8
8
8
(8 row(s) affected)
and:
select count( aid ) as "count"
from table_1
count
-----------
64
(1 row(s) affected)
Like it should.
Hmm, strange. I must be thinking of some other SQL Server then. Or, more
likely, some other error situation.
regards
Steve
--
Steve Holden +1 571 484 6266 +1 800 494 3119
Holden Web LLC http://www.holdenweb.com/ This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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