Thank you, I was doing this:
def encode_id(id):
...
number = id
charset = string.letters + string.digits
base = len(charset)
count = 0
while number:
digit = number % base
...
and kept getting an error, when I did this:
id = row[0]
encode_id(id)
I didn't realize that the L stood for long, anyhow this works:
id = int(row[0]
encode_id(id)
On Sep 30, 7:15 pm, Malcolm Tredinnick <[EMAIL PROTECTED]>
wrote:
> On Tue, 2008-09-30 at 15:08 -0700, Merrick wrote:
> > If I run the following query on psql:
>
> > SELECT nextval('redirect_link_id_seq');
>
> > it returns an integer, say 5
>
> > when I do the following with the django shell I get a different
> > result:
>
> > >>> def get_next_id():
> > ... cursor = connection.cursor()
> > ... cursor.execute("SELECT nextval('redirect_link_id_seq');")
> > ... next_id = cursor.fetchone()
> > ... return next_id
> > >>> row = get_next_id()
> > >>> row[0]
> > 5L
>
> > Of course I can cast to an integer
> > >>> int(row[0])
> > 5
>
> > But I was hoping someone could let me know why my return value is 5L
> > and not 5, sorry for such a noob question.
>
> Because that's what the db wrapper returns. In practice it makes
> effectively no difference. Historically, Python use to distinguish
> between longs and integers, but even in those days you could almost
> always just work with them as "numbers". Today, that's particularly
> true. Just use the value returned. Why do you need to it to have type
> "int" instead of type "long"?
>
> Regards,
> Malcolm
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