Hi Anthony,
Thanks for helping me solve this problem, I very much appreciate your help.
Annet
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You receiv
Just add the additional joins:
db((db.ntw_edge.outID == 1622) & 'ntw_edge.labelID = alias_edge.labelID').
select(
join=[db.ntw_edge.with_alias('alias_edge').on('ntw_edge.inID =
alias_edge.outID'),
db.vtx_vertex.on(db.ntw_edge.inID == db.vtx_vertex.id),
db.ntw_edge_label_se
HI Anthony,
Thanks for your reply.
> What is the SQL you are trying to produce?
>
SELECT ...
FROM ntw_edge
INNER JOIN ntw_edge_alias ON ntw_edge.inID=alias_edge.outID
INNER JOIN vtx_vertex ON ntw_edge.inID=vtx_vertex.id
INNER JOIN ntw_edge_label_set ON ntw_edge.labelID=ntw_edge_label_set.id
On Monday, February 19, 2018 at 3:12:19 AM UTC-5, Annet wrote:
>
> Hi Anthony,
>
> Thanks for your reply.
>
> I get the results I want. Two more questions, using this syntax, how would
> I join
>
> ntw_edge_inID on vtx_vertex.id
>
> and
>
> ntw_edge_labelID on ntw_edge_label_set.id
>
> without end
Hi Anthony,
Thanks for your reply.
I get the results I want. Two more questions, using this syntax, how would
I join
ntw_edge_inID on vtx_vertex.id
and
ntw_edge_labelID on ntw_edge_label_set.id
without ending up with more records than I have now.
Second, up to now I have been writing joins
This:
join = db.ntw_edge.with_alias('alias_edge').on('ntw_edge.inID =
alias_edge.outID')
db((db.ntw_edge.outID == 1622) & 'ntw_edge.labelID = alias_edge.labelID').
select(join=join)
will produce this:
SELECT "ntw_edge"."id", "ntw_edge"."outID", "ntw_edge"."inID", "ntw_edge".
"labelID", "ntw_edg
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