Re: how to create aggregate xml document in 8.3?
От | Matt Magoffin |
---|---|
Тема | Re: how to create aggregate xml document in 8.3? |
Дата | |
Msg-id | 51059.192.168.1.108.1197434964.squirrel@msqr.us обсуждение исходный текст |
Ответ на | Re: how to create aggregate xml document in 8.3? (Tom Lane <tgl@sss.pgh.pa.us>) |
Список | pgsql-general |
> "Matt Magoffin" <postgresql.org@msqr.us> writes: >> Thanks very much, that helps. Now I'm wondering if it's also possible to >> then fill in another nested element level in the XML output, from the >> rows >> that are aggregated into the <range> count. > > Something involving xmlagg in the sub-query, perhaps? No time to > experiment with it now. Thanks for the first tip, anyway. I got stuck with trying this out myself... I had ended up with select xmlelement( name "matchback-months", xmlattributes(1 as "count", 'true' as "multi"), xmlagg(ranges)) from ( select xmlelement(name "range", xmlattributes(m.range, count(s.id) as "sales-conv-from-lead"), xmlagg(sales)) from ( select xmlelement(name "sale", xmlattributes(ss.vin, ms.lead_id as "lead-id") ) as sales from mb_sale ss inner join mb_lead ms on ms.sale_id = ss.id where ss.sale_date >= date('2007-08-01') and ss.sale_date <= date('2007-08-30') and ss.sale_type = 'd' and ms.range = m.range order by ss.sale_date ) ssub ) as ranges from mb_sale s inner join mb_lead m on m.sale_id = s.id where s.sale_date >= date('2007-08-01') and s.sale_date <= date('2007-08-30') and s.sale_type = 'd' group by m.range order by m.range ) sub; but this does not compile: ERROR: syntax error at or near "from" LINE 20: from mb_sale s ^ If anyone has any suggestions, much appreciated. -- m@
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