如何让这个查询起作用?
SELECT weather.id,cities.name,weather.date,weather.degree FROM weather JOIN weather.city_id ON cities.id WHERE weather.date = '2011-04-30';
天气不是架构,它是一张桌子!@H_502_4@
也许:
SELECT weather.id,weather.degree FROM weather JOIN cities ON (weather.city_id = cities.id) WHERE weather.date = '2011-04-30';
postgres抱怨关于weather.city_id的加入,这被解释为模式’weather’中名为’city_id’的表/视图@H_502_4@
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