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';
错误:架构“天气”不存在。
天气不是架构,它是一张桌子!
答案 0 :(得分:10)
也许:
SELECT weather.id, cities.name, weather.date, weather.degree
FROM weather JOIN cities ON (weather.city_id = cities.id)
WHERE weather.date = '2011-04-30';
postgres抱怨weather.city_id
上的联接被解释为架构'天气'中名为'city_id'的表/视图