我需要像运行SQL查询一样
sql = 'SELECT * FROM users WHERE id != ' + self.id.to_s + ' AND id NOT IN (SELECT artner_id FROM encounters WHERE user_id = ' + self.id.to_s + ')' sql += ' AND id NOT IN (SELECT user_id FROM encounters WHERE partner_id = ' + self.id.to_s + ' AND predisposition = ' + Encounter::Negative.to_s + ')' sql += ' AND cfg_sex = ' + self.sex.to_s + ' AND cfg_country = ' + self.country.to_s + ' AND cfg_city = ' + self.city.to_s sql += ' ORDER BY rand() LIMIT 1'
它可以由AR.find_by_sql执行,但之前的代码可读性差.
是否有任何查询构建器,可以构建该查询？
例如,Kohana(它是PHP框架,我是PHP开发人员,但我想将这种孩子语言改为ruby / rails)有一个查询构建器,其工作方式如下：
$sql = DB::select('*')->from('users'); $sql->where('id','NOT_IN',DB::expr('SELECT partner_id FROM encounters WHERE user_id = '.$user->id)); $sql->where('id',DB::expr('SELECT user_id FROM encounters WHERE partner_id = '.$user->id.' AND predisposition = '.Encounter::Negative)); .... etc ...
使用像Kohana查询构建器这样的查询构建器构建的查询更具可读性和可理解性.
有什么宝石可以解决这个问题吗？
解决方法
你需要
squeel宝石.它使用块扩展AR,并且可以轻松地进行非常复杂的查询.
只有少数功能：
# not_in == cool! )
Product.where{id.not_in LineItem.select{product_id}}
# SELECT "products".* FROM "products" WHERE "products"."id" NOT IN
# (SELECT "line_items"."product_id" FROM "line_items" )
# outer joins on pure Ruby:
LineItem.joins{product.outer}
# LineItem Load (0.0ms) SELECT "line_items".* FROM "line_items"
# LEFT OUTER JOIN "products" ON "products"."id" = "line_items"."product_id"
# calcs,aliasing:
Product.select{[avg(price).as(middle)]}
# SELECT avg("products"."price") AS middle FROM "products"
# comparison
Product.where{id != 100500}
Product.where{price<10}
# logical OR
Product.where{(price<10) | (title.like '%rails%')}
# SELECT "products".* FROM "products" WHERE (("products"."price" < 10 OR
# "products"."title" LIKE '%rails%'))
# xxx_any feature (also available xxx_all)
Product.where{title.like_any %w[%ruby% %rails%]}
# SELECT "products".* FROM "products" WHERE (("products"."title" LIKE '%ruby%' OR
# "products"."title" LIKE '%rails%'))
注意使用块：{…}这里不是哈希.还要注意没有符号.
如果你决定选择它,请阅读以“这带来一个重要的含义”开头的部分.