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因为我对 Rails 的了解比 SQL 还新,所以我采取了捷径来构建一个相当复杂的(对我来说,无论如何)查询。

好消息:我的查询工作正常

坏消息:我被 hackish 代码困住,无法弄清楚如何将动态变量放入其中(如果它采用 Rails 友好格式,这应该很容易)。

这是我的查询(为便于阅读而分解):

ActiveRecord::Base.connection.select_all( "SELECT DISTINCT ps.* FROM merchants merch, product_sales ps, users u, user_merchant_relations umr, memberships m

INNER JOIN marketing_efforts me ON me.id = ps.marketing_effort_id

LEFT OUTER JOIN member_sales ms ON me.id = ms.marketing_effort_id

LEFT OUTER JOIN crm_sales cs ON me.id = cs.marketing_effort_id

WHERE ((me.start < '2013-05-26' AND me.end > '2013-05-26') OR (me.start < '2013-05-26' AND me.end IS NULL)) AND (me.num_avail IS NULL OR me.num_avail > 0) AND (me.gender IS NULL OR me.gender = u.gender OR u.gender IS NULL) AND ((u.birthdate > me.birthdate_start AND u.birthdate < me.birthdate_end) OR (u.birthdate > me.birthdate_start AND me.birthdate_end IS NULL) OR (me.birthdate_start IS NULL AND me.birthdate_end IS NULL) OR u.birthdate IS NULL) AND ((ms.member_id = m.member_id AND m.user_id = u.id) OR ms.member_id IS NULL) AND (cs.crm_tier_id = umr.crm_tier_id OR cs.crm_tier_id IS NULL) AND u.id = 3 AND merch.id = 1")

出现“2013-05-26”的任何地方我都需要访问 Time.now。最后的 u.id 和 merch.id 也需要接受变量。

我以为我可以简单地做: .S=?...Q=?....L=?...", value, value, value) 但这不起作用。

更新

我在这:

ProductSale.joins('INNER JOIN marketing_efforts ON marketing_efforts.id = product_sales.marketing_effort_id').joins('LEFT OUTER JOIN member_sales ON marketing_efforts.id = member_sales.marketing_effort_id').joins('LEFT OUTER JOIN crm_sales ON marketing_efforts.id = crm_sales.marketing_effort_id')

所以我正在处理所有的连接,但我遇到的问题是,使用原始 SQL,我可以轻松定义我需要访问的所有表......例如:

FROM merchants merch, product_sales ps, users u, user_merchant_relations umr, memberships m

而我不确定如何使用 Rails 做到这一点。结果,我得到FROM product_sales

ProductSale Load (0.3ms) SELECT "product_sales".* FROM "product_sales" INNER JOIN ....

因此,例如,当我尝试说时,users.id = 1我得到:

ActiveRecord::StatementInvalid: SQLite3::SQLException: no such column: users.id:

有什么想法吗?谢谢你。

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1 回答 1

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也许不是最好的方法,但它有效并解决了主要问题(无法传递动态变量):

ProductSale.find_by_sql [ " QUE=?...R=?..Y=? ", value, value, value]

于 2013-05-26T13:53:57.290 回答