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我正在尝试通过以下 SQL 查询中的左外连接上的字段对检索到的记录进行排序：

@companies = Company.scoped
@companies = @companies.where("companies.is_deleted = 'f' AND companies.state IN (?)", ["draft", "pending"])
@companies = @companies.includes(:events)
@companies = @companies.order("events.created_at DESC")

SELECT "companies"."id" AS t0_r0, "companies"."name" AS t0_r1, "companies"."reference" AS t0_r2, "companies"."state" AS t0_r3, "companies"."description" AS t0_r4, "companies"."remarks" AS t0_r5 "events"."id" AS t2_r0, "events"."eventable_type" AS t2_r1, "events"."eventable_id" AS t2_r2, "events"."event_type" AS t2_r3, "events"."creator_company_id" AS t2_r4, "events"."creator_user_id" AS t2_r5, "events"."created_at" AS t2_r6
FROM "companies"
LEFT OUTER JOIN "events" ON "events"."eventable_id" = "companies"."id" AND "events"."eventable_type" = 'company'
WHERE "companies"."is_deleted" = 'f' AND companies.state IN ('draft','pending')
ORDER BY events.created_at DESC

但是检索到的记录没有按预期正确排序（即按 events.created_at）

任何线索如何在保持性能高效的同时成功地做到这一点？

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1 回答 1

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我认为您不应该使用 LEFT JOIN，或者如果代码在这种情况下生成 LEFT JOIN，则可能是错误的。

表 A 和 B 的左外连接（或简称左连接）的结果始终包含“左”表 (A) 的所有记录，” -- http://en.wikipedia.org/wiki/Join_%28SQL %29#Left_outer_join

因此，如果您不希望获得 COMPANIES 的所有记录，即使 events 上没有匹配项，您也不应该在此处真正使用左连接。它将为右侧表生成具有 NULL 值的行。

于 2012-02-09T10:35:16.837 回答