对连接表中没有键的所有记录和具有匹配键的记录进行单个查询

发布于 2024-08-06 01:03:33 字数 934 浏览 6 评论 0原文

我有一次旅行，有很多驻地。我需要一个查询来返回未指定居住信息的所有行程。以及与指定居住地匹配的所有旅行。

我可以从此查询中获得第一个：

SELECT * FROM `trips` WHERE (((NOT EXISTS (SELECT id FROM residencies WHERE trips.id = residencies.trip_id))

但要获得第二个，我还需要添加这一点：

INNER JOIN `residencies` ON (`trips`.`id` = `residencies`.`trip_id`)

在 WHERE 子句之前添加联接会要求提供具有居住 ID 和不具有居住 ID 的结果。这显然没有返回任何结果。那么我该如何编写此代码才能在一个查询中获取完整的结果集呢？不允许使用存储过程。

我正在使用 Rails，因此如果答案是特定于 Rails 的，那么这是一个额外的好处（但绝对不是必需的）。如果有人能展示如何使用 searchlogic 插件来完成此操作，那就太好了。

目前，我的第一个要求是命名范围：

Trip.named_scope :residencies_empty, :conditions => ['NOT EXISTS (SELECT id FROM residencies WHERE trips.id = residencies.trip_id)']

第二个要求可通过 searchlogic 获得：

Trip.residences_id_equals(id)

理想的解决方案是 searchlogic 范围，如下所示：

Trip.residencies_null_or_residencies_id_equals(id)

原文

I have a trip that has many residencies. I need a single query that returns all trips where no residency information has been specified. And all trips that match a specified residency.

I can get the first from this query:

SELECT * FROM `trips` WHERE (((NOT EXISTS (SELECT id FROM residencies WHERE trips.id = residencies.trip_id))

But to get the second, I need to add this bit as well:

INNER JOIN `residencies` ON (`trips`.`id` = `residencies`.`trip_id`)

Adding the join before the WHERE clause is asking for results that have a residency ID and no residency IDs. That obviously returns nothing. So how can I write this to get the full result set in one query? Stored procedures aren't allowed on this.

I'm using Rails, so it's a bonus (but definitely not required) if the answer is Rails-specific. And a huge bonus if someone can show how this can be done with the searchlogic plugin.

Currently, I have the first requirement as a named scope:

Trip.named_scope :residencies_empty, :conditions => ['NOT EXISTS (SELECT id FROM residencies WHERE trips.id = residencies.trip_id)']

The second requirement is available through searchlogic:

Trip.residences_id_equals(id)

The ideal solution would be a searchlogic scope that looks like this:

Trip.residencies_null_or_residencies_id_equals(id)

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人生戏 2024-08-13 01:03:33

我建议对具有特定居住地的行程使用另一个“EXIST”：

SELECT * FROM `trips` WHERE 
  (NOT EXISTS (SELECT id FROM residencies WHERE trips.id = residencies.trip_id))
  OR
  (EXISTS (SELECT id FROM residencies WHERE trips.id = residencies.trip_id 
           AND other_criteria_for_specific_residency)
  )

这似乎是最具可读性的解决方案，但如果性能很重要，您应该检查 EXPAIN 以了解这是如何优化的（与 MySql 中最复杂的查询一样）。

I suggest using another "EXIST" for the trips with the specific residency:

SELECT * FROM `trips` WHERE 
  (NOT EXISTS (SELECT id FROM residencies WHERE trips.id = residencies.trip_id))
  OR
  (EXISTS (SELECT id FROM residencies WHERE trips.id = residencies.trip_id 
           AND other_criteria_for_specific_residency)
  )

This seems to be the most readable solution, but if performance is important, you should check EXPAIN to see how this is optimized (as with most complex queries in MySql).

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日暮斜阳 2024-08-13 01:03:33

尝试：

SELECT * FROM `trips`
    LEFT JOIN  residencies ON trips.id = residencies.trip_id

您将从 trips 中获取所有列中的数据，但数据只会填充到存在行的驻地的列中，如果不存在驻地行，则这些列将为空。

TRY:

SELECT * FROM `trips`
    LEFT JOIN  residencies ON trips.id = residencies.trip_id

You will get data in all of the columns from trips, but data will only be populated in columns from residencies where a row existed, if no residencies row existed those columns will be null.

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