Question

我的情况是属性表包含地址ID（来自g_addresses表），而申请人表也保存g_addresses中的地址ID。我想一起加入这些，但选择表格中的所有字段。

我知道使用'as'为字段创建别名，但有没有办法为整个表生成别名？

SELECT *
FROM (`reference`)
LEFT JOIN `applicants` ON `applicants`.`id` = `reference`.`applicant_id`
LEFT JOIN `g_people` applicant_person ON `applicant_person`.`id` = `applicants`.`person_id`
LEFT JOIN `g_addresses` applicant_address ON `applicant_address`.`id` = `applicants`.`address_id`
LEFT JOIN `properties` ON `properties`.`id` = `reference`.`property_id`
LEFT JOIN `g_addresses` property_address ON `property_address`.`id` = `properties`.`address_id`
WHERE `reference`.`id` = 4

这会产生一个只包含一个地址行而不是两者的结果，返回的行是最后一个连接的行而不是之前的行，表示它在返回时被覆盖。

Answer 1

在您的情况下，我认为您不应该使用屏蔽引用，例如*或`reference`.*，因为您最终可能会得到一个包含相同列名的行集（id ，address_id）。

如果要从连接表中提取所有列，则应该在SELECT子句中单独指定它们，并为每个列分配唯一的别名：

SELECT
  ref.`id` AS ref_id,
  ref.`…`  AS …,
  …
  app.`id` AS app_id,
  …
FROM `reference` AS ref
LEFT JOIN `applicants`  AS app ON app.`id` = ref.`applicant_id`
LEFT JOIN `g_people`    AS ape ON ape.`id` = app.`person_id`
LEFT JOIN `g_addresses` AS apa ON apa.`id` = app.`address_id`
LEFT JOIN `properties`  AS pro ON pro.`id` = ref.`property_id`
LEFT JOIN `g_addresses` AS pra ON pra.`id` = pro.`address_id`
WHERE ref.`id` = 4

Answer 2

更具体地说明您选择的列

SELECT 
  applicant_address.*,
  property_address.*,
  applicants.*,
  applicant_person.*,
  properties.*
FROM (`reference`)
LEFT JOIN `applicants` ON `applicants`.`id` = `reference`.`applicant_id`
LEFT JOIN `g_people` applicant_person ON `applicant_person`.`id` = `applicants`.`person_id`
LEFT JOIN `g_addresses` applicant_address ON `applicant_address`.`id` = `applicants`.`address_id`
LEFT JOIN `properties` ON `properties`.`id` = `reference`.`property_id`
LEFT JOIN `g_addresses` property_address ON `property_address`.`id` = `properties`.`address_id`
WHERE `reference`.`id` = 4

Mysql - 如何在左连接中对整个表进行别名

2 个答案: