Question

我有一个＆＃39; field_revision ...＆＃39;我试图从中获取信息的表格。使用我当前的查询，它返回与该值匹配的所有entity_id，即使revision_id小于该特定id的最高值。下面是一个示例表：

  factory :request do
    content { Faker::Lorem.paragraph }
    association :requester
    association :group

    trait :has_neighbor do
      association :responder
    end

    trait :is_fulfilled do
      is_fulfilled true
    end

    trait :created_one_day_ago do
      created_at { 1.day.ago }
    end

    factory :request_with_neighbor, traits: [:has_neighbor]
    factory :fulfilled_request , traits: [:has_neighbor, :is_fulfilled]
    factory :day_old_request, traits: [:created_one_day_ago]
  end

我需要一个返回所有entity_ids的查询，其中enable_value = true且revision_id最高。对于此示例表，查询需要返回781和786，因为对于最高版本ID，这两个都是真的。

Answer 1

SELECT T1.*
  FROM
  field_revision T1 INNER JOIN 
    (SELECT entity_id,MAX(revision_id) AS revision_id
  FROM
  field_revision
  GROUP BY entity_id) T2 ON T1.entity_id = T2.entity_id AND T1.revision_id = T2.revision_id
    WHERE T1.enable_value=1

编辑：OP

的说明

T1 - field_revision的别名

T2 - 查看field_revision entity_id明智且最大revision_id。

因此，在INNER JOINING中我们使用entity_id和max（revision_id）连接T2与T1，并在T1中打印相应的条目

希望这会有所帮助..

Answer 2

如果你只想要entity_id那么你可以试试这个（但我认为@Subin C Poonamgode的回答是正确的）：

SELECT DISTINCT aa.entity_id
FROM field_revision AS aa
INNER JOIN field_revision AS bb
ON aa.entity_id = bb.entity_id
WHERE aa.revision_id > bb.revision_id AND aa.enable_value = 1;

在Mysql中显示每个组中具有最大值的行

2 个答案: