如果您只需要每个应用程序的最新版本，一个相关的子查询就足够了：

select 
    a.*,
    (
        select d.version 
        from deployments d 
        where d.app_id = a.id 
        order by d.updated_at desc 
        limit 1
    ) last_version
from applications a
order by a.app_name

这应该是一个有效的解决方案，前提是您在上有一个索引 deployments(app_id, updated_at, version) .
如果您想从 deployments 表，则选项是 join 以及使用相关子查询进行过滤：

select a.app_name, d.version, d.updated_at, d.created_at
from applications a
inner join deployments d on d.app_id = a.id
where d.updated_at = (
    select max(d1.updated_at) from deployments d1 where d1.app_id = a.app_id
)
order by a.app_name

SELECT de.*
FROM application a
INNER JOIN
(SELECT d.app_id, MAX(d.updated_at) AS MaxDateTime FROM deployments d GROUP BY d.app_id) de
on de.app_id = a.id
order by a.name;

为多列编辑：mysql的旧版本允许在选择列表中使用非聚合列。在较新的版本（5.7+）中，可以删除sql模式“only\ full\ group\ by”以获得相同的效果。现在，可以在派生表中添加所有列。
如果需要可移植查询，则需要连接回以获得结果：

SELECT d.*
FROM application a
INNER JOIN
(SELECT d.app_id, MAX(d.updated_at) AS MaxDateTime FROM deployments d GROUP BY d.app_id) de
on de.app_id = a.id
INNER JOIN deployments d
ON d.updated_at = de.update_at
order by a.name;

如果您想要每个应用程序的最后一次部署，可以使用 IN 使用聚合子查询：

select * 
from applications a
join deployments d on d.app_id = a.id
where (d.app_id, d.updated_at) in
(
  select app_id, max(updated_at)
  from deployments
  group by app_id
)
order by a.name;