Question

我有一种情况需要

订购我的结果基于逻辑顺序
分组以仅获取订单后的第一个结果。

我有两张表objects和object_versions

每个对象可以有多个版本，两个表通过

objects.id = object_versions.id

我的订单逻辑是以下abs(object_versions.size - $someNumber) 这非常适合订购问题而没有组我仍然可以在结果中获得所有其他版本。查看方案和小提琴http://www.sqlfiddle.com/#!2/a967b/23

查询1（正确的结果，但没有分组）

SQL

SELECT 
  objects.id as object_id, 
  object_versions.id as object_versions_id, 
  object_versions.size as object_versions_size
FROM objects
INNER JOIN object_versions ON objects.id = object_versions.id
ORDER BY abs(object_versions.size - 1000) ASC;

口才

Object::join('object_versions', 'objects.id', '=', 'object_versions.id')
->orderByRaw("abs(width - ?) ASC", [1000])
->select('objects.id as object_id', 'object_versions.id as object_versions_id', 'object_versions.size as object_versions_size')->get();

结果

id  id  size
1   1   800
2   2   800
2   2   400
1   1   400
2   2   300
1   1   300

查询2（订单前分组产生的结果不佳）

SQL

SELECT 
  objects.id as object_id, 
  object_versions.id as object_versions_id, 
  object_versions.size as object_versions_size
FROM objects
INNER JOIN object_versions ON objects.id = object_versions.id
GROUP BY object_id
ORDER BY abs(object_versions.size - 1000) ASC;

口才

Object::join('object_versions', 'objects.id', '=', 'object_versions.id')
->orderByRaw("abs(width - ?) ASC", [1000])
->groupBy("objects.id")
->select('objects.id as object_id', 'object_versions.id as object_versions_id', 'object_versions.size as object_versions_size')->get();

结果

id  id  size
1   1   300
2   2   300

查询3（订单后的分组更正结果，但不知道如何以雄辩的方式生成此内容）

SQL

SELECT * FROM (
  SELECT 
    objects.id as object_id, 
    object_versions.id as object_versions_id, 
    object_versions.size as object_versions_size
  FROM objects
  INNER JOIN object_versions ON objects.id = object_versions.id
  ORDER BY abs(object_versions.size - 1000) ASC
) as result 
GROUP BY object_id;

口才我不知道该怎么做......

结果

1   1   800
2   2   800

在Eloquent中的ORDER_BY之后执行GROUP_BY

0 个答案: