我正在尝试获取用户的所有配置文件数据。并且用户可以为服务器上传多个文件，这可以给用户徽章。我的测试环境中有 2 个用户，其中 1 个用户有 3 个文件。现在，当我从 Controller 返回 JSON 时，我得到三个结果，但我想合并或以某种方式合并，以便一个用户可以将所有数据放在一个对象中。
这是我目前返回的 JSON
[
{
"id":1,
"avatar":"john.jpg",
"role":2,
"first_name":"John",
"last_name":"Doe",
"user_name":"johndoe",
"account_status":2,
"one_child":"6",
"introduction":"test",
"file_type":1,
"file":"1498725633.docx",
"status":2,
"lat":59.44,
"lng":24.74,
"rating":"5",
"recommended":"1",
"rating_count":2,
"distance":0.29988841983648
},
{
"id":1,
"avatar":"john.jpg",
"role":2,
"first_name":"Joe",
"last_name":"Doe",
"user_name":"johndoe",
"account_status":2,
"one_child":"6",
"introduction":"test",
"file_type":4,
"file":"118771941-merged.mp4",
"status":1,
"lat":59.44,
"lng":24.74,
"rating":"5",
"recommended":"1",
"rating_count":2,
"distance":0.29988841983648
},
{
"id":4,
"avatar":"capture.JPG",
"role":2,
"first_name":"Jane",
"last_name":"Doe",
"user_name":"janedoe",
"account_status":2,
"one_child":"4",
"introduction":"Test",
"file_type":2,
"file":"1498732136.docx",
"status":1,
"lat":59.46,
"lng":24.83,
"rating":"3",
"recommended":"2",
"rating_count":2,
"distance":5.5651349391591
}
]
如您所见，john doe 有两个文件，但我希望所有文件和文件类型都在一个对象下，这样就只有一个 John 和一个 Jane 对象 atm。
这是我从 laravel 中提取的原始 sql
select
`users`.`id` as `id`,
`users`.`avatar`,
`users`.`role`,
`users`.`first_name`,
`users`.`last_name`,
`users`.`user_name`,
`users`.`account_status`,
`user_wage_preferences`.`one_child`,
`user_introductions`.`introduction`,
`user_files`.`file_type`,
`user_files`.`file`,
`user_files`.`status` as `status`,
`user_contact_informations`.`lat`,
`user_contact_informations`.`lng`,
ROUND(AVG( user_reviews.rating )) AS rating,
SUM(user_reviews.recommended) as recommended,
COUNT(user_reviews.id) as rating_count,
`user_files`.`status`,
(
6371 * acos( cos( radians(59.4424504) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(24.7377842) ) + sin( radians(59.4424504) ) * sin( radians(lat) ) )
)
AS distance
from
`users`
inner join
`user_files`
on `users`.`id` = `user_files`.`user_id`
left join
`user_reviews`
on `users`.`id` = `user_reviews`.`nanny_id`
inner join
`user_introductions`
on `users`.`id` = `user_introductions`.`user_id`
inner join
`user_wage_preferences`
on `users`.`id` = `user_wage_preferences`.`user_id`
inner join
`user_contact_informations`
on `users`.`id` = `user_contact_informations`.`user_id`
where
`users`.`role` = ?
and `users`.`account_status` = ?
group by
`users`.`id`,
`users`.`avatar`,
`users`.`role`,
`users`.`first_name`,
`users`.`last_name`,
`users`.`user_name`,
`user_files`.`status`,
`users`.`id`,
`user_contact_informations`.`lat`,
`user_contact_informations`.`lng`,
`user_wage_preferences`.`one_child`,
`user_introductions`.`introduction`,
`user_files`.`file_type`,
`user_files`.`file`
having
`recommended` > ?
order by
`distance` asc jquery - 3.2.1.min.js:2 Uncaught TypeError: Cannot use 'in' operator to search for 'length' in
select
`users`.`id` as `id`,
`users`.`avatar`,
`users`.`role`,
`users`.`first_name`,
`users`.`last_name`,
`users`.`user_name`,
`users`.`account_status`,
`user_wage_preferences`.`one_child`,
`user_introductions`.`introduction`,
`user_files`.`file_type`,
`user_files`.`file`,
`user_files`.`status` as `status`,
`user_contact_informations`.`lat`,
`user_contact_informations`.`lng`,
ROUND(AVG( user_reviews.rating )) AS rating,
SUM(user_reviews.recommended) as recommended,
COUNT(user_reviews.id) as rating_count,
`user_files`.`status`,
(
6371 * acos( cos( radians(59.4424504) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(24.7377842) ) + sin( radians(59.4424504) ) * sin( radians(lat) ) )
)
AS distance
from
`users`
inner join
`user_files`
on `users`.`id` = `user_files`.`user_id`
left join
`user_reviews`
on `users`.`id` = `user_reviews`.`nanny_id`
inner join
`user_introductions`
on `users`.`id` = `user_introductions`.`user_id`
inner join
`user_wage_preferences`
on `users`.`id` = `user_wage_preferences`.`user_id`
inner join
`user_contact_informations`
on `users`.`id` = `user_contact_informations`.`user_id`
where
`users`.`role` = ?
and `users`.`account_status` = ?
group by
`users`.`id`,
`users`.`avatar`,
`users`.`role`,
`users`.`first_name`,
`users`.`last_name`,
`users`.`user_name`,
`user_files`.`status`,
`users`.`id`,
`user_contact_informations`.`lat`,
`user_contact_informations`.`lng`,
`user_wage_preferences`.`one_child`,
`user_introductions`.`introduction`,
`user_files`.`file_type`,
`user_files`.`file`
having
`recommended` > ?
order by
`distance` asc
我在迁移中建立了所有连接，因此 user_id 引用了 users 表 中的 id。如何获取一个用户对象下的所有文件？
最佳答案
您可以添加 relation，而不是在查询中加入 users 和 user_files 表给他们。
在您的情况下，users 模型将具有如下函数:
public function files()
{
return $this->hasMany('App\UserFiles', 'user_id');
//Where UserFiles is the name of the model of the user_files table
//and user_id is the foreign key
}
一旦你建立了这个关系，你就可以使用 with 将 files 关系添加到你的查询中:
Users::with('files')->where(...)->orderBy(...)->get();
这会将文件加载到用户对象中，如下所示:
{
"id":1,
"avatar":"john.jpg",
"role":2,
"first_name":"John",
"last_name":"Doe",
"user_name":"johndoe",
"account_status":2,
"one_child":"6",
"introduction":"test",
"files": [
{
//File Object
},
{
//File Object
}
],
"lat":59.44,
"lng":24.74,
"rating":"5",
"recommended":"1",
"rating_count":2,
"distance":0.29988841983648
}
关于php - 如何在 Laravel Eloquent 查询中收集项目？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44928437/