Laravel LeftJoin从第一列返回重复值(mysql)

Laravel LeftJoin从第一列返回重复值(mysql),mysql,laravel,left-join,Mysql,Laravel,Left Join,专栏工作者: 专栏建议: 我的问题是: $getRecommendationWorker = Worker::select('workers.id', 'workers.name', 'workers.photo', 'workers.description', 'workers.profile_worker', 'recommendations.recommendation', 'recommendations.author', 'recommendations.work_author',

专栏工作者:

专栏建议:

我的问题是:

$getRecommendationWorker = Worker::select('workers.id', 'workers.name', 'workers.photo', 'workers.description', 'workers.profile_worker', 'recommendations.recommendation', 'recommendations.author', 'recommendations.work_author', 'recommendations.profile_author')
->leftJoin('recommendations', 'workers.id', '=', 'recommendations.worker_id')
->get();
拉维返回:

{"id":5,"name":"Piotr Debowski","photo":"3c29641e2a.jpeg","description":"sdasdasd","profile_worker":"https:\/\/codepen.io\/","recommendation":"pierwsza rekomendacja","author":"Jan Kowalski","work_author":"IT spec","profile_author":"https:\/\/stackoverflow.com\/questions\/43136250\/eloquent-join-table-with-two-conditions"}
{"id":5,"name":"Piotr Debowski","photo":"3c29641e2a.jpeg","description":"sdasdasd","profile_worker":"https:\/\/codepen.io\/","recommendation":"druga rekomendacja","author":"Jan Wo\u017aniak","work_author":"dsad","profile_author":"https:\/\/nczas.com\/2018\/08\/21\/wreszcie-zatrzymali-t"}
我从表
workers
返回了
id
name
和其他值两次。 我想让你看起来像这样:

{"id":5,"name":"Piotr Debowski","photo":"3c29641e2a.jpeg","description":"sdasdasd","profile_worker":"https:\/\/codepen.io\/","recommendation":"pierwsza rekomendacja","author":"Jan Kowalski","work_author":"IT spec","profile_author":"https:\/\/stackoverflow.com\/questions\/43136250\/eloquent-join-table-with-two-conditions","recommendation":"druga rekomendacja","author":"Jan Wo\u017aniak","work_author":"dsad","profile_author":"https:\/\/nczas.com\/2018\/08\/21\/wreszcie-zatrzymali-t"}
我希望从表
workers
中只返回一行,并从表
建议中返回与第一个表相关的所有记录。
当我使用
groupBy
(例如:
..->->-groupBy('workers.id')
)时,只返回表
建议中的第一条记录。我不能在其中使用
,因为所有值都将显示在同一页上。
列
worker\u id
与表
workers

编辑:

工人模型:

public function recommendations()
{
    return $this->hasMany('App\Recommendation');
}
推荐模式:

public function workers()
{
    return $this->belongsTo('App\Worker');
}

最简单的方法是建立
推荐关系

$workers = Worker::with('recommendations')->get();
然后通过循环访问这些建议,例如

@foreach($workers as $worker)

    @foreach($worker->recommendations as $recommendation)

        //

    @endforeach

@endforeach
但是,如果您希望行的格式与问题中的示例相同,则可以利用以下事实:


连接就是这样工作的<代码>使用groupBy时,仅返回表建议中的第一条记录
。。。这不是你想要的吗?@Devon No from table
Recommensions
我想要得到与table
workers
相关的所有记录为什么给我负号,我的查询有什么问题!!是的,但是你有多个针对单个工作人员的建议,这就是为什么你需要对它们进行分组…@Devon当我使用groupBy('Recommensions.worker_id')
Laravel时,从表
Recommensions
中只返回一条记录,但我想从
Recommensions
表中返回所有记录,它们与表
workers
相关,如果您有不同的建议值,则不应出现这种情况。worker\u id.转储查询的SQL(使用调试栏或
toSql()
$workers = Worker::with('recommendations')->get()
    ->flatMap(function ($worker) {

        $recommendations = $worker->recommendations->isEmpty()
            ? collect([new \App\Recommendation])
            : $worker->recommendations;

        return $recommendations->map(function ($recommendation) use ($worker) {
            return (object)[
                'id'             => $worker->id,
                'name'           => $worker->name,
                'photo'          => $worker->photo,
                'description'    => $worker->description,
                'profile_worker' => $worker->profile_worker,
                'recommendation' => $recommendation->recommendation,
                'author'         => $recommendation->author,
                'work_author'    => $recommendation->work_author,
                'profile_author' => $recommendation->profile_author,
            ];
        });
    });