Laravel LeftJoin从第一列返回重复值（mysql）

Question

专栏工作者：

列建议：

我的查询：

$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();

Laravel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"}

我两次返回了id，name和表workers中的其他值。 我想要退货看起来像这样：

{"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"}

我要从表workers中仅返回一行，并从表recommendations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，仅返回表recommendations中的第一条记录。我不能使用where，因为所有值都将显示在同一页面上。 列worker_id与表workers

相关

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');
}

Answer 1

最简单的方法是load flex关系：

recommendations

，然后通过遍历建议的方式访问建议，例如

$workers = Worker::with('recommendations')->get();

但是，如果您希望像问题中的示例一样对行进行格式化，则可以利用它们是collections的事实：

@foreach($workers as $worker)

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

        //

    @endforeach

@endforeach