Question

我具有以下表格结构

Users table
    id- integer
    name-string

Casefiles table
    id- integer
    user_id- foreign_key
    name-string

Followups table
    id- integer
    date- date
    casefile_id-foreign_key

我正在使用以下模型之间的关系 User.php

public function casefiles()
    {
      if ($this->hasRole(['sales_executive'])) {
        return Casefile::where('user_id', $this->id);
      }
    }

Casefile.php

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

    public function followups()
    {
      return $this->hasMany('App\FollowUp');
    }

Followup.php

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

我想直接获取用户跟进信息。我怎样才能做到这一点？

Answer 1

您需要在hasManyThrough()中使用User.php，然后才能添加它，

 public function followUps()
 {
   return $this->hasManyThrough('App\FollowUp','App\Casefile');
 }

那么您通常可以使用User::with('followUps')->find(1)->followUps

进行跟进

我注意到您正在检查您的关系$this->hasRole(['sales_executive']中的角色，这可能会发生错误，因为如果该语句为假，则不是。我认为您也采取了另一种方法，并且也处理了该问题。有关hasManyThrough的更多信息，请选择此link

Answer 2

HasManyThrough关系就是这种情况。

您的情况应该是

// Followups migration
Schema::create('followups', function (Blueprint $table) {
    $table->bigIncrements('id');
    $table->date('date');
    $table->unsignedBigInteger('casefile_id')->nullable();
    $table->timestamps();

    $table->foreign('casefile_id')
        ->references('id')
        ->on('casefiles');
});

// Casefile migration
Schema::create('casefiles', function (Blueprint $table) {
    $table->bigIncrements('id');
    $table->string('name');
    $table->unsignedBigInteger('user_id')->nullable();
    $table->timestamps();

    $table->foreign('user_id')
       ->references('id')
       ->on('users');
});

// User.php

public function followups() {
    return $this->hasManyThrough(FollowUp::class, Casefile::class);
}

// YourController.php
$user = User::all()->first();
dd($user->followups);

Answer 3

更改User.php模型中与casefile和后续文件的关系：

public function casefiles()
{
  if ($this->hasRole(['sales_executive'])) {
    return $this->hasMany(Casefile::class);
  }
}

public function followups() {
    return $this->hasManyThrough(FollowUp::class, Casefile::class);
}

获取关系模型数据

3 个答案: