php - Laravel: hasManyThrough 关系

Question

我浪费了一整天的时间，但无法弄清楚 hasManyThrough 关系。

我有这样的关系:

# get related users
public function users()
{
    return $this->hasManyThrough(User::class, FuneralHomeUser::class, 'user_id', 'id');
}

这会生成这个查询:

SELECT `users`.*, `funeral_homes_users`.`user_id`
FROM `users`
INNER JOIN `funeral_homes_users` ON `funeral_homes_users`.`id` = `users`.`id`
WHERE `funeral_homes_users`.`user_id` = '4'

一切都很好，除了 ON funeral_homes_users.id = users.id 应该是 ON funeral_homes_users.user_id = users.id。所以我唯一想得到的是 id，它应该有 user_id 用于 funeral_homes_users.id(例如它应该是 funeral_homes_users.user_id) 但我无法弄清楚。

以下是供引用的表格:

// funeral_homes
Schema::create('funeral_homes', function (Blueprint $table) {
    $table->increments('id');
    $table->string('name', 255);
    $table->timestamps();
});

// funeral_homes_users
Schema::create('funeral_homes_users', function (Blueprint $table) {
    $table->increments('id');
    $table->integer('funeral_home_id')->unsigned();
    $table->integer('user_id')->unsigned();
    $table->timestamps();
});

// users
Schema::create('users', function (Blueprint $table) {
    $table->increments('id');
    $table->string('email')->unique();
    $table->string('password', 60);
    $table->rememberToken();
    $table->string('first_name');
    $table->string('last_name');
    $table->timestamps();
});

如有任何帮助，我们将不胜感激。谢谢!!!

Answer 1

如果我没理解错的话，你的情况是:

USER 有很多 FUNERAL_HOME

FUNERAL_HOME 属于许多 USER

如果那是正确的，你的关系方法应该返回这样的东西:

User.php

public function FuneralHomes()
{
    return $this->belongsToMany(FuneralHome::class, 'funeral_homes_users');
}

---

FuneralHome.php

public function Users()
{
    return $this->belongsToMany(User::class, 'funeral_homes_users');
}

看看doku