Php 使用Laravel中的两列表获取名称

我是初学者laravel-开发者

我有以下迁移：

Schema::create('clients', function (Blueprint $table) {
            $table->id();
            $table->string('email');
            $table->integer('type')->comment('1 - klient indywidualny, 2 - firma');
            $table->string('crm_number')->unique();
            $table->string('password');
            $table->string('verify_token');
            $table->rememberToken();
            $table->timestamp('password_assigned_at')->nullable();
            $table->timestamp('email_verified_at')->nullable();
            $table->timestamps();
            $table->deactivation();
            $table->softDeletes();

            $table->unique(['email', 'deleted_at']);
        });

Schema::create('client_infos', function (Blueprint $table) {
            $table->id();
            $table->foreignId('client_id')->constrained('clients')->onDelete('cascade');
            $table->string('first_name');
            $table->string('last_name');
            $table->string('phone_nr')->nullable();
            $table->timestamps();
        });
Schema::create('client_company_infos', function (Blueprint $table) {
            $table->id();
            $table->foreignId('client_id')->constrained('clients')->onDelete('cascade');
            $table->string('name');
            $table->string('nip')->nullable();
            $table->string('phone_nr')->nullable();
            $table->string('contact_email')->nullable();
            $table->string('invoice_email')->nullable();
            $table->timestamps();
        });

和型号：

类客户端扩展Authenticatable实现可呈现 { 使用停用， 须呈报， 软删除

const INDIVIDUAL = 1;
const COMPANY = 2;


protected $fillable = [
    'email',
    'crm_number',
    'email_verified_at',
    'password_assigned_at',
    'verify_token',
    'password',
    'type'
];

protected $hidden = [
    'password',
    'verify_token',
    'remember_token',
];

protected $dates = [
    'password_assigned_at',
    'email_verified_at',
    'created_at',
    'updated_at',
    'deactivated_at',
    'deleted_at',
];


public function getAdminUrlAttribute(): AbstractAdminUrlPresenter
{
    return new ClientUrlPresenter($this);
}

public function info()
{
    return $this->hasOne(ClientInfo::class);
}

public function companyInfo()
{
    return $this->hasOne(ClientCompanyInfo::class);
}

public function contractInfo()
{
    return $this->hasOne(ClientContractInfo::class);
}

public function getHasAssignedPasswordAttribute(): bool
{
    return !is_null($this->password_assigned_at);
}

public function getNameAttribute(): string
{
    return $this->isCompany()
        ? data_get($this, 'companyInfo.name', "")
        : data_get($this, 'info.full_name', "");
}

public function isIndividualClient()
{
    return $this->type == self::INDIVIDUAL;
}

public function isCompany()
{
    return $this->type == self::COMPANY;
}

}

现在我需要选择客户端名称为的所有记录：

如果公司->客户\公司\信息名称

如果是个人->客户名和客户名

我需要使用join或selectRaw

我怎样才能做到

请帮助我

在这里尝试一下：

  $clients = Client::all()->map(function($client) {
    
    if($client->type === Client::INDIVIDUAL) {
        $client->name = $client->info->first_name . ' ' . $client->info->last_name;
        
    } else if($client->type === Client::COMPANY) {
        $client->name = $client->companyInfo->name;
    }
    
    return $client;
    
})
  

在这里试试这个：

  $clients = Client::all()->map(function($client) {
    
    if($client->type === Client::INDIVIDUAL) {
        $client->name = $client->info->first_name . ' ' . $client->info->last_name;
        
    } else if($client->type === Client::COMPANY) {
        $client->name = $client->companyInfo->name;
    }
    
    return $client;
    
})
  

---

我有错误：“消息”：“尝试获取非对象的属性‘名称’，：（重构信息和公司信息中的关系…您应该在关系中分配一个表名称。我有错误：“消息”：“尝试获取非对象的属性‘名称’，：（重构您在信息和公司信息中的关系…您应该在关系中分配一个表名。到目前为止您尝试了什么？到目前为止您尝试了什么？