Question

我有三个表：

subset

这是我的模特： 1. ApiCredential

subset(df, grepl(paste(String_Patterns, collapse= "|"), Column_Text))

ApiProvider

SET NAMES utf8;
SET time_zone = '+00:00';

DROP TABLE IF EXISTS `api_credentials`;
CREATE TABLE `api_credentials` (
    `id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
    `user_id` bigint(20) unsigned NOT NULL,
    `api_provider_id` bigint(20) unsigned NOT NULL,
    `access` text COLLATE utf8_unicode_ci,
    `created_at` timestamp NULL DEFAULT NULL,
    `updated_at` timestamp NULL DEFAULT NULL,
    PRIMARY KEY (`id`),
    KEY `api_credentials_user_id_foreign` (`user_id`),
    KEY `api_credentials_api_provider_id_index` (`api_provider_id`),
    CONSTRAINT `api_credentials_user_id_foreign` FOREIGN KEY (`user_id`) REFERENCES `users` (`id`) ON DELETE CASCADE,
    CONSTRAINT `api_credentials_api_provider_id_foreign` FOREIGN KEY (`api_provider_id`) REFERENCES `api_providers` (`id`) ON DELETE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

DROP TABLE IF EXISTS `api_providers`;
CREATE TABLE `api_providers` (
    `id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
    `name` char(255) COLLATE utf8_unicode_ci NOT NULL,
    `table_prefix` char(255) COLLATE utf8_unicode_ci NOT NULL,
    `created_at` timestamp NULL DEFAULT NULL,
    `updated_at` timestamp NULL DEFAULT NULL,
    PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

DROP TABLE IF EXISTS `users`;
CREATE TABLE `users` (
    `id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
    `name` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
    `email` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
    `email_verified_at` timestamp NULL DEFAULT NULL,
    `password` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
    `remember_token` varchar(100) COLLATE utf8_unicode_ci DEFAULT NULL,
    `created_at` timestamp NULL DEFAULT NULL,
    `updated_at` timestamp NULL DEFAULT NULL,
    PRIMARY KEY (`id`),
    UNIQUE KEY `users_email_unique` (`email`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

-- 2019-12-05 13:38:06

用户模型

<?php
namespace App\Models;

use Illuminate\Database\Eloquent\Model;
use Illuminate\Database\Eloquent\Relations\Pivot;

class ApiCredential extends Pivot
{

}

现在我想要什么： 我想从基于<?php namespace App\Models; use Illuminate\Database\Eloquent\Model; class ApiProvider extends Model { protected $fillable = [ 'name', 'table_prefix' ]; public function users() { return $this->belongsToMany('App\Models\User')->using('App\Models\ApiCredential'); } }和namespace App\Models; use Illuminate\Notifications\Notifiable; use Illuminate\Contracts\Auth\MustVerifyEmail; use Illuminate\Foundation\Auth\User as Authenticatable; use App\Models\ApiProvider; use App\Models\ApiCredential; class User extends Authenticatable { use Notifiable; /** * The attributes that are mass assignable. * * @var array */ protected $fillable = [ 'name', 'email', 'password', ]; /** * The attributes that should be hidden for arrays. * * @var array */ protected $hidden = [ 'password', 'remember_token', ]; public function apiProvider() { return $this->belongsToMany('App\Models\ApiProvider','api_providers','user_id','api_provider_id'); } public function hasCredentials(String $apiProvider) { return $this->apiProvider()->where('name', $apiProvider)->get(); } }的{{1}}表中获取记录。假设我在api_credentials中有3条记录（名称是adscane，cpalead，offertorro）。

我正在运行此函数以获取user的api_credential。

api_provider

但是它没有给出预期的结果。

有人可以帮助我解决此问题。

Answer 1

我相信

# User model
public function apiProvider()
{  
    return $this->belongsToMany('App\Models\ApiProvider','api_providers','user_id','api_provider_id');
}

应该是

# User model
public function apiProvider()
{  
    return $this->belongsToMany('App\Models\ApiProvider','api_credentials','user_id','api_provider_id');
}

编辑：访问api_credentials：

使用hasMany关系

# User model
public function apiCredentials()
{
    return $this->hasMany('App\Models\ApiCredential')
}

# ApiProvider model
public function apiCredentials()
{
    return $this->hasMany('App\Models\ApiCredential');
}

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

public function apiProvider()
{
    return $this->belongsTo('App\Models\ApiProvider');
}

然后，我们重新定义hasCredentials()函数，使其工作如下：

# User model
public function hasCredentials(String $apiProvider)
{
    $this->apiCredentials()->whereHas('apiProvider', function ($query) use ($apiProvider) {
         $query->where('name', '=', $apiProvider);
    })
    ->get();
}

使用数据如下所示：

$user = User::find($id);

foreach($user->hasCredentials('api_provider_name') as $apiCredential)
{
    // $apiCredential->name
    // $apiCredential->table_prefix
}

Answer 2

像这样使用

namespace App\Model;


use Illuminate\Database\Eloquent\Model;

class ApiCredential extends Model
{

    public function user()
    {
        return $this->belongsTo(User::class);
    }

    public function apiProvider()
    {
        return $this->belongsTo(ApiProvider::class);
    }
}

和

namespace App\Model;


use Illuminate\Database\Eloquent\Model;

class ApiProvider extends Model
{

    public function ApiCredentials()
    {
        return $this->hasMany(ApiCredential::class);
    }

}

和

namespace App\Model;

use Illuminate\Database\Eloquent\Relations\BelongsTo;
use Illuminate\Database\Eloquent\Relations\HasMany;
use Illuminate\Foundation\Auth\User as Authenticatable;
use Illuminate\Notifications\Notifiable;
use Laravel\Passport\HasApiTokens;
use Spatie\Activitylog\Traits\LogsActivity;

class User extends Authenticatable
{
    use Notifiable;

    /**
     * The attributes that are mass assignable.
     *
     * @var array
     */
    protected $fillable = [
        'name',
        'email',
        'password',
    ];

    /**
     * The attributes that should be hidden for arrays.
     *
     * @var array
     */
    protected $hidden = [
        'password',
        'remember_token',
    ];

    public function ApiCredentials()
    {
        return $this->hasMany(ApiCredential::class);
    }

}

现在这样查询

    User::with('apiCredentials.apiProvider',)
        ->where('id',1)
        ->whereHas('apiCredentials.apiProvider', function ($query)  {
        $query->where('name','cpalead' );
    })->get();

如何使用laravel关系在laravel中的3个表之间建立关系？

2 个答案: