Laravel 9从上级目录调用Seeder报错，求排查解决

场景说明 #

我的Laravel 9项目中，Seeder的存放路径如下：

• Database/Seeders/Production/Permissions/UserManagement/UserPermissionsTableSeeder
• Database/Seeders/Production/Permissions/PermissionsGeneratorTableSeeder

我需要在UserPermissionsTableSeeder中调用PermissionsGeneratorTableSeeder，编写的代码如下：

<?php

namespace Database\Seeders\Production\Permissions\UserManagement;

use Illuminate\Database\Console\Seeds\WithoutModelEvents;
use Illuminate\Database\Seeder;

class UserPermissionsTableSeeder extends Seeder
{
    /**
     * Run the database seeds.
     *
     * @return void
     */
    public function run($companyId = null)
    {
        $roles = [
            'super_admin' => [
                'user_index',
                'user_show',
                'user_store',
                'user_update',
                'user_destroy'
            ],
            'admin' => [
                'user_index',
                'user_show',
                'user_store',
                'user_update'
            ]
        ];

        $this->call(
            
          Database\Seeders\Production\Permissions\PermissionsGeneratorTableSeeder::class,
            false,
            ['roles' => $roles]
        );
    }
}

调用时出现以下错误：

Target class [Database\Seeders\Production\Permissions\UserManagement\Database\Seeders\Production\Permissions\PermissionsGeneratorTableSeeder] does not exist.

更新内容 #

以下是PermissionsGeneratorTableSeeder的代码：

<?php

namespace Database\Seeders\Production\Permissions;

use Illuminate\Database\Console\Seeds\WithoutModelEvents;
use Illuminate\Database\Seeder;
use Spatie\Permission\Models\Role;
use Spatie\Permission\Models\Permission;

class PermissionsGeneratorTableSeeder extends Seeder
{
    /**
     * Run the database seeds.
     *
     * @return void
     */
    public function run($roles = [])
    {
        if (!$roles) {
            return;
        }

        foreach ($roles as $key => $permissions) {
            $role = Role::query();
            $role = $role->where('name', $key);

            if ($key != 'super_admin') {
                $role = $role->where('company_id', $companyId);
            }

            $role = $role->first();

            if (!$role) {
                continue;
            }

            foreach ($permissions as $permission) {
                $discoveredPermission = Permission::where('name', $permission)->first();

                if ($discoveredPermission) {
                    $discoveredPermission->assignRole($role);
                    continue;
                }

                $permissionCreated = Permission::create([
                    'name' => $permission,
                    'guard_name' => config('auth.defaults.guard')
                ]);

                $permissionCreated->assignRole($role);
            }
        }
    }
}

问题原因与解决方法 #

核心错误原因 #

当前UserPermissionsTableSeeder的命名空间是Database\Seeders\Production\Permissions\UserManagement，调用call方法时直接写完整类名，Laravel会将其视为相对命名空间，自动拼接当前类的命名空间，导致生成了错误的类路径（两个命名空间叠加）。

两种解决方式 #

方式一：通过use引入目标类

在UserPermissionsTableSeeder的顶部添加use语句，引入目标Seeder的命名空间：

<?php

namespace Database\Seeders\Production\Permissions\UserManagement;

use Illuminate\Database\Console\Seeds\WithoutModelEvents;
use Illuminate\Database\Seeder;
// 新增引入语句
use Database\Seeders\Production\Permissions\PermissionsGeneratorTableSeeder;

class UserPermissionsTableSeeder extends Seeder
{
    // ... 其他代码不变
    public function run($companyId = null)
    {
        // ... 其他代码不变
        $this->call(
          PermissionsGeneratorTableSeeder::class,
            false,
            ['roles' => $roles]
        );
    }
}

方式二：使用绝对命名空间

在目标类名前添加反斜杠\，告诉Laravel从根命名空间开始查找，避免拼接当前命名空间：

$this->call(
    \Database\Seeders\Production\Permissions\PermissionsGeneratorTableSeeder::class,
    false,
    ['roles' => $roles]
);

额外注意事项 #

在PermissionsGeneratorTableSeeder的代码中，run方法里用到了$companyId变量，但方法参数未定义该变量，会触发变量未定义错误。需要修改run方法的参数定义：

public function run($roles = [], $companyId = null)
{
    // ... 后续代码不变
}

内容的提问来源于stack exchange，提问作者Ryan H