Json 响应按来自拉拉维尔中另一个表的 ID 筛选

Question

目前，我的数据库中有这2个不同的。views

第一个是列出所有项目及其详细信息的地方view_project_percentage

以及每个特定项目的任务列表view_projtask

view_project_percentage

view_projtask

我有这个工作代码

我的模型

Project.php

 <?php

 namespace App\Models;

use Illuminate\Database\Eloquent\Model;

class Project extends Model

{

protected $table = "view_project_percentage";

public function tasks()

{

    return $this->hasMany(ProjectTask::class, 'projCode', 'proj_code')

    ->where('deleted',0);

}

}

和

ProjectTask.php

<?php

namespace App\Models;

use Illuminate\Database\Eloquent\Model;

class ProjectTask extends Model

{

protected $table = "view_projtask";

public function project()

{

    return $this->belongsTo(Project::class, 'proj_code', 'projCode')

    ->where('deleted',0)

    ->where('taskDeleted',0);

} 

}

在我的我有这个Controller

 public function get_all_projtask()

{

    return response()->json(Project::with('tasks')->get());

}

Answer 1

您可以为员工项目创建另一个模型<?php...class EmployeeProject extends Model{protected $table = 'tbl_emp_proj';public function projects(){&nbsp; &nbsp; return $this->hasMany(Project::class, 'projCode');}在您的控制器中，您可以使用它们的 empployee id 和延迟负载关系来过滤它。顺便说一句，在这里阅读更多文档public function get_all_projtask(){&nbsp; &nbsp; $useProject = EmployeeProject::where(['emp_id' =>auth()->user()->company_id])-firstOrFail();&nbsp; &nbsp; return response()->json($useProject->load('projects.tasks'));}