Php Laravel多表数据过滤器

我使用的是Laravel 5，我有3个数据库表

项目

，

应用程序

，

用户

，它们的表结构如下：

项目

Schema::create('projects', function (Blueprint $table) {
        $table->increments('id');
        $table->string('name');
        $table->text('description');
        $table->char('status',1);
        $table->timestamps();
});

Schema::create('applications', function (Blueprint $table) {
        $table->increments('id');
        $table->integer('project_id')->unsigned()->index();
        $table->integer('user_id')->unsigned()->index();
        $table->char('status');
        $table->timestamps();
});

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

应用程序

Schema::create('projects', function (Blueprint $table) {
        $table->increments('id');
        $table->string('name');
        $table->text('description');
        $table->char('status',1);
        $table->timestamps();
});

Schema::create('applications', function (Blueprint $table) {
        $table->increments('id');
        $table->integer('project_id')->unsigned()->index();
        $table->integer('user_id')->unsigned()->index();
        $table->char('status');
        $table->timestamps();
});

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

用户

Schema::create('projects', function (Blueprint $table) {
        $table->increments('id');
        $table->string('name');
        $table->text('description');
        $table->char('status',1);
        $table->timestamps();
});

Schema::create('applications', function (Blueprint $table) {
        $table->increments('id');
        $table->integer('project_id')->unsigned()->index();
        $table->integer('user_id')->unsigned()->index();
        $table->char('status');
        $table->timestamps();
});

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

现在，我想执行一个查询，返回特定用户尚未向该项目提交任何应用程序的所有项目。到目前为止，这是我能得到的最接近的结果，但是对于没有应用程序的每个项目，它将在

projects.id

中返回

NULL

DB::table('projects')
        ->leftjoin('applications', 'applications.project_id', '=', 'projects.id')
        ->where('applications.user_id', '!=', $user->id)
        ->orWhereNull('applications.id')
        ->get();

查询结果

{
    "total":1,
    "per_page":5,
    "current_page":1,
    "last_page":1,
    "next_page_url":null,
    "prev_page_url":null,
    "from":1,"to":1,
    "data":[{
        "id":null,
        "name":"Project2",
        "description":"Project2 detail",
        "status":null,
        "created_at":null,
        "updated_at":null,
        "project_id":null,
        "user_id":null
    }]
}

$projects_not_applied = DB::table('projects')
        ->whereNotIn('id',$user_applications)
        ->get();

有人能解决这个问题吗

DB::table('projects')
    ->leftjoin('applications', function($join){
        $join->on('applications.project_id', '=', 'projects.id')
            ->where('applications.user_id', '=', $user->id); 
    })
    ->whereNull('applications.id')
    ->get();

---

我想这就是你想要的

我只是想通过使用两个不同的查询并将它们组合在一起来解决这个问题

1）选择所有用户应用程序的项目id

$user_applications = Application::where('user_id', '=', $user->id)
        ->select('project_id as id')
        ->get();

2）选择项目不在上一列表中的项目

{
    "total":1,
    "per_page":5,
    "current_page":1,
    "last_page":1,
    "next_page_url":null,
    "prev_page_url":null,
    "from":1,"to":1,
    "data":[{
        "id":null,
        "name":"Project2",
        "description":"Project2 detail",
        "status":null,
        "created_at":null,
        "updated_at":null,
        "project_id":null,
        "user_id":null
    }]
}

$projects_not_applied = DB::table('projects')
        ->whereNotIn('id',$user_applications)
        ->get();

查询结果（用户未申请项目2）：

---

这个查询仍然给了我与

project.id

=

NULL

相同的结果：（您是如何试图检索该值的？）对不起，这是什么意思？我不太明白您的问题