(Laravel)从一个表获取数据，该表的ID与另一个链接到该表的表相对应

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问题描述

我当时正在玩我的第一个Laravel项目，上面的问题可能有点令人困惑，但让我解释一下:-

I was playing around with my first Laravel project and my question above might be a little bit confusing but let me explain:-

我有3张桌子(实际上我有更多张桌子，但让我们忽略它)，我有standards，stddetails& sections如下所示:-

I have 3 tables (actually I have more, but lets ignore it), I have standards, stddetails & sections as shown:-

因此，外键对应如下:-

So the foreign key corresponds are below:-

standards

列stdPK =表中的列stdFKstddetails

column stdPK from table standards = column stdFK from tablestddetails

列ssctionPK =
中的列ssctionFK表stddetails

场景是这样的:-

让我们说我在表standards 中的stdPK上具有一个$id要匹配. 使用该$id ，我需要从表sections 中获取所有数据.问题是，我找不到正确的查询，因为standards和sections表都只与stddetails表链接.

Lets say I have an $id to be match on stdPK from table standards. With that $id, I am required to get all data from table sections. The problem is, I can't find a right query for that as both standards and sections tables only linked with stddetails table.

目前我在web.php中的查询是:-

Currently my query in my web.php is this:-

Route::get('getstddtl/{id?}', function ($id) {

     $stdsec = Section::
            leftJoin('stddetails', 'ssctionFK', '=', 'ssctionPK')
            ->join('standards', function($join) use ($id){
                $join->on('stdPK', '=', 'stdFK')
                        ->where('stdFK', $id);
            });

    return view('standarddtl', [
        'stdsec' => $stdsec

    ]);
});

我认为这应该很容易，男孩...我错了...我希望有人可以帮助我，因为我的大脑思维能力非常有限.

I thought it should be easy, boy... I was wrong... I hoped someone can help me with this because my brain have very limited thinking capacity.

更新1:-

我在每个模型中都设置了雄辩的关系，并使用Laravel的Eloquent方法检索数据:-

I have set the eloquent relationship in each model and uses Laravel's Eloquent method in retrieving the data:-

$stdsec = Stddetail::with('section')->find($id);

所有数据均从stddetails& sections表，现在的问题是在显示页面的sections表中的sections表中的列ssctionName中显示数据很困难，因为它返回错误.

All data is retrieved from both stddetails & sections tables, the problem now is difficulty on displaying data from column ssctionName in sections table in a display page as it return an error.

显示页面上的相关代码如下:-

The related code on the display page is below:-

@foreach ($stdsec as $task)
       {{strtoupper($task->ssctionName)}}
@endforeach

显示的错误:-

我认为雄辩的方法很好，现在显示部分给我带来麻烦.关于如何解决这个问题有什么想法吗?

I think the eloquent method is good, now the display part is giving me trouble. Any ideas on how to solve this?

更新2:-

这是每个表格的模型:-

Here's the models for each table:-

表standards作为标准模型:-

namespace App;

use Illuminate\Database\Eloquent\Model;

class Standard extends Model
{

    public $primaryKey = 'stdPK';

    public function stddetail()
    {
        return $this->hasMany(Stddetail::class, 'stdFK', 'stdPK');
    }
}

表stddetails作为 Stddetail 模型:-

namespace App;

use Illuminate\Database\Eloquent\Model;

class Stddetail extends Model
{
    public $table = "stddetails";
    public $primaryKey = 'sdtlPK';
    protected $fillable = ['stdFK', 'sdtlPgNo', 'sdtlLnNo', 'sdtlText', 'sdtlShrtnote', 'sdtlSchm', 'svrFK', 'ssctionFK', 'sdtlRefLbl'];

    public function standard()
    {   
        return $this->belongsTo(Standard::class, 'stdFK');
    }

    public function section()
    {   
        return $this->belongsTo(Section::class, 'ssctionFK');
    }

}

表sections作为部分模型:-

namespace App;

use Illuminate\Database\Eloquent\Model;

class Section extends Model
{
    public $primaryKey = 'ssctionPK';
    public function stddetails()
    {
        return $this->hasMany(Stddetail::class, 'ssctionFK', 'ssctionPK');
    }
}

sections

(Laravel)从一个表获取数据，该表的ID与另一个链接到该表的表相对应

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