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Php 显示名称而不是id.Laravel_Php_Laravel_Laravel 5_Eloquent_Laravel 5.2 - Fatal编程技术网

Php 显示名称而不是id.Laravel

Php 显示名称而不是id.Laravel,php,laravel,laravel-5,eloquent,laravel-5.2,Php,Laravel,Laravel 5,Eloquent,Laravel 5.2,我有两张桌子;学生和年级。学生包含外键等级\参考等级id。 对于查看学生,我使用了以下代码:; 控制器: public function viewStudent() { $students=Student::all(); return $students('grade_id'); $grade = Grade::find($id, ['id', 'grade_name']); return view('students.v

我有两张桌子;学生和年级。学生包含外键等级\参考等级id。 对于查看学生,我使用了以下代码:; 控制器:

public function viewStudent()
    {

       $students=Student::all();
        return $students('grade_id');
        $grade = Grade::find($id, ['id', 'grade_name']);

        return view('students.view',compact('students'));
    }
视图:


它显示了我所说的等级id。但我要的是你的名字而不是那个。有人能帮我吗?

首先,您需要在学生模型中定义与
成绩的关系:

class student extends Model
{
  public function grade() {
    return $this->belongsTo(Grade::class);
  }
}
然后,你应该能够急切地为学生加载分数

$students = Student::with('grade')->get();
并使用访问学生的年级名称

$student->grade->name

首先,您需要在学生模型中定义与
成绩
的关系:

class student extends Model
{
  public function grade() {
    return $this->belongsTo(Grade::class);
  }
}
然后,你应该能够急切地为学生加载分数

$students = Student::with('grade')->get();
并使用访问学生的年级名称

$student->grade->name
模型中

class student extends Model
{

protected $fillable = ['first_name', 'middle_name', 'last_name', 'grade_id'];

  public function grade()
  {
      return $this->belongsTo('Grade');
  }
}
内部控制器

$students = Student::with('grade')->get();
鉴于

@foreach ($students as $student)
    $studentGradeName = $student->grade->name;
@endforeach
模型中

class student extends Model
{

protected $fillable = ['first_name', 'middle_name', 'last_name', 'grade_id'];

  public function grade()
  {
      return $this->belongsTo('Grade');
  }
}
内部控制器

$students = Student::with('grade')->get();
鉴于

@foreach ($students as $student)
    $studentGradeName = $student->grade->name;
@endforeach