Question

嗨，我想将一种形式的数据存储到2个db表中，但无法执行：在我的代码中，我希望老师可以添加多个学生，将老师数据保存到教师表中，并将学生数据保存到学生表中我的数据库表的教师ID屏幕截图为：https://ibb.co/ctr7FDP https://ibb.co/nwmcnPX，这是视图https://ibb.co/CBytRCR，根据此视图，我想在学生表中插入多个具有其教师ID的学生姓名，，任何准则和Note ,,, m使用资源路由

教师控制代码：

        at com.hazelcast.internal.serialization.impl.JavaDefaultSerializers$JavaSerializer.write(JavaDefaultSerializers.java:107)
        at com.hazelcast.internal.serialization.impl.StreamSerializerAdapter.write(StreamSerializerAdapter.java:43)
        at com.hazelcast.internal.serialization.impl.AbstractSerializationService.toBytes(AbstractSerializationService.java:154)
        at com.hazelcast.internal.serialization.impl.AbstractSerializationService.toBytes(AbstractSerializationService.java:129)
        at com.hazelcast.spi.impl.operationservice.impl.OutboundResponseHandler.toNormalResponsePacket(OutboundResponseHandler.java:156)
        at com.hazelcast.spi.impl.operationservice.impl.OutboundResponseHandler.sendNormalResponse(OutboundResponseHandler.java:128)
        at com.hazelcast.spi.impl.operationservice.impl.OutboundResponseHandler.sendResponse(OutboundResponseHandler.java:103)
        at com.hazelcast.spi.Operation.sendResponse(Operation.java:416)
        at com.hazelcast.executor.impl.DistributedExecutorService$CallableProcessor.sendResponse(DistributedExecutorService.java:314)
        at com.hazelcast.executor.impl.DistributedExecutorService$CallableProcessor.run(DistributedExecutorService.java:297)
        at com.hazelcast.util.executor.CachedExecutorServiceDelegate$Worker.run(CachedExecutorServiceDelegate.java:227)
        at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142)
        at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617)
        at java.lang.Thread.run(Thread.java:745)
        at com.hazelcast.util.executor.HazelcastManagedThread.executeRun(HazelcastManagedThread.java:64)
        at com.hazelcast.util.executor.HazelcastManagedThread.run(HazelcastManagedThread.java:80)

刀片文件：

  public function store(Request $request)
{
    $request->validate([
        'teacher_name'    =>  'required',
    ]);

     $teacher = new Teacher;
     $teacher->teacher_name = $request->teacher_name;
     $teacher->save();

     return redirect('teacher');
}

老师模型：

    <form method="POST" action="{{ route('teacher.store') }}">
  @csrf
  <div class="form-group">
    <label for="">Teacher Name</label>
    <input type="text" name="teacher_name" class="form-control input {{ $errors->has('teacher_name') ? 'is-danger' : ''}}" id="" placeholder="Teacher Name">
  </div>

  <div class="form-group">
    <label class="text-right">Enter Student Name</label>
    <div class="col-md-8">
      <div class="wrapper">
        <div><input type="text" name="student_name[]" class="form-control input-lg" placeholder="Enter Student Name" /></div>
        </div>
        <p><button class="add_fields">Add More Students</button></p>
    </div>
  </div>

  <input type="submit" name="submit" class="btn btn-primary" value="submit">

  @include('errors')

</form>

学生模型：

  public function student()
{
    $this->hasMany('App\Student');
}

Answer 1

public function students()   //student to students
{
    $this->hasMany('App\Student');
}

插入代码应为

$teacher->students()->createMany([[$student1],[$student2]]);

Answer 2

假设您要从一个名为student_id的变量中以表格的形式引入多个学生，则可以在保存attach()之后使用Laravel的teacher方法附加他们。。

$teacher->student()->attach($request->get('student_id'));

由于您有很多学生要教，我建议您将 Teacher 模态上的方法重命名为复数（students）。

但是，如果您的FK是我的话，我建议您查看表单student_id[]而不是名字-让name来自学生领域，但为了要进行连接，您需要阵列中的ids，而不是name。

如何从Laravel中的单个表格存储多个表中的数据

2 个答案: