Laravel 6 - `Undefined variable: value` 第一次只有错误

Question

第一次从table保存数据到数据库时出现这个错误。

ErrorException Undefined variable: value

所以我必须手动输入来自tinker或mysql

的数据

我的控制器

public function store(Request $request)

{

    $validateData = $request->validate([

        'name_device_type' => 'required|max:255',

        'signature' => 'Nullable'

    ]);



    $id = DeviceType::getidDeviceTypes();

    foreach ($id as $value);   // Error happend in this line.

    $lastdevicetypeId = $value->id;

    $newdevicetypeId = $lastdevicetypeId + 1;

    $GetnewdevicetypeId = sprintf('DT%04d', $newdevicetypeId);



    $devicetypes = new DeviceType();

    $devicetypes->idDeviceType = $GetnewdevicetypeId;

    $devicetypes->name_device_type = $request->input('name_device_type');

    $devicetypes->signature = $request->input('signature');

    $devicetypes->save();



    return redirect('/devicetypes')->with('success', 'New Device Type is added');

}

我的迁移table

public function up()
    {
        Schema::create('device_types', function (Blueprint $table) {
            $table->bigIncrements('id');
            $table->string('idDeviceType');
            $table->string('name_device_type');
            $table->mediumText('signature');
            $table->timestamps();
        });
    }

我的create.blade.php

{!! Form::open(['action' => 'DeviceTypesController@store', 'method' => 'POST']) !!}
        <div class="form-group">
            {!! Form::label('name_device_type', 'Type Device'); !!}
            {!! Form::text('name_device_type', '', ['class' => 'form-control', 'placeholder' => 'Type Device']); !!}
        </div>
        <div class="form-group">
            {!! Form::label('signature', 'Signature (Optional)'); !!}
            {!! Form::textarea('signature', '', ['id' => 'classic-ckeditor5', 'class' => 'form-control', 'placeholder' => 'Signature']); !!}
        </div>
        {{ Form::button('<i class="far fa-save"></i> Submit', ['type' => 'submit', 'class' => 'btn btn-info'] )  }}
    {!! Form::close() !!}

模特

class DeviceType extends Model
{
    // Table Name
    protected $table = 'device_types';
    // Primary Key
    protected $primaryKey = 'idDeviceTypes';
    // Timestamps
    public $timestamps = true;

    public $incrementing = false;

    public static function getidDeviceType(){
        return $getidDeviceType = DB::table('device_types')->orderBy('id','desc')->take(1)->get();
    }
}

但是如果 table 有数据，这个错误就会消失，如果我删除所有数据并使 table 为空，这个错误也会再次出现。

Answer 1

您在 foreach 循环定义之后有一个分号：foreach ($id as $value); 并且您正在下一行 $lastdevicetypeId = $value->id; 中使用 $value，这超出了您的循环范围。

您应该删除 foreach 循环之后的 ; 并将其更改为 : 并在要结束循环的位置添加 endforeach;。

示例：

foreach ($id as $value):
    $lastdevicetypeId = $value->id;
    $newdevicetypeId = $lastdevicetypeId + 1;
    $GetnewdevicetypeId = sprintf('DT%04d', $newdevicetypeId);

    $devicetypes = new DeviceType();
    $devicetypes->idDeviceType = $GetnewdevicetypeId;
    $devicetypes->name_device_type = $request->input('name_device_type');
    $devicetypes->signature = $request->input('signature');
    $devicetypes->save();
endforeach;

或者您可以将此代码写为：

foreach ($id as $value){
    $lastdevicetypeId = $value->id;
    $newdevicetypeId = $lastdevicetypeId + 1;
    $GetnewdevicetypeId = sprintf('DT%04d', $newdevicetypeId);

    $devicetypes = new DeviceType();
    $devicetypes->idDeviceType = $GetnewdevicetypeId;
    $devicetypes->name_device_type = $request->input('name_device_type');
    $devicetypes->signature = $request->input('signature');
    $devicetypes->save();
}

Answer 2

您可以抓取 table 的最大值 'id' 并向其加 1，而无需抓取整个记录：

...
// validation

$newdevicetypeId = DeviceType::max('id') + 1;

$GetnewdevicetypeId = sprintf('DT%04d', $newdevicetypeId);

// new DeviceType
...

可以选择在创建记录后设置此特定字段的模型事件，以便它有自己的 ID 供使用。

Laravel 6 - `Undefined variable: value` 第一次只有错误

Laravel 6 - `Undefined variable: value` only error in the first time

php

laravel

laravel-6