php - 我的代码中未定义的属性:stdClass::$ account_id错误

我遇到了以下错误
未定义的属性：stdClass:：$account_id（视图：
C:\xampp\htdocs\laravel\awsconfig\app\views\search.blade.php）
以下是导致此错误的代码：

$resource_types = array();
$resource_types['AWS::EC2::Instance'] = 'EC2Instance';
$resource_types['AWS::EC2::NetworkInterface'] = 'EC2NetworkInterface';
$resource_types['AWS::EC2::VPC'] = 'VPC';
$resource_types['AWS::EC2::Volume'] = 'Volume';
$resource_types['AWS::EC2::SecurityGroup'] = 'EC2SecurityGroup';
$resource_types['AWS::EC2::Subnet'] = 'Subnet';
$resource_types['AWS::EC2::RouteTable'] = 'RouteTable';
$resource_types['AWS::EC2::EIP'] = 'EIP';
$resource_types['AWS::EC2::NetworkAcl'] = 'NetworkAcl';
$resource_types['AWS::EC2::InternetGateway'] = 'InternetGateway';


 $accounts = DB::table('aws_account')->get();
 $account_list = array();




foreach(glob('../app/views/*.json') as $filename)
{
    //echo $filename;
    $data = file_get_contents($filename);

    if($data!=null)

    {


            $decoded=json_decode($data,true);


            if(isset($decoded["Message"]))
            {
            //echo "found message<br>";
                $message= json_decode($decoded["Message"]);
if(isset($message->configurationItem))
{
//  echo"found cfi<br>";
            $insert_array = array();
                $cfi = $message->configurationItem;
                switch ($cfi->configurationItemStatus)
                {

                    case "ResourceDiscovered":
                //echo"found Resource Discovered<br>";
                        if (array_key_exists($cfi->resourceType,$resource_types))

                        {
                            //var_dump($cfi->resourceType);
                            $resource = new $resource_types[$cfi->resourceType];

                            foreach ($cfi->configuration as $key => $value)
                            {
                                if (in_array($key,$resource->fields))
                                {
                                    $insert_array[from_camel_case($key)] = $value;
                                }
                            }




                            $resource->populate($insert_array);
                            if (!$resource->checkExists())
                            {
                                $resource->save();

                                    if(isset($cfi->configuration->tags))
                                    {
                                       foreach ($cfi->configuration->tags as $t )

                                        {

                                        $tag= new Tag;
                                    $tag->resource_type = "instance";

                                    $tag->resource_id = $resource->id;
                                    $tag->key = $t->key;
                                    $tag->value = $t->value;

                                    $tag->save();


                                        if(isset($cfi->awsAccountId))
                                    {
                                     foreach ($accounts as $a)

                                     {
                                       $account_list[] = $a->account_id;
                                     }


                                    if (!in_array($account_id,$account_list))
                                    {



                                    $account_id = new Account;
                                    $account_id->aws_account_id = $cfi->awsAccountId;

                                    $account_list[] = $account_id;
                                     $account_id->save();

                                        }






                                }
                                }
}




                            }

                        }

                        else
                        {
                            echo "Creating ".$cfi["resourceType"]." not yet supported<br>";
                        }





                    break;

我知道这会是件很愚蠢的事我很感激大家一如既往的帮助谢谢

最佳答案

基于您的代码，这是一个我正在解释的简单演示，DB::select将返回一个包含多个对象（可能不止一个）的数组，然后将其分配给$this->food。
记住，$this->food看起来像
Array ( [0] => stdClass Object ( [name] => 'Beef' ))
实际上，$this->food->name试图访问未定义的属性。
解决方案1：
您应该使用$this->food[0]->name来访问它。
我觉得很奇怪，但很有效。
解决方案2：
为什么不调用Food::find($id)来获取对象而不是$food = new food($id)
您可以通过阅读此http://laravel.com/docs/eloquent
希望这能帮你解决问题

关于php - 我的代码中未定义的属性:stdClass::$ account_id错误，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/29942824/