Question

将带有以太坊地址的数组传递给控制器会告诉我试图获取非对象的属性“地址”的错误

我正在使用Laravel和区块链以太坊开发Dapp，我已经配置了自己的以太坊节点

我有一个带有公共函数的BaseController，该公共函数将JSONRPC调用到URL API并向RegisterController返回一个地址，但是它出现了试图在代码行“地址”中获取非对象的属性“地址”的错误。 => $ taddress->地址

有人可以帮我吗？

我留下每个控制器的代码

BaseController

public function get_address()
{
    $fields = [
        "jsonrpc" => '2.0',
        "method"  => 'eth_accounts',
        //  "method" => 'personal_newAccount',
        //  "method" => 'createAddressForToken',
        "params"  => ['password'],
        "id"      => '1',
    ];
    $fields = json_encode($fields);
    $headers = [
        'Content-Type: application/json',
    ];
    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, APIURL);
    curl_setopt($ch, CURLOPT_POST, true);
    curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
    curl_setopt($ch, CURLOPT_POSTFIELDS, $fields);
    $result = curl_exec($ch);
    curl_close($ch);
    $response = json_decode($result);
    if (isset($response->error->code)) {
        return $address = 0;
    } else {
        return $address = $response->result;
    }
}

RegisterController

protected function create(array $data)
{
    $settings = AdminSettings::first();

    // Verify Settings Admin
    if ($settings->email_verification == 1) {

        $confirmation_code = str_random(100);
        $status = 'pending';

        //send verification mail to user         
        $_username = $data['name'];
        $_email_user = $data['email'];
        $_title_site = $settings->title;
        $_email_noreply = $settings->email_no_reply;

        Mail::send('emails.verify', ['confirmation_code' => $confirmation_code, 'title_site' => $_title_site],
            function ($message) use (
                $_username,
                $_email_user,
                $_title_site,
                $_email_noreply
            ) {
                $message->from($_email_noreply, $_title_site);
                $message->subject(trans('users.title_email_verify'));
                $message->to($_email_user, $_username);
            });

    } else {
        $confirmation_code = '';
        $status = 'active';
    }

    $token = str_random(75);

    $taddress = $this->get_address();
    json_encode($taddress);
    var_dump($taddress);

    return User::create([
        'name'              => $data['name'],
        'email'             => $data['email'],
        'password'          => bcrypt($data['password']),
        'countries_id'      => $data['countries_id'],
        'avatar'            => 'default.jpg',
        'status'            => $status,
        'role'              => 'normal',
        'token'             => $token,
        'confirmation_code' => $confirmation_code,
        'address'           => $taddress->address,
        //'address' => $taddress,
        'private_key'       => $taddress->password,
    ]);
}

感谢您的帮助：）

Answer 1

您的return语句中有错误。您将返回变量赋值的结果，而应该只返回变量。

这导致行$taddress = $this->get_address();不是对象，而是很可能是布尔值。这会导致您的错误。

if (isset($response->error->code)) {
    return $address = 0;
} else {
    return $address = $response->result;
}

应更改为此：

if (isset($response->error->code)) {
    return 0;
} else {
    return $response->result;
}

Answer 2

非常感谢您的回答！

我做了更改，就像他们在BaseController中这样告诉我：

if(isset($response->error->code))
    {
        return $address=0;  
    }
    else
    {   
        return $response->result;
        //return $address=$response->result;
    }

但是它在代码行的RegisterController中始终显示相同的错误：

'address' => $taddress->address

那会是什么

Laravel控制器显示错误尝试获取非对象的属性``地址''

2 个答案: