请帮助我了解如何在我的服务中注入buzz http包以从我的服务发送请求 .
我的services.yml
parameters:
app_bundle.webUrl: https://url.com/
app_bundle.Url: https://test.com
app_bundle.token: rerwe9888rewrjjewrwj
services:
app_bundle.send_message:
class: AppBundle\Utils\SendMessage
arguments: ["%app_bundle.webUrl%, %app_bundle.Url%, %app_bundle.token%, @buzz"]
我的AppBundle \ Utils \ SendMessage
<?php
namespace AppBundle\Utils;
class SendMessage
{
/**
* SendMessage constructor.
*
* @param $webUrl
* @param $Url
* @param $token
* @param Browser $buzz
*/
public function __construct($webUrl, $Url, $token, Browser $buzz)
{
$this->webUrl = $webUrl;
$this->Url = $Url;
$this->token = $token;
$this->buzz = $buzz;
}
/**
* @param $action
* @param null $data
* @return mixed
*/
private function sendRequest($action, $data = NULL)
{
$headers = array(
'Content-Type' => 'application/json',
);
$response = $this->buzz->post($this->Url . $this->token . '/' . $action, $headers, json_encode($data));
return $response;
}
}
但这导致了错误：
request.CRITICAL：未捕获PHP异常Symfony \ Component \ Debug \ Exception \ FatalThrowableError：“类型错误：传递给AppBundle \ Utils \ SendMessage :: __ construct（）的参数4必须是Buzz \ Browser的实例，没有给定，调用第270行的/app/var/cache/prod/appProdProjectContainer.php在/app/src/AppBundle/Utils/SendMessage.php第21行{“exception”：“[object]
1 回答
服务配置文件中定义的每个参数必须用双引号括起来并用逗号分隔，如上例所示：
您只为构造函数提供了一个字符串参数：
"%app_bundle.webUrl%, %app_bundle.Url%, %app_bundle.token%, @buzz"