Symfony服务

Question

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：未捕获的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，在/中调用第270行的app / var / cache / prod / appProdProjectContainer.php“/app/src/AppBundle/Utils/SendMessage.php第21行{”exception“：”[object]

Answer 1

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：

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"]

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"