Php 如何在Yii中的另一个命名空间中实例化对象？

我有以下几点

<?php
namespace app\commands;
use \Keyword as GoogleKeyword;

class KwController extends \yii\console\Controller
{

  public function actionTest() {
      $keyword = new GoogleKeyword();
  }

我不明白，因为它过去是有用的

这就是它的定义

$ grep Keyword vendor/googleads/googleads-php-lib/src/Google/Api/Ads/AdWords/v201509/CampaignCriterionService.php
if (!class_exists("Keyword", false)) {
  class Keyword extends Criterion {
    const XSI_TYPE = "Keyword";

这里是

composer.json

{
    "require": {
        "googleads/googleads-php-lib": "~6.5"

---

下列任何一项都有效

use \Keyword as GoogleKeyword; // name clash
require 'vendor/googleads/googleads-php-lib/src/Google/Api/Ads/AdWords/v201509/CampaignCriterionService.php'; # for Keyword

或

---

我认为

GetService

本身就有一些神奇的加载功能。

试着查看供应商内部/googleads/googleads php lib/src/Google/Api/Ads/AdWords/v201509/ActivityCriterionService.php并查看名称空间

use \Keyword as GoogleKeyword; // name clash
require 'vendor/googleads/googleads-php-lib/src/Google/Api/Ads/AdWords/v201509/CampaignCriterionService.php'; # for Keyword

  public function actionTest() {
    $gaw = new GoogleAdWords();
    $user = $gaw->getUser(); # returns an AdWordsUser
    $campaignCriterionService = $user->GetService('CampaignCriterionService', ADWORDS_VERSION);
    $keyword = new GoogleKeyword();