json_encode（$ response）; 如何处理Json API

Question

我坚持使用JSON。我可以使用Guzzle通过API GET响应接收数据。我正在尝试使用Laravel 5.7将数据提取为key：value格式，以便将其保存为5.55版的mysql db（不了解JSON格式）并使用能够向最终用户展示的Blade进行处理。 use GuzzleHttp\Psr7\Response;///得到响应///  $data = json_decode( (string) $response->getBody() );  echo $response->getBody() ; JSON响应格式为：{  "type": "fi.prh.opendata.bis",  "version": "1",  "totalResults": -1,  "resultsFrom": 0,  "previousResultsUri": null,  "nextResultsUri": null,  "exceptionNoticeUri": null,  "results": [ // <- IN this part has company information    {      "businessId": "0856064-3",      "name": "Company Ltd",      "registrationDate": "1991-09-18",      "companyForm": "Ltd",      "detailsUri": null,      "liquidations": [      ],      "names": [ // <- other array, which has history of Company names        {          "order": 0,          "version": 1,          "name": "Company Ltd",          "registrationDate": "1991-08-14",          "endDate": null,          "source": 1        }      ]    }  ]}1）试图从阵列中提取公司信息“结果”：echo $response->getBody()->getContents([results]); 刚收到错误：类App \ PRHData不存在，与具有代码的文件相同。2）试图从阵列中提取公司信息“结果”： foreach($data[result[0]]['businessId'] as $i => $v) { echo $v['businessId'].'<br/>'; }我收到错误消息： Use of undefined constant result - assumed 'result' (this will  throw an Error in a future version of PHP)我不知道该错误消息是什么。use GuzzleHttp\Client;use GuzzleHttp\Psr7\Request as GuzzleRequest;use GuzzleHttp\Psr7\Response;use GuzzleHttp\Psr7;class PRHData extends Model{public static function getJson(){///发送请求///这部分有效，我正在通过API获取数据。$client = new Client([   'base_uri' => 'https://avoindata.prh.fi/bis/v1/',   'defaults'=>[     'timeout'  => 2.0,     'cookies' => true,     'headers'  => ['content-type' => 'application/json']]  ]);  $response = $client->request('GET','0856064-3'); //<- number is                              parameter for GET request to API call