Question

我无法从json_decode数组中获取所需的标题数据，得分，我已经尝试了stackoverflow中已经讨论过的所有方法。任何人都可以帮助我..

$myKEY = "xyz";
$url_req= 'google api request here';
$results= checkPageSpeed($url_req_d);
$googleapi = json_decode($results,true);

Google api在var_dump（$ googleapi）时发送这样的数据，我需要从数组中获取标题和得分值。请回复建议的代码以提取标题和分数值，即“xyz”和“73”

{
 "kind": "pagespeedonline#result",
 "id": "www xyz com/",
 "responseCode": 200,
 "title": "xyz",
 "ruleGroups": {
  "SPEED": {
   "score": 73
  }
 },
 "pageStats": {
  "numberResources": 67,
  "numberHosts": 15,
  "totalRequestBytes": "9354",
  "numberStaticResources": 48,
  "htmlResponseBytes": "129210",
  "textResponseBytes": "5647",
  "cssResponseBytes": "142839",
  "imageResponseBytes": "411466",
  "javascriptResponseBytes": "635453",
  "otherResponseBytes": "94639",
  "numberJsResources": 17,
  "numberCssResources": 6
 },  .........

Answer 1

$googleapi['title']和$googleapi['ruleGroups']['SPEED']['score']应该可以解决问题。检查documentation以获取有关如何从多维数组访问元素的更多信息。

无法通过PHP中的json_decode从Google API获取所需数据

1 个答案: