无法通过PHP中的json_decode从Google API获取所需数据

问题描述：

I am unable to fetch desired data of title, score from the json_decode array, I have tried all the ways which are already discussed in *. Can anyone help me..

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

Google api send the data like this when var_dump($googleapi) and I need to fetch title and score values from the array. Please reply suggested code to extract title and score values i.e "xyz" and "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
 },  .........

我无法获取所需的标题数据，从json_decode数组得分，我已经尝试了所有方法已在*中讨论过。任何人都可以帮助我.. p>

  $ myKEY =“xyz”; 
 $ url_req ='google api request here'; 
 $ results = checkPageSpeed（$ url_req_d）;  
 $ googleapi = json_decode（$ results，true）;  
  code>  pre> 
 
  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” p> 
 
 
  {
“kind”：“pagespeedonline＃results”，
“id”  ：“www xyz com /”，
“”responseCode“：200，
”title“：”xyz“，
”ruleGroups“：{
”SPEED“：{
”得分“：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 
}，......... 
  code>  pre> 
  div>

答

$googleapi['title'] and $googleapi['ruleGroups']['SPEED']['score'] should do the trick. Check the documentation for more information on how you can access elements from multidimensional arrays.

无法通过PHP中的json_decode从Google API获取所需数据

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