Google 검색 콘솔 API - UrlCrawlErrorCountsPerType

Question

Google 검색 콘솔 API로 결과를 얻지 못했습니다. SITEURL 'SITEURL/테스트 AAP/SRC/구글/서비스/Resource.php : 165

$timestamp = strtotime('1 month ago'); 
      $query = new Google_Service_Webmasters_UrlCrawlErrorCountsPerType(); 
      $query->setCategory("authPermissions");  
      $filter = new Google_Service_Webmasters_UrlCrawlErrorCount(); 
      $filter->setCount('5'); 
      $filter->setTimestamp($timestamp); 

      $check = $query->setEntries(array($filter)); 
      $query->setPlatform("web"); 
      $getSearchresponseOb = $service->urlcrawlerrorscounts->query('siteurl','soft404','true','web'); 
      $getSearchresponse  = $getSearchresponseOb->toSimpleObject(); 
      print_r($getSearchresponse); 

Answer 1

I : 그 catch되지 않은 예외 메시지'Google_Exception '주는 기능을 호출하는 동안'(쿼리) PARAM 필요 누락 ' 오류 메시지가 분명하다고 생각하십시오. (. 예를 들어 http://www.example.com/ 일) :

$check = $query->setEntries(array($filter)); 
$query->setPlatform("web"); 
$getSearchresponseOb = $service->urlcrawlerrorscounts->query('**http://www.example.com**','soft404','true','web'); 
$getSearchresponse = $getSearchresponseOb->toSimpleObject(); 
print_r($getSearchresponse); 

구글의 문서 https://developers.google.com/webmaster-tools/v3/urlcrawlerrorscounts/query은 "SITEURL"URL을해야한다에 따르면 때문에이 같은 일을해야한다