好像这应该有效
$urls = array_map(function($result) {
return $result->url;
}, $rez->responseData->results);
题
非工作代码:
注意:尝试获取非对象的财产
$i = 1;
var_dump($rez->responseData->results[$i]->url);
致命错误:不能将类型stdclass的对象用作数组
$rez = (array) $rez;
var_dump($rez['responseData']['results']['1']['url']);
工作代码的结果:
如何仅扩展URL并将URL列表变成数组?
stdClass Object
(
[responseData] => stdClass Object
(
[results] => Array
(
[0] => stdClass Object
(
[GsearchResultClass] => GwebSearch
[unescapedUrl] => http://www.rottentomatoes.com/top/
[url] => http://www.rottentomatoes.com/top/
[visibleUrl] => www.rottentomatoes.com
...........................................................................
工作代码:
function google_search_api($args, $referer = 'http://localhost/test/', $endpoint = 'web'){
$url = "http://ajax.googleapis.com/ajax/services/search/".$endpoint;
if ( !array_key_exists('v', $args) )
$args['v'] = '1.0';
$url .= '?'.http_build_query($args, '', '&');
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
// note that the referer *must* be set
curl_setopt($ch, CURLOPT_REFERER, $referer);
$body = curl_exec($ch);
curl_close($ch);
//decode and return the response
return json_decode($body);
}
$rez = google_search_api(array(
'q' => 'movie list', 'start' => '5',
));
echo '<pre>';
print_r($rez);
echo '</pre>';
?>
我尝试了上述多种变体,但只是没有弄清楚。可能很简单,因此任何评论都将不胜感激。
解决方案
好像这应该有效
$urls = array_map(function($result) {
return $result->url;
}, $rez->responseData->results);