我是API的新人,Curl和Json。我需要将我的appilcation链接到API。
我要连接的API有一个CURL示例:
curl -v -H "Accept: application/json" -H "Content-type: application/json" -X GET https://www.api.com -u 'user:token'
我做了一些研究,并提出了这样的要求:
$url = "https://www.api.com";
$headers = array(
-H "Accept: application/json",
-H "Content-type: application/json"
-u 'Authorization: Basic '. base64_encode("user:token") // <---
);
//initializing curl object
$curl = curl_init();
//adding fields to the curl object to enter the site
curl_setopt($curl, CURLOPT_URL, $url);
curl_setopt($curl, CURLOPT_HTTPHEADER, $headers);
curl_setopt($curl, CURLOPT_SSL_VERIFYPEER, 0);
curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);
curl_setopt($curl, CURLOPT_ENCODING, 'gzip,deflate');
//executing the curl call and getting data back
$json = curl_exec($curl);
curl_close($curl); // close the connection
//echo $json;
return $json;
我收到了Json请求,但我仍然不理解API提供的示例。我甚至不知道我是否正在好好卷曲。
有人可以向我解释,如果我做得好,我怎么能解释Curl的例子呢?
答案 0 :(得分:0)
这不是有效的PHP语法:
$headers = array(
-H "Accept: application/json",
-H "Content-type: application/json"
-u 'Authorization: Basic '. base64_encode("user:token") // <---
);
你应该改用:
$headers = array(
'Accept: application/json',
'Content-type: application/json',
);
对于auth,请使用:
curl_setopt($curl, CURLOPT_USERPWD, $username . ':' . $password);