[英]Using CURL to post JSON with PHP variables
我正在尝试使用ZenDesk的API通过我的网站上的付款表单来创建帐户。 他们给出的示例代码是:
curl -v -u {email_address}:{password} https://{subdomain}.zendesk.com/api/v2/users.json \
-H "Content-Type: application/json" -X POST -d '{"user": {"name": "Roger Wilco", "email": "roge@example.org"}}'
由于我需要包括PHP变量,因此我尝试使用此变量:
$data = array("name" => $entry["1"], "email" => $entry["3"], "role" => "end-user");
$data_string = json_encode($data);
$ch = curl_init('https://xxxx.zendesk.com/api/v2/users.json');
curl_setopt($ch, CURLOPT_USERPWD, "xxxx@example.com:xxxx");
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
curl_setopt($ch, CURLOPT_POSTFIELDS, $data_string);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_HTTPHEADER, array(
'Content-Type: application/json',
'Content-Length: ' . strlen($data_string))
);
$result = curl_exec($ch);
但是,它不起作用。 就复制第一个代码段的功能而言,我的代码是否正确?
我找到了ZenDesk API的另一个示例,并能够提出以下建议:
<?PHP
define("ZDAPIKEY", "SECRETKEYGOESHERE");
define("ZDUSER", "me@mysite.com");
define("ZDURL", "https://mysite.zendesk.com/api/v2");
/* Note: do not put a trailing slash at the end of v2 */
function curlWrap($url, $json, $action)
{
$ch = curl_init();
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
curl_setopt($ch, CURLOPT_MAXREDIRS, 10 );
curl_setopt($ch, CURLOPT_URL, ZDURL.$url);
curl_setopt($ch, CURLOPT_USERPWD, ZDUSER."/token:".ZDAPIKEY);
switch($action){
case "POST":
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
curl_setopt($ch, CURLOPT_POSTFIELDS, $json);
break;
case "GET":
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "GET");
break;
case "PUT":
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "PUT");
curl_setopt($ch, CURLOPT_POSTFIELDS, $json);
default:
break;
}
curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-type: application/json'));
curl_setopt($ch, CURLOPT_USERAGENT, "MozillaXYZ/1.0");
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_TIMEOUT, 10);
$output = curl_exec($ch);
curl_close($ch);
$decoded = json_decode($output);
return $decoded;
}
$arr = array("z_name"=>$namevariable,
"z_email"=>$emailvariable,
"z_role"=>"end_user",
"z_verified"=>"yes"
);
$create = json_encode(array('user' => array('name' => $arr['z_name'], 'email' => $arr['z_email'], 'role' => $arr['z_role'])), JSON_FORCE_OBJECT);
$data = curlWrap("/users.json", $create, "POST");
var_dump($data);
?>
它似乎正在单独工作,因此可以回答这里存在的问题。
谢谢大家的帮助 :)
我知道这个问题已经解决了,但是由于我在遇到相同问题时发现了它,并且由于它没有解决我的问题,所以我认为我会发表自己的想法。 希望它可以帮助其他人。
这是在必须通过PUT提交json数据的情况下对我有用的组合:
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_FRESH_CONNECT, TRUE);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, TRUE);
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
curl_setopt($ch, CURLOPT_TIMEOUT, 90);
curl_setopt($ch, CURLOPT_POSTFIELDS, $json);
curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-type: application/json', 'Content-Length: ' . strlen($json), 'X-HTTP-Method-Override: PUT'));
请注意,它不需要CURLOPT_CUSTOMREQUEST
或CURLOPT_PUT
因为X-HTTP-Method-Override: PUT
参数可以解决这个问题。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.