如何在php中使用curl发布图像
我想更改配置文件映像另一个应用程序curl请求我正在尝试下面的代码有人能帮我吗如何在php中使用curl发布图像,php,curl,zend-framework,socialengine,Php,Curl,Zend Framework,Socialengine,我想更改配置文件映像另一个应用程序curl请求我正在尝试下面的代码有人能帮我吗 $viewer = Engine_Api::_()->user()->getViewer(); $apiData = array( "email" => $viewer->email, "profile_image_file" => $_FILES['Filedata']['name'], ); $apiHost = "https://tenant.thetenantsnet.
$viewer = Engine_Api::_()->user()->getViewer();
$apiData = array(
"email" => $viewer->email,
"profile_image_file" => $_FILES['Filedata']['name'],
);
$apiHost = "https://tenant.thetenantsnet.co.uk/api/api/save_profile_image";
$response = $this->callRiseAPI2($apiData,$apiHost);
private function callRiseAPI2($apiData,$apiHost){
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $apiHost);
curl_setopt($ch, CURLOPT_POST, count($apiData));
curl_setopt($ch, CURLOPT_POSTFIELDS, http_build_query($apiData));
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
$jsonData = curl_exec($ch);
if (false === $jsonData) {
throw new \Exception("Error: _makeOAuthCall() - cURL error: " . curl_error($ch));
}
curl_close($ch);
//return the API response
return json_decode($jsonData);
}
正如Anoxy所说,您需要在标题中输入内容类型:
$viewer = Engine_Api::_()->user()->getViewer();
$apiData = array(
"email" => $viewer->email,
"profile_image_file" => $_FILES['Filedata']['name'],
);
$apiHost = "https://tenant.thetenantsnet.co.uk/api/api/save_profile_image";
$response = $this->callRiseAPI2($apiData,$apiHost);
private function callRiseAPI2($apiData,$apiHost){
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $apiHost);
curl_setopt($ch, CURLOPT_POST, count($apiData));
curl_setopt($ch, CURLOPT_POSTFIELDS, http_build_query($apiData));
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt($ch, CURLOPT_HTTPHEADER, 'Content-Type: multipart/form-data');
$jsonData = curl_exec($ch);
if (false === $jsonData) {
throw new \Exception("Error: _makeOAuthCall() - cURL error: " . curl_error($ch));
}
curl_close($ch);
//return the API response
return json_decode($jsonData);
}
你很接近。您需要指定标题:“内容类型:多部分/表单数据”是的,我放置了标题,但它没有发布电子邮件,然后:(