Kann man das in einem CURL-Durchgang erledigen? Oder muss ich dazu einen zweiten Curl(function) anlegen?
- - - - - - - - - - Beitrag nachträglich erweitert - - - - - - - - - -
Ich glaube, ich habe noch einen Denkfehler dabei?
PHP-Code:
function verystream_upload_url()
{
$curl = curl_init();
curl_setopt($curl, CURLOPT_URL, "https://api.verystream.com/file/ul?login=xxxxx&key=xxxx");
curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($curl, CURLOPT_POST, 1);
/*curl_setopt($curl, CURLOPT_CUSTOMREQUEST, 'GET');*/
$curl_exec = curl_exec($curl);
curl_close($curl);
$res = json_decode($curl_exec, true);
return $res;
}
function verystream_upload($url, $file)
{
$header = array('Content-Type: multipart/form-data');
$fields = array('file' => '@/' . base64_encode($file));
$curl = curl_init();
curl_setopt($curl, CURLOPT_URL, $url);
curl_setopt($curl, CURLOPT_HTTPHEADER, $header);
curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($curl, CURLOPT_POST, 1);
curl_setopt($curl, CURLOPT_POSTFIELDS, $fields);
/*curl_setopt($curl, CURLOPT_CUSTOMREQUEST, 'GET');*/
$curl_exec = curl_exec($curl);
curl_close($curl);
$res = json_decode($curl_exec, true);
return $res;
}
$upload_url = verystream_upload_url();
$upload = verystream_upload($upload_url['result']['url'], 'test.zip');
echo "Test2576767";
echo "<hr />";
echo $upload_url['result']['url'];
echo "<hr />";
echo '<pre>' . var_export($upload_url, true) . '</pre>';
echo "<hr />";
echo '<pre>' . var_export($upload, true) . '</pre>';
Ich schicke das file an ausgegebene url "$upload_url['result']['url']" ?