使用PHP cURL上传Wistia

Question

有谁知道如何使用PHP cURL从表单中指定的文件上传输入中将视频上传到Wistia？ 这是我的代码，但是我似乎无法使其与当前的API一起使用。 我看过一些与此主题相似的帖子，但仍然没有运气...

<?php
$pathToFile = $_FILES['fileName']['tmp_name']; /* /tmp/filename.mov */
$nameOfFile = $_FILES['fileName']['name']; /* filename.mov */

$data = array(
'api_password' => '********',
'file' => '@'.$pathToFile,
'project_id' => '********'
);

$ch = curl_init();
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_URL, "https://upload.wistia.com" );
curl_setopt($ch, CURLOPT_POSTFIELDS, $data); 
$result = curl_exec($ch);
curl_close($ch);
?>

更新-完整代码（仍然无法正常工作）

当使用“'file'=>'@'。$ pathToFile”时，cURL对于curl_exec和curl_error都返回false，但是当使用“'url'=' http://example.com/file.mov '...时，它可以正常工作。我没有正确处理表格吗？下面是完整的代码：

<?php
if ($_POST['submit']) {
    $filePath = $_FILES['fileUploaded']['tmp_name'];
    $fileName = $_FILES['fileUploaded']['name'];
    $fileSize = $_FILES['fileUploaded']['size'];

    echo("File Path: ");
    echo $filePath;
    echo '<br>';

    $data = array(
        'api_password'  => '******',
        /* 'url'            => 'http://example.com/file.mov', WORKS */
        'file'          => '@'.$filePath, /* DOES NOT WORK */
        'project_id'    => '******'
    );

    // WORKING CMD LINE
    // curl -i -F api_password=****** -F file=@file.mov https://upload.wistia.com/

    /* cURL */
    $chss = curl_init('https://upload.wistia.com');
    curl_setopt_array($chss, array(
        CURLOPT_POST => TRUE,
        CURLOPT_RETURNTRANSFER => TRUE,
        CURLOPT_POSTFIELDS => http_build_query($data)
    ));

    // Send the request
    $KReresponse = curl_exec($chss);
    $KError = curl_error($chss);

    // Decode the response
    $KReresponseData = json_decode($KReresponse, TRUE);

    echo '<br>';
    echo("Response:");
    print_r($KReresponseData);

    echo '<br>';
    echo("Error:");
    print_r($KError);
}
?>
<form name="upload-form" method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>" enctype="multipart/form-data">
    <input type="file" name="fileUploaded">
    <input type="submit" name="submit" value="Upload">
</form>

任何帮助将非常感激！

Answer 1

尝试这个：

$filePath = "@{$_FILES['fileUploaded']['tmp_name']};filename={$_FILES['fileUploaded']['name']};type={$_FILES['fileUploaded']['type']}";