Php Ajax大文件上传失败

Question

我目前使用REST有一个php api，它有以下方法：

private function createVideoCast(){
            if($this->get_request_method() != "POST"){
                $error = array('status' => "Failed", "msg" => "Invalid Request Method");
                $this->response($this->json($error), 406);
            }
            else{
                $vid    = (int)$this->getLastVideoCastId() + 1;
                $date   = $this->_request['date'];
                $video  = $this->_files['video'];

                if(!empty($video['name']) && !empty($date)){
                    $vorder = (int)$this->getLastVideoCastVorder($date) + 1;

                    $vidExts = array("video/mp4");

                    if(!in_array($video['type'], $vidExts)){
                        $error = array('status' => "Failed", "msg" => "Video Must Be Of Type MP4");
                        $this->response($this->json($error), 400);
                    }
                    if($video['size'] > 1073741824){
                        $error = array('status' => "Failed", "msg" => "Video Maximum Size Is 1GB");
                        $this->response($this->json($error), 400);
                    }

                    $desired_base_dir   = "videocasts/".$date;
                    if(is_dir($desired_base_dir) == false){
                        mkdir($desired_base_dir);
                        $video_name = $vorder.".mp4";

                        if(is_dir($desired_base_dir."/".$video_name) == false){
                            move_uploaded_file($video['tmp_name'], $desired_base_dir."/".$video_name);
                        }
                        else{
                            unlink($desired_base_dir."/".$video_name);
                            move_uploaded_file($video['tmp_name'], $desired_base_dir."/".$video_name);
                        }
                        $video_path = $desired_base_dir."/".$video_name;

                        $sql = "INSERT QUERY TO MY DATABASE";

                        if(mysql_query($sql, $this->db)){
                            $success = array('status' => "Success", "msg" => "Video Successfully Uploaded.");
                            $this->response($this->json($success), 200);
                        }
                        else{
                            $error = array('status' => "Failed", "msg" => mysql_error());
                            $this->response($this->json($error), 400);
                        }
                    }
                }
                else{
                    $error = array('status' => "Failed", "msg" => "Date & Video Required");
                    $this->response($this->json($error), 400);
                }
            }
        }

我有一个通过ajax提交的html表单，如下所示：

<script type="text/javascript">
        function _(el){
            return document.getElementById(el);
        }
        function createVideoCast(){
            var date  = _("date").value;
            var video = _("video").files[0];

            var formdata = new FormData();
            formdata.append("date", date);
            formdata.append("video", video);

            var ajax = new XMLHttpRequest();
            ajax.upload.addEventListener("progress", progressHandler, false);
            ajax.addEventListener("load", completeHandler, false);
            ajax.addEventListener("error", errorHandler, false);
            ajax.addEventListener("abort", abortHandler, false);
            ajax.open("POST", "../newsApi.php?rquest=createVideoCast");
            ajax.send(formdata);
        }
        function progressHandler(event){
            var percent = (event.loaded / event.total) * 100;
            _("progressBar").value = Math.round(percent);
            _("upload-status").innerHTML = Math.round(percent)+"%";
        }
        function completeHandler(event){
            console.log(event);
            var message = JSON.parse(event.target.responseText);
            _("message").innerHTML = message.msg;
            _("progressBar").value = 0;

            if(message.msg == "VideoCast Successfully Created."){
                _("vc-form").reset();
            }
        }
        function errorHandler(event){
            console.log(event);
            _("message").innerHTML = "Upload Failed";
        }
        function abortHandler(event){
            _("message").innerHTML = "Upload Aborted";
        }
    </script>

适用于大约250MB以下的视频，但一旦视频超过该尺寸，我就会收到以下错误：

＆＃34;＆lt; - SHTML Wrapper - 500服务器错误 - ＆gt;↵[处理此指令时出错]↵＆＃34;

我的php.ini文件设置设置为适合1GB文件。我希望有人可以提前为此提供帮助。