用angularjs+php+mysql上传文件

嘿，伙计们，我做了简单的上传，一切都很好，但我有一个问题，上传一个文件，并添加到表与其他信息。

app.directive("fileInput", function ($parse) {
    return {
        link: function ($scope, element, attrs) {
            element.on("change", function (event) {
                var files = event.target.files;
                $parse(attrs.fileInput).assign($scope, element[0].files);
                $scope.$apply();
            });
        }
    }
});
$scope.uploadFile = function () {
        var form_data = new FormData();
        var server_id = $scope.info_mainServerID;

        console.log(server_id); // value that I need

        angular.forEach($scope.files, function (file, data) {
            form_data.append('file', file);
            form_data.append('data', JSON.stringify(data));
        });

        $http.post('php/upload.php', form_data,
            {
                transformRequest: angular.identity,
                headers: {'Content-Type': undefined, 'Process-Data': false},
                // data: angular.toJson($scope.info_mainServerID)
                server_id : $scope.info_mainServerID
            })

            .success(function (response) {
                console.log(response);
                alertify.notify(response, 'success', 2.5);
            })
            .error(function (response) {
                console.log(response);
                alertify.notify(response, 'error', 2.5);
            });
    };

以及upload.php

$data = json_decode(file_get_contents("php://input"));
$server_id = $data->server_id;

if(count($data) > 0){
    if(!empty($_FILES)){
        $path = '../documents/' . $_FILES['file']['name'];

        if(move_uploaded_file($_FILES['file']['tmp_name'], $path)){
            $insertQuery =
                "
                    INSERT INTO tbl_document(server_id,file_name)
                    VALUES ($server_id,'".$_FILES['file']['name']."');
                ";

            if(mysqli_query($connect, $insertQuery)){
                echo 'File uploaded';
            }
            else{
                echo 'File uploaded but not saved';
            }

        }
    }
    else{
        echo("Error description: " . mysqli_error($connect));
    }
}

我不知道把数据放在哪里 server_id ，或者它为什么给我发送错误
试图在5行的upload.php中获取非对象的属性
如果我只在tbl\u文档中添加文件，一切都会很好

$insertQuery = " INSERT INTO tbl_document(file_name) VALUES ('".$_FILES['file']['name']."'); ";