Recipe 23.7. Uploading Files

Problem

You want to allow users to upload files.

Solution

Use the upload( ) method of the FileReference object.

Discussion

The upload( ) method of a FileReference object allows you to upload a file to a server by using a server-side script that is configured to accept uploads using HTTP(S). At a minimum, the upload( ) method requires one parameter as a URLRequest object specifying the URL of the script to which you want to send the file data:

var urlRequest:URLRequest = new URLRequest("uploadScript.cgi");
fileReference.upload(urlRequest);

All uploads use POST with a Content-Type of multipart/form-data. By default, the Content-Disposition is set to Filedata. The script needs to know the Content-Disposition value so it can read the file data. If the script needs to use a Content-Disposition other than the default, you can specify a value as the second, optional parameter you pass to the upload( ) method, as follows:

fileReference.upload(urlRequest, "UploadFile");

You can only upload a file if the user has selected the file using browse( ). If the user selects a file using a FileReference object's browse( ) method, then you can call upload( ) after the select event occurs. If the user selects files using FileReferenceList, then you must call the upload( ) method for each file in the object's fileList property. The fileList property of a FileReferenceList object is an array of FileReference objects corresponding to each of the files selected by the user.

The upload( ) method can throw errors. The possible errors thrown by upload( ) are identical to those thrown by download( ). Additionally, like download( ), the upload( ) method can dispatch security error events and IO error events. See Recipe 23.1 for more details on handling these errors and error events.