WCF RESTful服务-上载两个文件

Question

I need to upload two files with WCF REST. This WCF Restful service file upload with multi-platform support works for me when uploading single file, but when I try something like

[WebInvoke(Method = "POST", UriTemplate = "FileUpload/{file1Name};{file2Name}")]

it fails. So the question is: How to put two arguments into Uri?

Thanks in advance, JP

Answer 1

The answer to the linked question deals with uploading a single file - and that's fairly simple, because the service will read the entire HTTP request body, and that will be bound to the stream parameter. When you're dealing with two (or more files) this doesn't work - you need some delimiter to be able to know when one file ends and the other starts. There are a few media types which handle this for that (multipart/* being the most common), but they're not supported natively in WCF (you'd either need to create a custom formatter to deal with it, or take the whole body as a stream and split the files in the FileUpload operation).

Now assuming that you have a way to receive two files in the request body, we can go to your specific question. The URI template feature allows you to pass multiple parameters, but they cannot be in the same "part" of the URI path. One option would be to split them in multiple parts:

[WebInvoke(UriTemplate = "FileUpload/{file1Name}/{file2Name}")]

Or use query parameters for that:

[WebInvoke(UriTemplate = "FileUpload?file1={file1Name}&file2={file2Name}")]

That should do the job. Another alternative, since you already have to split the files on the input stream anyway, would be to use a format which allows you to specify the names in the stream itself. For example, this would be one alternative:

<length of first file name, in bytes>
<first fileName>
<length of the first file, in bytes>
<bytes corresponding to the first file>
<length of second file name, in bytes>
<second fileName>
<length of the second file, in bytes>
<bytes corresponding to the second file>
00 (or any other value indicating the end of the files)

There are many others (including multipart/*), and in all of them you'll need to somehow parse the input to take the multiple files apart.