#MultipartForm How to get the original file name? - jax-rs

Question

I am using jboss's rest-easy multipart provider for importing a file. I read here http://docs.jboss.org/resteasy/docs/1.0.0.GA/userguide/html/Content_Marshalling_Providers.html#multipartform_annotation regarding #MultipartForm because I can exactly map it with my POJO.
Below is my POJO
public class SoftwarePackageForm {
#FormParam("softwarePackage")
private File file;
private String contentDisposition;
public File getFile() {
return file;
}
public void setFile(File file) {
this.file = file;
}
public String getContentDisposition() {
return contentDisposition;
}
public void setContentDisposition(String contentDisposition) {
this.contentDisposition = contentDisposition;
}
}
Then I got the file object and printed its absolute path and it returned a file name of type file. The extension and uploaded file name are lost. My client is trying to upload a archive file(zip,tar,z)
I need this information at the server side so that I can apply the un-archive program properly.
The original file name is sent to the server in content-disposition header.
How can I get this information? Or atleast how can I say jboss to save the file with the uploaded file name and extension? Is it configurable from my application?

Answer 1

After looking around a bit for Resteasy examples including this one, it seems like there is no way to retrieve the original filename and extension information when using a POJO class with the #MultipartForm annotation.
The examples I have seen so far retrieve the filename from the Content-Disposition header from the "file" part of the submitted multiparts form data via HTTP POST, which essentially, looks something like:
Content-Disposition: form-data; name="file"; filename="your_file.zip"
Content-Type: application/zip
You will have to update your file upload REST service class to extract this header like this:
#POST
#Path("/upload")
#Consumes("multipart/form-data")
public Response uploadFile(MultipartFormDataInput input) {
String fileName = "";
Map<String, List<InputPart>> formParts = input.getFormDataMap();
List<InputPart> inPart = formParts.get("file"); // "file" should match the name attribute of your HTML file input
for (InputPart inputPart : inPart) {
try {
// Retrieve headers, read the Content-Disposition header to obtain the original name of the file
MultivaluedMap<String, String> headers = inputPart.getHeaders();
String[] contentDispositionHeader = headers.getFirst("Content-Disposition").split(";");
for (String name : contentDispositionHeader) {
if ((name.trim().startsWith("filename"))) {
String[] tmp = name.split("=");
fileName = tmp[1].trim().replaceAll("\"","");
}
}
// Handle the body of that part with an InputStream
InputStream istream = inputPart.getBody(InputStream.class,null);
/* ..etc.. */
}
catch (IOException e) {
e.printStackTrace();
}
}
String msgOutput = "Successfully uploaded file " + fileName;
return Response.status(200).entity(msgOutput).build();
}
Hope this helps.

Answer 2

You could use #PartFilename but unfortunately this is currently only used for writing forms, not reading forms: RESTEASY-1069.
Till this issue is fixed you could use MultipartFormDataInput as parameter for your resource method.

Answer 3

If you user MultipartFile class, than you can do something like:
MultipartFile multipartFile;
multipartFile.getOriginalFilename();

Answer 4

It seems that Isim is right, but there is a workaround.
Create a hidden field in your form and update its value with the selected file's name. When the form is submitted, the filename will be submitted as a #FormParam.
Here is some code you could need (jquery required).
<input id="the-file" type="file" name="file">
<input id="the-filename" type="hidden" name="filename">
<script>
$('#the-file').on('change', function(e) {
var filename = $(this).val();
var lastIndex = filename.lastIndexOf('\\');
if (lastIndex < 0) {
lastIndex = filename.lastIndexOf('/');
}
if (lastIndex >= 0) {
filename = filename.substring(lastIndex + 1);
}
$('#the-filename').val(filename);
});
</script>